Post on 26-Feb-2018
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 172
Chapter 5
Stereochemistrythe study of the 3-Dthe study of the 3-D
structure of moleculesstructure of molecules
Organic Chemistry 8th Edition
L G Wade Jr
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 272
Stereochemistry
bull For pharmaceuticals slight
differences in the 3D spatial
arrangement can make the difference
between targeted treatment and
undesired side-effects
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 372
ypes of somers
Structural or Constitutional sameformula different bonding se$uence
CH3CH3
CH3CH3
CH3
CH3
CH3CH2OHH3C-O-CH3
and
and
C amp (
C ) amp
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472
ypes of somers
Stereoisomers same formulasame bonding se$uence but
different spatial arrangement of the
atoms +eometric or cis-trans
) nantiomers mirror image isomers
3 Diastereomers not mirror images ofeach other
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572
Stereoisomersbull Same bonding se$uence
bull Different arrangement in space
bull ample ampC-CampCamp-Camp
has two geometric cis-trans0 isomers
OHOH
O
O
OHOH
H H
O O
fumaric acid mp )12C maleic acid mp 31C
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672
dentify the following pairs as
either constitutional isomers
stereoisomers or identical
)
3
(
5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 272
Stereochemistry
bull For pharmaceuticals slight
differences in the 3D spatial
arrangement can make the difference
between targeted treatment and
undesired side-effects
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 372
ypes of somers
Structural or Constitutional sameformula different bonding se$uence
CH3CH3
CH3CH3
CH3
CH3
CH3CH2OHH3C-O-CH3
and
and
C amp (
C ) amp
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472
ypes of somers
Stereoisomers same formulasame bonding se$uence but
different spatial arrangement of the
atoms +eometric or cis-trans
) nantiomers mirror image isomers
3 Diastereomers not mirror images ofeach other
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572
Stereoisomersbull Same bonding se$uence
bull Different arrangement in space
bull ample ampC-CampCamp-Camp
has two geometric cis-trans0 isomers
OHOH
O
O
OHOH
H H
O O
fumaric acid mp )12C maleic acid mp 31C
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672
dentify the following pairs as
either constitutional isomers
stereoisomers or identical
)
3
(
5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 372
ypes of somers
Structural or Constitutional sameformula different bonding se$uence
CH3CH3
CH3CH3
CH3
CH3
CH3CH2OHH3C-O-CH3
and
and
C amp (
C ) amp
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472
ypes of somers
Stereoisomers same formulasame bonding se$uence but
different spatial arrangement of the
atoms +eometric or cis-trans
) nantiomers mirror image isomers
3 Diastereomers not mirror images ofeach other
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572
Stereoisomersbull Same bonding se$uence
bull Different arrangement in space
bull ample ampC-CampCamp-Camp
has two geometric cis-trans0 isomers
OHOH
O
O
OHOH
H H
O O
fumaric acid mp )12C maleic acid mp 31C
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672
dentify the following pairs as
either constitutional isomers
stereoisomers or identical
)
3
(
5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472
ypes of somers
Stereoisomers same formulasame bonding se$uence but
different spatial arrangement of the
atoms +eometric or cis-trans
) nantiomers mirror image isomers
3 Diastereomers not mirror images ofeach other
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572
Stereoisomersbull Same bonding se$uence
bull Different arrangement in space
bull ample ampC-CampCamp-Camp
has two geometric cis-trans0 isomers
OHOH
O
O
OHOH
H H
O O
fumaric acid mp )12C maleic acid mp 31C
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672
dentify the following pairs as
either constitutional isomers
stereoisomers or identical
)
3
(
5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572
Stereoisomersbull Same bonding se$uence
bull Different arrangement in space
bull ample ampC-CampCamp-Camp
has two geometric cis-trans0 isomers
OHOH
O
O
OHOH
H H
O O
fumaric acid mp )12C maleic acid mp 31C
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672
dentify the following pairs as
either constitutional isomers
stereoisomers or identical
)
3
(
5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672
dentify the following pairs as
either constitutional isomers
stereoisomers or identical
)
3
(
5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772
Chirality
bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from
4 nonsuperimposable0 the original ob9ect
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872
hich ob9ects are chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972
Chirality in 8olecules
bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral
bull nantiomers nonsuperimposable mirrorimages different molecules
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072
ampelpful 4rulelt
bull if both hal6es of a molecule are
symmetric about an 4internal
mirror plane then the molecule is
achiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172
=rob 5-) Draw the mirror image of the original
structure gt determine whether the mirror image is
the same cmpd abel each structure as chiral
or achiral and label each pair of enantiomers
c )-bromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272
=rob 5-) Continued
CH3
H
H
CH3
CH3
CH3
H
H
O
a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane
e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372
Chiral Carbonsbull etrahedral carbons with ( different attached groups
are chiral bull f there7s only one chiral carbon in a molecule its
mirror image will be a different compound enantiomer0
bull Chiral carbons are designated with a
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472
H Cl
CH3
OH
H
H
Sol6ed =rob 5- Star each
asymmetric carbon atom
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572
=rob 5-3 Draw structures and star
asymmetric carbon atoms
a0 )-pentanol b0 3-pentanol c0 alinine
g0 h0
CH3 CH3
OH
CH3 CH3
OH
CH3 OH
O
NH2
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672
8irror =lanes of Symmetry
bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror
plane of symmetry
bull A molecule with an internal mirror plane cannot be
chiral
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772
=rob 5-5 Determine whether
each compound has an internal
mirror plane of symmetry
CH3
CH3CH3
h0
Bo
asymmetric
carbons
g0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872
Baming nantiomersusing the Cahn-ngold-=relog con6ention
bull nantiomers are different moleculestherefore they must ha6e different
names
bull sually only one enantiomer will bebiologically acti6e
bull Configuration around the
chiral carbon is specifiedwith E0 or S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072
Assigning =riorities
CH3
H
NH2
OHOCH3 Br
Br
CH2CH2Br
Br CH3
Halinine
1
2
3
4
3-dibromobutane
1
2
3
4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172
Assigning E0 or S0
bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0
bull Draw an arrow from highest to lowest prioritygroup
bull Clockwise E0 Counterclockwise S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272
he steering wheel analogylt
bull urn the steering wheel clockwiseand the car turns to the right E0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372
Sol6ed =rob 5-) Draw the
enantiomers of 3-dibromobutane
and label them as E0 or S0
H2CH2C CH3
Br Br
HC
3-dibromobutane
CBr Br
CH3 H3C
HH
CH2CH2Br CH2CH2Br
1 1
2 2
3 3
4 4
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472
Sol6ed =rob 5-3 he structure of one of the
enantiomers of car6one is shown below Find
the asymmetric carbon atom determine
whether it has the E0 or S0 configuration
car6one
1
23
4
(E ) S0
O
C
CH2H3C
H
s it or
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572
hat to do when
group ( is in front
bull Eotate the
molecule in
space so that
group (
points away
from you E
S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672
hat to do when
group ( is in front
bull Draw the arrow
but then apply the
rule backwards
eg now clock-
wise is S and
counterclockwise
is E Elt S0
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772
hat to do when
group ( is in front
bull Switch group ( with whate6er group is
pointed away from you Bow determine
configuration as normal he originalcompound is the new compound7s
enantiomer
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872
ample Amphetamine
E0S0 E0
NH2
CH3
H
NH2
H
CH3
NH2
CH3 H
mirror
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072
Answer From Solutions
8anual to =rob 5-
Cl H
ClH
HCl
ClH
(d)
HH
ClCl
(e)
H H
ClCl
S0
S0
S0
E0
Assume lower groups
are back and away
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172
=roblem Are these pairs of
compounds enantiomers or
diastereomers
CH3CH3
CH3
CH3
CH3 CH3
CH3
CH3
S
S
E E
CH3
CH3
OH
Cl
CH3Cl
OH
CH3
E E E S
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372
+ood and Had nantiomers
he errible Case of halidomide
bull A +erman drug company launched thalidomide
in I52bull t was found to act as a
bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness
bull Current knowledge held that drugs could not
pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but
thalidomide came in from o6erseas and for use in
clinical trials
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872
Eemember
bull E0 and S0 are names
nomenclature terms0
bull d0 and l0 or K0 and -0 are physical properties and must
be determined bymeasurement
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072
Eacemic 8itures
bull $ual $uantities of d- and l- enantiomers
a 5J5J mi0
bull Botation dl0 or plusmn 0
bull Bo optical acti6ity
bull he miture may ha6e different bp and
mp from the enantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172
Eacemic =roducts
f optically inacti6e reagents combineto form a chiral molecule a racemic
miture of enantiomers is formed
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272
ptical =urity
bull Also called enantiomeric ecess
bull Amount of pure enantiomer in ecess ofthe racemic miture
bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer
bull 8iture composition would be 25-)5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372
