Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and...
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Transcript of Fischer-Rosanoff Convention Before 1951, only relative configurations could be known. Sugars and...
Fischer-Rosanoff Convention• Before 1951, only relative configurations could be known.
• Sugars and amino acids with same relative configuration as (+)-glyceraldehyde were assigned D and same as (-)-glyceraldehyde were assigned L.
• With X-ray crystallography, now know absolute configurations: D is (R) and L is (S).
• No relationship to dextro- or levorotatory. =>
D and L Assignments
CHO
H OH
CH2OH
D-(+)-glyceraldehyde
*CHO
H OH
HO H
H OH
H OH
CH2OHD-(+)-glucose
*
COOH
H2N H
CH2CH2COOH
L-(+)-glutamic acid
*=>
Properties of Diastereomers
• Diastereomers have different physical properties: m.p., b.p.
• They can be separated easily.• Enantiomers differ only in reaction with
other chiral molecules and the direction in which polarized light is rotated.
• Enantiomers are difficult to separate. =>
Resolution of Enantiomers
React a racemic mixture with a chiral compound to form diastereomers, which can be separated.
=>
ChromatographicResolution of Enantiomers
=>
Chapter 6Alkyl Halides: Nucleophilic Substitution and Elimination
Organic Chemistry, 5th EditionL. G. Wade, Jr.
Classes of Halides• Alkyl: Halogen, X, is directly bonded to sp3
carbon.
C
H
H
H
C
H
H
Br
alkyl halide
C CH
H
H
Cl
vinyl halide
I
aryl halide
=>
•Vinyl: X is bonded to sp2 carbon of alkene.
•Aryl: X is bonded to sp2 carbon on benzene ring.
• Examples:
Polarity and Reactivity• Halogens are more electronegative than C.
• Carbon-halogen bond is polar, so carbon has partial positive charge.
• Carbon can be attacked by a nucleophile.
• Halogen can leave with the electron pair. =>
CH
HH
Br
+ -
Classes of Alkyl Halides
• Methyl halides: only one C, CH3X• Primary: C to which X is bonded has only
one C-C bond.• Secondary: C to which X is bonded has two
C-C bonds.• Tertiary: C to which X is bonded has three
C-C bonds. =>
Classify These:
CH3 CH CH3
Cl
CH3CH2F
(CH3)3CBr CH3I =>
Dihalides• Geminal dihalide: two halogen atoms are
bonded to the same carbon
C
H
H
H
C
H
Br
Br
geminal dihalide
C
H
H
Br
C
H
H
Br
vicinal dihalide
=>
•Vicinal dihalide: two halogen atoms are bonded
to adjacent carbons.
IUPAC Nomenclature
• Name as haloalkane.
• Choose the longest carbon chain, even if the halogen is not bonded to any of those C’s.
• Use lowest possible numbers for position.
CH3 CH CH2CH3
Cl CH3(CH2)2CH(CH2)2CH3
CH2CH2Br
2-chlorobutane 4-(2-bromoethyl)heptane=>
Systematic Common Names
• Name as alkyl halide.
• Useful only for small alkyl groups.
• Name these:
CH3 CH CH2CH3
Cl
(CH3)3CBr
CH3 CH
CH3
CH2F =>
“Trivial” Names• CH2X2 called methylene halide..
•CHX3 is a haloform.
•CX4 is carbon tetrahalide.
•Examples: –CH2Cl2 is methylene chloride
–CHCl3 is chloroform -CHI3 is iodoform
–CCl4 is carbon tetrachloride
Preparation of RX
• Free radical halogenation (Chapter 4) - REVIEW
• Free radical allylic halogenation– produces alkyl halide with double bond on the
neighboring carbon. LATER =>
Substitution Reactions
• The halogen atom on the alkyl halide is replaced with another group.
C C
H X
+ Nuc:-C C
H Nuc
+ X:-
•Since the halogen is more electronegative than
carbon, the C-X bond breaks heterolytically and
X- leaves.
The group replacing X- is a nucleophile. =>
Elimination Reactions
• The alkyl halide loses halogen as a halide ion, and also loses H+ on the adjacent carbon to a base.
C C
H X
+ B:- + X:- + HB C C
•The alkyl halide loses halogen as a halide ion, and also loses H+ on the adjacent carbon to a base.
•A pi bond is formed. Product is alkene.Also called dehydrohalogenation (-HX).
