MCAT Prep Organic Equation Sheet

8/11/2019 MCAT Prep Organic Equation Sheet

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MOLECULAR STRUCTURE

Hybridization: an atom is SP3 hybridized if it contains only single bonds (tetrahedralgeometry) 109.5o

an atom is SP2 hybridized if it contains 1 double bond (trigonal geometry) 120o

an atom is SP hybridized if it contains 2 double bonds or 1 triple bond (linear,180o)***NOTE*** this only works for neutral atoms a carbocation is sp2 hybridized

Determining formal charge: # of electrons an atom wants (4 for C, 5 or N, 6for O) – (# of bonds and each electron in a lone pair)

a single bond has 1 bond

a double bond has 1 bond and 1 bond

a triple bond has 1 bond and 2 bonds

Resonance:

- occurs when there’s ,-unsaturation next to an atom with a charge or anatom with a lone pair next to a carbocation- to draw the resonance contributor always move electrons (ie. Either thedouble bond in the case of a cation and lone pair of electrons in the case of anegative charge)

CH 3H

H H

+H CH 3

HH

+

O O

. .

. .

. .

. .

.

.

. .

.

.

-

-

OH

R

R

++ O

R

RH

Naming alkenes: E or ZWhen an alkene is tri- or tetra- substituted, E/Z nomenclature is used. To dothis, each substituent across the double bond is assigned a priority accordingto the Cahn-Ingold-Prelog rules:(i) Rank according to atomic number of attached atom (Br>Cl>O>N>C>H)(ii) If the above rule does not solve the ranking, look at 2nd, 3rd, 4th, atoms awayto try and find a difference in atomic number(iii) Multiple-bonded atoms are equivalent to the same number of single-bonded

atoms

High

Low

High

Low

Low

High

High

Low

Z- on the same side E- on opposite sides

Assigning stereochemistry to chiral centers: substituents are ranked according to the Cahn-Ingold-Prelog rules (E/Zalkenes)

Once substituents have been ranked, the lowest ranked is aimed awayfrom the viewer (i.e. into the page).- If the remaining 1st, 2nd and 3rd ranked substituents are arranged:

(i) Clockwise: R Stereochemistry(ii) Counter-clockwise S Stereochemistry

Racemic mixtures:

equal (50:50) mixtures of two enantiomers; often denoted by (+/-)

Enantiomers have identical physical properties but diastereomers do not.

Meso compounds- molecules that have at least 2 stereocentres BUT areachiral

because they have a plane of symmetry.

Fischer Projections: A

C

B D

A

C

DB=

- most highly oxidized group at the top (position A)- longest carbon chain is vertical (A to C)- Can rotate 180° but not 90 or 270°- Can hold one substituent in place then rotate others either clockwise orcounter-c.w.

isome s

(have the same molecular formula)

structural isomers(also known as constitutional isomers)

- have a different bonding arrangement of atoms

ex.

H3C

CH2

H2C

CH3

H3C

CH

CH3

CH3

sterioisomers

(have the same bonding arrangement of a

but a different 3-D arrangement of atoms

geometric(also know as cis/trans)

or Z / E

Cis/Z Trans/E

***NOTE*** a conformation is a rotation of a bond

or a flip of the chair in cyclohexane

- double bonds in a ring can'thave cis/trans isomers

- double bonds with 2 of the

same substituents on the

same carbon can't have

cis/trans isomers

ex.

can't have cis/trans isomers

chiral centers(also known as steri

or stereogenic cente

- occurs at an sp3 hy

carbon with 4 diffe

substituents

Cl

HF

CH3

FH3C

R

enantiomer

- C=O's don't have

cis/trans isomers

definitions:

- enantiomers are

sterioisomers whi

mirror images

- diasteriomers are

sterioisomers whi

mirror images

- meso compounds

or more chiral cen

a plane of symme

them achiral

ex. if you have a m

with 2 chiral center

are R,R

- it's enantiomer wo

- it's diasteriomers w

R,S or S,R

total stereoisomers = 2n (where n equals the number of chiral centers and double

bonds which can have cis/trans isomers)



Important functional groups:

R OH R

O

R R H2N

R

HN

R

R

NR

R

R H

O

R

O

R R

O

OH

R

O

O

RR

O

NH2

alkane alkene alkyne

alcohol ether amine

aldehyde ketone carboxylic acid

ester amide

Acidity Summary:

HI HBr HClO

OHR

OH

R H> > >>> > >>O

HR

> >

H

H

RNH2 RCH2ROH

H2OCH3R

O

>

pka less than 1 5 10 15 20 25 30 35 35 50

Poorer leaving groupsBetter leaving groups

St ongest acids Weakest acids

Strongest conjugate basesWeakest conjugate bases

HYDROCARBONS, ALCOHOLS, AND

SUBSTITUTIONS\ELIMINATIONS

Cyclohexane:

C

A (a x ia l)

B

D

B

C

D

A (e q u a to ri a l)

R ing- f li p i n cyc lohexane

1,3-Diaxial interactions (occur between A-C, C-D and A-D, on the left above)

- Result from steric strain between axial substituents 3 carbonsapart on a ring

- Equatorial position generally preferred to avoid 1,3-diaxialinteractions

Radical Reactions:

Homolytic bond breaking

An example of a radical reaction mechanism

Cl Cl+

Cl

hv

H CH3 Cl H +

CH3Cl Cl CH3

Step 1: Initiation

Step 2: Propagation

Step 3: Termination

Cl Cl

Alkenes Alkenes may be synthesized via alcohol dehydration

OH

ajo minor

+Zaitsev's Rule:form most substituted doub

E1

H2SO4 o

H3PO4

or dehydrohalogenation (similar mechanism).

Addition reactions:H a l o h y d r i n

R

R H

H

B r 2B r

H OH 2 O

A n t i a n dM a r k o v n ik o v

R R

HH

R

R H

H

H O H 2

H 2 S O 4

M a r k o v n ik o v

H y d r a t i o n : I n d u s t r i a l

H

H O

R R

HH

H g ( O A c ) 2

R

R H

HH 2 O ,

th e n N a B H 4

H

H

RR H

H O

M a r k o v n ik o v

H y d r a t i o n : O x y m e r c u r a t i o n

R

R

H

H2 . H 2 O 2 / N a O H H

H

O H

RR

H

1 . B H 3 / T H F

n o t e : s y n a d d i t io n A n t i- M a r k o v n ik o v

H y d r a t i o n : H y d r o b o r a t i o n

R H

HR

(i) O 3 /-78

R R H

O O

R H

RR

KMnO 4 /H+

R R

OP e r m a n g a n a te C l e a v a g e

( i i) Z n / H+

+

+

O z o n o l y s is

R

O

Substitution/Elimination Summary:SN2 1o > benzylic/allylic > 2o > (3o does not work)

Needs a stong nucleophile



SN1 3o > benzylic/allylic > 2o > (1o does not work)Weak nucleophiles will do since any nucleophile will attack a

carbocationE2 Doesn’t matter what the carbon with the nucleophile is since you’re

pulling an HNeed a strong base

E1 3o > benzylic/allylic > 2o > (1o does not work)Weak bases will do since any base will attack a carbocation

C Br

H3C HOH

C

H3C H

Br HOC

CH3H

HO

(R)-confuguration(S)-configuration

opticallypure

opticallypure

SN2

SN1

E1

C Br C

H2O H2O

C OH CHO+

H2Oethanol

nucleophile can

attack from eitherface

(R)-configuration

50% (R) 50% (S)

racemicoptically

pure

H+ loss

slow fast

E2H

I

CH3

H

H3C

O

H3C CH3

H

anti-periplanarrelationship of H and X

(E)-alkeneonly

C Br C C

H

CH3OH

+ CH3OH2+

slow fast

Carbocation Stability: 3o > benzylic/allylic > 2

o > 1

o > methyl

Lv CH 3

H

ben zylic ca rbo n

ben zylic le aving g rou p

CH3

Lvallylic leaving group

allylic carbon

When the atom with the lone pair can act as a base or nucleophile: SN

2 and E

2 will compete and SN

1 and E

1 will compete.

To determine the mechanism:

- Look at degree of substitution of the halide, the more substituted, the greathe chance

that it will undergo an SN1 or an E1- Look at what the halide is reacting with:

- a Nucleophile (SN2 or SN1)- a Base (E2 or E1)

- If you have a molecule which can act as a base or a nucleophile look at tdouble bond

that would be formed in the elimination mechanism. If it is conjugated orhighly

substituted elimination will be favoured over substitution.