Chirality of Conformers
bull f e$uilibrium eists between two chiral
conformers molecule is not chiral
bull Mudge chirality by looking at the most
symmetrical conformer
bull Cycloheane can be considered to be
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472
8obile Conformers
Bonsuperimposable mirror
images but e$ual energyand intercon6ertible
So consider it
planar on a6erage
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572
ip earn to Draw Cylics in
=lanar 4Approimations
H
H
HH
HH
H
HHH
H H
O
H
H
HH
HH
HH
H H
H
H
H
HH
H
H
H
H
H O
H
H
H
HH
H
H
H
H
H
H
H H
H
H
H
H
HH
H
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772
=rob 5-( Draw each compound in its most
symmetric conformation and determine
whether it is capable of showing optical acti6ity
c )-dichloroproane
Br
H
H
Br
d cis-3-
dibromocycloheane
Br
H
H
Br
f trans-(-
dibromocycloheane
Cl
Cl
H3C H
HH
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872
Bonmobile Conformers
f the conformer is sterically hinderedit may eist as enantiomers
Allenes
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972
Allenes
bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with
ad9acent double bonds -CCC-
Allene is achiral
MAKE THE MODELS
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072
Allenes ill He Chiral f heir
nd +roups Are Different
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272
Draw the
carbon chain
6ertically
Fi h = 9 ti E l
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372
Fischer =ro9ection Eules
bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top
bull Eotation of 1J deg
in plane doesn7t
change molecule
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472
Fischer =ro9ection Eules
bull Do not rotate IJ deg
bull Do not turn o6er out of plane ie noflipping like a pancake
Fi h 8i
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572
Fischer 8irror magesbull asy to draw easy to find enantiomers
easy to find internal mirror planes
bull ample )S3E0-)3-dibromobutane
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672
Fischer E0 and S0
bull owest priority usually amp0 comes forwardso assignment rules are backwards
bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0
bull ample
(S )
(S)
Br
Br
CH3
CH3
H
H
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772
ip Draw the Fischer=ro9ection and then
determine whether it is E0or S0 f it is not the one
you want 9ust switch the
horriGontal groups
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872
=rob 5-2 Draw a Fischer
=ro9ection for each compound
CH3
OH
OH
CH3
Br
OH
a S0-propane-)-diol
b E0-)-bromobutan--ol
= b 5 1 8 k Fi h
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972
=rob 5-1 8ake a Fischer
=ro9ection Draw the mirror
image Determine whether themirror image is the same as or
different from the original
H
CH2OH
CHO
OH
CH2OH
Br H
CH2OH
Br
CH2Br
Br
CH3
a b c
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072
=rob 5-I abel each asymmetric
carbon atom as E0 or S0 in each
Fischer =ro9ection
H
H
CH2OH
CH2
OH
OH
OH
CH3
COOH
HH2N
d h i
Br
CH2OH
CH3
Cl
C i f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172
bull Are cisOtrans isomers enantiomers or
diastereomers
Categories of somers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272
Diastereomers
bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0
bull 8olecules with ) or more chiral carbons
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372
Diastereomer Alkenes
Cis-trans isomers are not mirrorimages so they are diastereomers
CH3
H
CH3
H
CH3
CH3
H
H
Diastereomeric Eing
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472
Diastereomeric Eing
Compoundsbull Cis-trans isomers possible
bull 8ay also ha6e enantiomers
bull ample trans-)-dimethylcyclopentane
8 C d
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572
8eso Compoundsbull An achiral compound containing two or
more chiral carbonsOchirality centers
bull Also contains an internal mirror plane of
symmetry
H
H
CH2OH
CH2OH
OH
OH
S t f ti f
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672
wo Sets of nantiomers of
the Same Compound are
Diastereomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772
=rob 5-)J For each pair gi6e
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872
=rob 5 )J For each pair gi6e
the relationship between the
two compoundsCl
Cl
ande OHH
OHH
CH2OH
CHO
HHO
OHH
CH2OH
CHO
andf
Br Br
and
d
= ti f Di t
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972
=roperties of Diastereomersbull Diastereomers ha6e different physical
properties mp bpbull hey can be separated easily by
distillation
crystalliGation
chromatography bull nantiomers differ only in reaction with other
chiral molecules and the direction in which polariGed light is rotated
bull nantiomers are difficult to separate
E l ti f ti
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072
Eesolution of nantiomers
Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated
Chromatographic Eesolution
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172
C o atog ap c eso ut o
of nantiomers
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5
7252019 Wade PowerpointLecture for Chapter 5
httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272
nd of Chapter 5