Ingold
Sir Christopher
Father of Physical Organic Chemistry
Coined such names and symbols as:
SN1, SN2, E1, E2, nucleophile, electrophileresonance effect, inductive effect/
In print, he often attacked enemies vigorously and sometimes in vitrolic manner.
Ingold
SN2 Mechanism
• Rate is first order in each reactant
CH
Br
HH
H O CHO Br
H
HH
CHO
H
HH
+ Br-
•Both reactants are involved in RDSNote: one-step reaction with no intermediate
•Bimolecular nuleophilic substitution.•Concerted reaction: new bond forming
and old bond breaking at same time
INVERSION OF CONFIGURATION
SN2 Energy Diagram
• One-step reaction.
• Transition state is highest in energy. =>
Uses for SN2 Reactions• Synthesis of other classes of compounds.
• Halogen exchange reaction.
=>
SN2: Nucleophilic Strength• Stronger nucleophiles react faster.• Strong bases are strong nucleophiles, but not all strong nucleophiles are basic.
=>
Trends in Nuc. Strength
• Of a conjugate acid-base pair, the base is stronger: OH- > H2O, NH2
- > NH3
•Decreases left to right on Periodic Table. Moreelectronegative atoms less likely to form new bond:
OH- > F-, NH3 > H2O
Increases down Periodic Table, as size and polarizability increase: I- > Br- > Cl-
Polarizability Effect
=>
Bulky Nucleophiles
Sterically hindered for attack on carbon, so weaker nucleophiles.
CH3 CH2 O ethoxide (unhindered)weaker base, but stronger nucleophile
C
CH3
H3C
CH3
O
t-butoxide (hindered)stronger base, but weaker nucleophile
=>
Solvent Effects (1)Polar protic solvents (O-H or N-H) reduce the
strength of the nucleophile. Hydrogen bonds must be broken before nucleophile can attack the carbon.
=>
Solvent Effects (2)
• Polar aprotic solvents (no O-H or N-H) do not form hydrogen bonds with nucleophile
• Examples:
CH3 C Nacetonitrile
dimethylformamide (DMF)
CH
O
NCH3
CH3
C
O
H3C CH3
acetone =>
Crown Ethers• Solvate the cation, so
nucleophilic strength of the anion increases.
O
O
O
O
OO
K+
18-crown-6
CH2Cl
KF, (18-crown-6)
CH3CN
CH2F
=>
Fluoride anion becomes a good nucleophile
SN2: Reactivity of Substrate
Carbon must be partially positive.
=>
Must have a good leaving group
Carbon must not be sterically hindered.
Leaving Group Ability• Electron-withdrawing
=>
•Stable once it has left (not a strong base)
•Polarizable to stabilize the transition state.
Structure of Substrate
• Relative rates for SN2: CH3X > 1° > 2° >> 3°
• Tertiary halides do not react via the SN2 mechanism, due to steric hindrance. =>
Effect of Beta Branching
Br
Br
Br
> >
Propyl Isobutylbromide
Neopentylbromide
bromide
NOTE: ALL ABOVE ARE PRIMARY
Miscellaneous Substrate
X
X X
All have sterically hinderedbacksides - No SN2 reactivity
Stereochemistry of SN2
Walden inversion
=>
EXAMPLES
Me Br
cis
DMSO
NC-
Me
CN
trans
MeBr
CN
Br
CH3
D
H
DMSO
NC-
H
CH3
D
NC
SR
EXAMPLES 2
N3-
DMSO
CH3
HF
HN3
CH3
CH3
HBr
CNH
CH3
CH3
HF
ClH
CH3
DMSO
CN-
CH3
CNH
CNH
CH3
MESO
EXAMPLE 3
I
Br
H
CH3S-
Acetone
I
H
SCH3
I
Cl -O CH3
O
CH3CN
I
O
O
CH3
EXAMPLE 4
How would you prepare the following from an alkyl halide?N3
TARGET
XN3-
DMSOretrosyntheticanalysis
SN1 Reaction
• Unimolecular nucleophilic substitution.
•Two step reaction with carbocation intermediate.
•Rate is first order in the alkyl halide, zero order in the nucleophile.
•Racemization occurs.
SN1 Mechanism (1)
Formation of carbocation (slow)
(CH3)3C Br (CH3)3C+
+ Br-
=>
SN1 Mechanism (2)• Nucleophilic attack
(CH3)3C+
+ H O H (CH3)3C O H
H
(CH3)3C O H
H
H O H+ (CH3)3C O H + H3O+
=>
• Loss of H+ (if needed)
SN1 Energy Diagram
• Forming the carbocation is endothermic
• Carbocation intermediate is in an energy well.
