MCAT Organic Summary Sheet

equal (50:50) (50:50) mixtures of tw o enantiomers; of ten denoted by (+/-)

MOLECULAR STRUCTU S TRUCTURE RE

Enantiomers have identical physical properties but diastereomers do not. Hybridization: an atom is SP3 hybridized if it contains only single bonds (tetrahedral geometry) 109.5o an atom is SP2 hybridized if it contains 1 double bond (trigonal geometry) 120o an atom is SP hybridized 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

Meso compounds - molecules that have at least 2 stereocentres BUT are achiral because they have a plane of symmetry. Fischer Projections:  A

Determining formal charge: charge: # of electrons an atom wants (4 for C, 5 or N, 6 for O) – (# of bonds and each electron in a lone pair) a single bond has 1 σ bond a double bond has 1 σ bond and 1 π bond a triple bond has 1 σ bond and 2 π bonds

B

CH

+

3

 . .. . . .-

H

H

H

CH

+ H

D C

isomers (have the same molecular formula)

H

sterioisomers (have the same bonding arrangement of atoms  but a differen differentt 3-D arrangement arrangement of atoms) atoms)

structural isomers (also known as constitutional isomers)

-

+

R

- have a different bonding arrangement of atoms ex.

.O. +

R

CH3

H2 C

H3C H

R

C H2

CH3

High

High

High

Low

Low

Low

Low

High

Racemic mixtures: mixtures:

chiral centers (also known as steriocenters or stereogenic centers) - occurs at an sp3 hybridized carbon with 4 different substituents

CH3

Cl

Cis/Z

Trans/E

enantiomers

Cl

- double bonds in a ring can't F F H H have cis/trans isomers CH3 H3C - double bonds with 2 of the R S same substituents on the definitions: same carbon can't have - enantiomers are cis/trans isomers sterioisomers which are ex. mirror images images - diasteriomers are sterioisomers which aren't mirror images images - meso compounds have 2 can't have cis/trans isomers or more chiral centers and a plane of symmetry making - C=O's don't have them achiral cis/trans isomers ex. if you have a molecule with 2 chiral centers which are R,R  - it's enantiomer would be S,S - it's diasteriomers would be R,S or S,R 

E- on opposite s ides

 Ass ig ni ng st ereo chem is tr y to chi ral cent ers :  substituents are ranked according to the Cahn-Ingold-Prelog rules (E/Z alkenes) Once substituents have been ranked, the lowest ranked is aimed away from the viewer (i.e. into the page). - If the remaining 1st, 2nd and 3rd ranked substituents are arranged: (i) Clockwise: R Stereochemistr Stereochemistr y (ii) Counter-clockwise S Stereochemistry

geometric (also know as cis/trans) or Z / E

CH H3C

Naming alkenes: alkenes : E or Z When an alkene is tri- or tetra- substituted, substituted, E/Z nomenclature is used. To do this, each substituent across the double bond is assigned a priority according to 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 away to try and find a difference in atomic number (iii) Multiple-bonded atoms are equivalent to the same number of singlebonded atoms

Z- on the same side

D C

.. .. OH ..

B

3

O

O

R

=

- 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 or counter-c.w.

Resonance: - occurs when there’s α,β-unsaturation next to an atom with a charge or an atom with a lone pair next to a carbocation - to draw the resonance contributor always move electrons (ie. Either the double bond in the case of a cation and lone pair of electrons i n the case of a negative charge) H

 A

***NOTE*** ***NOTE*** a conformation is a rotat ion of a bond or a flip of the chair in cyclohexane n

total stereoisomers = 2  (where n equals the number of chiral centers and double  bonds  bonds which which can have cis/trans cis/trans isomers) isomers)

Important functional groups:

hv Cl Cl

Step 1: Initiation

 

alkane

 

alkene

alkyne R H N

O R

OH

R

 

alcohol

R

R

H2N R

ether 

R

 

R

OH

carboxylic acid 

H2SO4 or  H3PO4

minor 

 Add it io n r eact io ns: H a lo h y d r i n

 Aci di ty Summ ary : Strongest acids

R

H

R

H

R

R

B r 2

B r 

Weakest acids

Better leaving groups

H

Poorer leaving groups

Weakest conjugate bases

2O

HO

O

O

> ROH > H2O

> OH

R

10

15

H >

>

R

CH3

20

25

> RNH2 > RCH2

H

30

H

35

35

R

H

R

H

H

50

O H H

HYDROCARBONS, ALCOHOLS, AND SUBSTITUTIONS\ELIMINATIONS

2

R

H

R

H

H

H

R

2

O ,

th e n N a B H

H

R

H

HO 4

H M a r k o v n ik o v

H y d r a t io n : H y d r o b o r a ti o n

 A (e q u a to ri a l)