Nucleophiles: are atoms with a lone pair of electrons. In nucleophili

substitution they donate the pair of electrons to form a new covalentbond.(factors listed from most important to least important)I->Br

->Cl

->F

-

CN->OH

->F

-

H2S>H2O- the best nucleophiles are negatively charged (ie. OH

- > H2O)

- the larger the atom the better the nucleophile (ie. I- > Br

- > Cl

- > F

-)

- smaller molecules are better nucleophiles than larger ones

OH > OCH3 O

CH3

CH3

- -- - >> O

CH3

CH3

CH3

- the lower electronegativity of the atom with the lone pair, the strongethe Nu (ie. CN->OH

->F)

Leaving Groups: groups that best stabilize a negative charge (tosylaiodide, bromide, chloride, acetate)(factors listed from most important to least important)- good leaving groups leave neutral - good leaving groups are stable anions (resonance stablilized)- larger the atom bearing the negative charge the better the Lv grou(I

- > Br

- > Cl

- > F

-)

- more electronegative the atom bearing the negative charge the bethe leavinggroup (F

-> O

- > N

- > C

-)

S

O

O

O S

O

O

OR S

O

O

OR

R

Tosyl groups are very stable as they delocalize the negative char

over 3 different oxygens

Note: halogens are good leaving groups too as they are veryelectronegative, but they don’t have resonance stabilization like tosyl group

Bases: Are atoms with a lone pair of electrons. An atom with a lone of electons can be a base or a nucleophile. By definition if an atom wlone pair attacks at the carbon it is a nucleophile. If it pulls a proton a base.- good bases are negatively charged- the bigger the molecule with the lone pair the better the base andpoorer the Nu

Nucleophile

s m a l l N ul a r g e b a s e

n e g a t i v e s t r o n g

n e u t r a l w e a k

Grignards:

RX + Mg RMgBr = R -



A Grignard will act as a base 1st if there’s an acidic proton around (ROH, SH,RCO2H) otherwise it acts as a Nucleophile.

R

O

R

R 3C

OH

R

R

R 3C Mg X

1.

2. H3O+

acting as a nucleophile

RCO 2H + RMgX RCO 2-

+ RH + MgX+

acting as a base

Is it an Oxidation or Reduction? - oxidation is a gain of oxygen or loss of hydrogen- reduction is gain of hydrogen or loss of oxygen***NOTE*** treat S, I, Br, Cl and OH groups as oxygen***NOTE*** addition of HI, HBr, HCl or H-OH is not an oxidation or reductionsince

you’re adding an O and an H

Alcohols Alkyl Halides from Alcohols

O H

H

O H

H

C l

H

B r

H

S O C l2P y r i d i n e

P B r 3E t h e r

Pinacol rearrangement

R R'

HO OHheat

O

H

R'

R

H+

Tosylation

pyr

R OH

CH3

Tos ClR OTos

CH3

Nu-

Sn2

Nu R

CH3

Epoxide Formation

H

R'R

H R R'

O

HH

RCO3H

R

O

R'

+ HBr ROH + RBr

Alkyl halide formation from ether

Oxidations:

NP C C = C r O3 + HC l +

o r K 2C r 2O 7 /H 2 SO 4 /a ce to n e (s tro n g o xid iz e

or CrO 3 /H 2S O 4 / ace tone

Chrom ic Ac id = H 2C rO 4 / ace tone

(weak ox id izer )

OH

OH

RCH 2OH

O

RCH 2OH

H

O

OH

O

P C C o r chromic ac id

P C C o r chromic ac id

3o

2o

1o

no react ion

1o

chromic ac id

PCC

Reductions:

R NR

O

R OH

O

R OR

O LiAlH 4

LiAlH 4

LiAlH 4

RCH2OH

RCH 2OH

RCH 2NR

R R

OH

R R

ONaBH4 or

LiAlH4 NaBH4 and LiAlH4 = H

-

mild strong(reducing agent

NaBH4 not strong enough to do this



CARBONYLS AND AMINES

Carbonyl reactions:

Nucleophilic addition reactions

O

nucleophilesreact here

These reactions produce an alcohol. Nucleophiles include H2O, CN

- or RM

Wittig Reaction

+-O CH 2 P

Ph

Ph

Ph

R

R

Conjugate addition

O

HN O

Ne t h a n o l

OO

Cu Li)( 2



R

O

H

OH

H

OOH

H

H

R

mild base

The aldol condensation between two aldehydes to give a -hydoxy aldehyde

O

HR

O

HR

OHaldol products can dehydrate under acidicor basic conditions to give the conjugated product