=>
Rates of SN1 Reactions• 3° > 2° > 1° >> CH3X
–Order follows stability of carbocations (opposite to SN2)
–More stable ion requires less energy to form
•Better leaving group, faster reaction (like SN2)
Polar protic solvent best: It solvates ions strongly with hydrogen bonding
Stereochemistry of SN1Racemization:
inversion and retention
=>
Rearrangements
• Carbocations can rearrange to form a more stable carbocation.
•Methyl shift: CH3- moves from adjacent carbon
if no H’s are available.
Hydride shift: H- on adjacent carbon bonds with C+.
Hydride Shift
CH3 C
Br
H
C
H
CH3
CH3CH3 C
H
C
H
CH3
CH3
CH3 C
H
C
H
CH3
CH3CH3 C
H
C
CH3
CH3
H
CH3 C
H
C
CH3
CH3
HNuc
CH3 C
H
C
CH3
CH3
H Nuc
=>
Methyl Shift
CH3 C
Br
H
C
CH3
CH3
CH3CH3 C
H
C
CH3
CH3
CH3
CH3 C
H
C
CH3
CH3
CH3CH3 C
H
C
CH3
CH3
CH3
CH3 C
H
C
CH3
CH3
CH3
NucCH3 C
H
C
CH3
CH3
CH3 Nuc
=>
Interesting Rearrangement = PUSH PULL
H3C CH3
Br
Ag
push pull
H3C
C
CH2
CH3
+
CH3
MeO
H
-AgBrBr
H3C CH3 Ag+
MeOH
H3C CH2
CH3MeO
H+
CH3
-H+H3C
CH3MeO
CH3
SN2 or SN1?• Primary or methyl • Tertiary
•Strong nucleophile•Polar aprotic solvent•Rate = k[halide][Nuc]
•Inversion at chiral carbon•No rearrangements
•Weak nucleophile (may also be solvent)
•Polar protic solvent, Ag+
•Rate = k[halide]
•Racemization of optically active compound
Rearranged products
E1 Reaction
• Unimolecular elimination
•Two groups lost (usually X- and H+)
•Nucleophile acts as base)
Also have SN1 products (mixture
SN1 and E1 have common first step.
E1 EXAMPLES
Br
CH2-H
heat
MeOH CH2
E-1
+
OMe
CH2-H
SN-1
H3C
CH3
CH3
CH3
Br
H
CH3CH2OH
H3C
CH3
CH3
H
CH2
+H3C
CH3
CH3
CH3
E1 Mechanism
• Halide ion leaves, forming carbocation.
H C
H
H
C
CH3
CH3
Br
C
H
H
H
C CH3
CH3
O
H
H
C
H
H
H
C CH3
CH3
C C
H
CH3
CH3
H+ H3O+
•Base removes H+ from adjacent carbon.Pi bond forms.
A Closer LookO
H
H
C
H
H
H
C CH3
CH3
C C
H
CH3
CH3
H+ H3O+
=>
E1 Energy Diagram
• Note: first step is same as SN1=>
E2 Reaction
• Bimolecular elimination
•Requires a strong base
•Halide leaving and proton abstraction happens
simultaneously - no intermediate
E2 Examples
E2 Mechanism
H C
H
H
C
CH3
CH3
Br
C C
H
CH3
CH3
H
O
H+ H2O Br
-+
=>
Saytzeff’s Rule• If more than one elimination product is possible, the most-
substituted alkene is the major product (most stable).
• R2C=CR2>R2C=CHR>RHC=CHR>H2C=CHR tetra > tri > di > mono
C C
Br
H
C
H
CH3
H
H
H
CH3
OH-
C CH
HC
H H
CH3
CH3
C
H
H
H
C
H
CCH3
CH3
+
=>minor major
E2 Stereochemistry
=>
E1 or E2?• Tertiary > Secondary
• Weak base
• Good ionizing solvent
• Rate = k[halide]
• Saytzeff product
• No required geometry
• Rearranged products
• Tertiary > Secondary
• Strong base required
• Solvent polarity not important
• Rate = k[halide][base]
• Saytzeff product
• Coplanar leaving groups (usually anti)
• No rearrangements
=>
End of Chapter 6