O

H

O

H3C

O

HO

H

1

2

3

4

5 12

3

4 5

aldol reactions can occur intramoleuclarly to givecyclic products (favouring 5 or 6 membered rings)

R

O

OEt

OH

OEt

OO

OEtR

mild base

Claisen condensation is similar to the aldol reaction, with esters as the starting material

Michael Addition

OOLDA

THF

-78oC

O-

O- Li

+

O

H2O

OH

O

O

O

Carboxylic Acid

R

O

OH

R OH

R

O

R

CN

H

CrO 3

CrO 3

H+/ H2O

LiAlH4

R

O

O

R

O

O

R

O

NH

R

O

R'

R'

H+/heat

R'OHH+

R'NH2

H+

H2O

H2O

H2O

+

+

+

alcohols

aldyhydes

nitriles

anhydrides

esters

amides

carboxylicacids

R

O

Cl

SOCl2

acid chlorides

R

O

Nu

Nu Nucleophilic acylsubstitutions

AminesBasicity Review

A base is an atom with a lone pair of electrons. The best bases arenegatively charged.

RCH2

_ _ R 2 N> R 3 N> RNH2>

N

NH2

> >

O

R NR2

> ROH>OH > _

OH+

N

R

Reductive Aminolysis

+ H2 NR

1o amine+ H2O

(know this forwaand backwards)

an imine or Schiff ase

O

HN

N

H +

a s e c o n d a r ya m in e

a n e n a m i n e

Hoffmann Elimination:

CH3(CH2)5 NH2

CH3I

(excess)CH3(CH2)5 N(CH3)3I

Ag2O

H2O, heatCH3(CH2)3CH CH2 +

1-Hexene (60%Hexyltrimethyl-ammonium iodide

Hexylamine

Diazotization Reaction:

NH2

+ HNO2 + H2SO4

N N

HSO4 + 2 H2O

O

NH 2NH 2

KOHW ol f f-K ischner reduc t ion

Diazonium Coupling Reaction:

N N HSO4 Y N

An azo compoundwhere Y = OH or NR 2

N

(E)



BIOCHEMISTRY AND LAB TECHNIQUES

Lipids:hydrolysis

O

O

O

C

C

C

O

O

O

R2

R1

R

1) OH -

in H2O, heat

2) H3O+

OH

OH

OH

+

C

O

RHO

C

O

R1

HO

C

O

R2

HOGlycerol

Fatty acids

Amino Acids:

HO

O

NH2

R

HO

O

NH2

R' HO

O HN

R

O

NH2

R'

++ H2O

amino acids dipeptide

CarbohydratesH

O

H OH

HHO

O HH

O HH

C H2OH

O

H O

H

H

OH

OH

H

H

OH

H

OH

D-glucose

The human body can assimilate only D-fructose and D-glucose and cannotassimilate L-fructose and L-glucose.

An Approach To Structure Determination:

1. Determine the units of unsaturation2. Gather information from the IR spectrum

From an IR spectrum you should be able to tell if there is a C=O, O-H,CO2H, N-H, nitrile, C=C or alkyne

- an IR is good for determining functional groups present when thereare heteroatoms in the molecular formula- ex. If there is an O in the molecular formula the IR can tell you if itis a ketone or aldehyde, carboxylic acid or alcohol. If none of thesepeaks are observed then it is probably an ether

ex. If there is an N in the molecular formula the IR can tell

you if it is an N-H or nitrile. If neither of these peaks areobserved the N may be a tertiary amine or amide. If itwas an amide you would observe a C=O peak in the IR.

3. Gather information from the NMR spectrum- Easy things to spot in the NMR are aromatic ring, aldehyde,carboxylic acid and alkene.- If there is 4 or more units of unsaturation immediately look to see ifthere is an aromatic ring in the structure (peak in the NMR spectrumbetween 6.8-8 ppm)

- Then look for the number of CH3 peaks there are (integrate forprotons)- Then use all the data you’ve learned from the molecular formuIR and NMR to draw possible structures. Then look at eachstructure and compare them to the number of chemical shifts in NMR and the splitting and integration observed in the NMR. Th

structure should match the observed NMR data perfectly. If itdoesn’t it is not the correct structure. Eliminate it and look at thenext possible structure.

MCAT Prep Organic Equation Sheet

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Transcript of MCAT Prep Organic Equation Sheet