C

H O

H g ( O A c )2 H

D

B

4

H

H y d r a ti o n : O x y m e r c u ra ti o n

R i n g - f li p i n c y c l o h e x a n e

C

SO

R

R

2

M a r k o v n ik o v

Cyclohexane:  A (a x ia l)

H

H y d r a ti o n : I n d u s tr ia l

H

O

> R

H

A nti and  M a r k o v n ik o v

Strongest conjugate bases

OH

D

Zaitsev's Rule: form most substituted double bond 

NH2

amide

5

E1

or dehydrohalogenation (similar mechanism).

ester

 pka less than 1

Cl CH3

+ major 

R

R

CH3

OH

R O

HI > HBr > HCl >

+

CH3

Cl

Step 3: Termination

O

R

H

 Al kenes  Alkenes may be synthesized via alcohol dehydration

ketone

O

Cl

N R

amine

R

 

aldehyde

Cl

O

R

H

CH3

+

R

O

O

R

H

Cl

Step 2: Propagation

Cl

R

H

H 1. BH

B R

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 carbons apart on a ring Equatorial position generally preferred to avoid 1,3-diaxial interactions

2. H H

2

3

O

O H

/T H F 2

H R H 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

R

/N a O H

O z o n o l y s is R

H

R

Radical Reactions:  Homolytic bond breaking  An example of a radical reaction mechanism

H

O

(i) O

O

3 /-78

( i i) Z n / H

+

+

R

R

H

H

P e r m a n g a n a te C l e av a g e R

H

R

R

KMnO

4 /H

O

O

+

+ R

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

R

R

OH

SN1

3o > benzylic/allylic > 2o > (1o does not work) Weak nucleophiles will do since any nucleophile will attack a carbocation E2 Doesn’t matter what the carbon with the nucleophile is since you’re pulling an H Need a strong base E1 3o > benzylic/allylic > 2o > (1o does not work) Weak bases will do since any base will attack a carbocation OH

H3C H

H CH3

H3C H

C Br 

HO

SN2

C

HO C Br 

•

(R)-confuguration

(S)-configuration

optically pure

optically pure

SN1

H2O ethanol

H+ loss

C

slow

C Br 

H2O (R)-configuration

O

H H3C

OH

H2O

50% (R)

>

H CH3 H (E)-alkene only

slow

C

C

S

+

+ CH3OH2

O

H R 

O

O

•

Carbocation Stability: 3o > benzylic/allylic > 2 o > 1o > methyl H Lv

CH 3

allylic carbon  ben zyli c ca rbo n CH 3

allylic leaving group

When the atom with t he lone pair can act as a base or nucleophile: SN2 and E2 will compete and SN 1 and E1 will compete. To determine the mechanism:

O

R

S

O

R

S

O

O

O

Tosyl groups are very stable as they delocalize the negative charge over 3 different oxygens

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

CH3OH

 ben zyli c le avi ng g rou p

CH3

fast

C Br 

CH3

Leaving Groups: groups that best stabilize a negative charge (tosylate, iodide, 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 group (I  > Br  > Cl  > F ) - more electronegative the atom bearing the negative charge the better the leaving group (F > O  > N  > C ) O

E1

-O

CH3

•

anti-periplanar relationship of H and X

- > -O

OCH3

CH3

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

CH3 H3C

->

CH3

50% (S) racemic

I

Nucleophiles:  are atoms with a lone pair of electrons. In nucleophilic substitution they donate the pair of electrons to form a new covalent bond. (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

C OH + HO C

fast

nucleophile can attack from either face

optically pure

E2

- Look at degree of substitution of the halide, the more substituted, the greater the chance that it will undergo an S N1 or an E1 - Look at what the halide is reacting with: - a Nucleophile (S N2 or SN1) - a Base (E2 or E1) - If you have a molecule which can act as a base or a nucleophile look at the double bond that would be formed in the elimination mechanism. If it is conjugated or highly substituted elimination will be favoured over substitution.

Lv

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

Nucleophile

s m a ll large n e g a tiv e neutral

Nu  b a s e stro n g weak 

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. O

R 3C

Mg

X

1.

R

OH R

R

R 3C

2. H3O+

RCO 2 H + RMgX

RCO 2

-

acting as a nucleophile

P C C = C r O 3  + H C l +

C h r o m i c A c i d = H 2 C rO 4 / a c e t o n e o r K 2 C r 2 O 7 /H 2 S O 4 /a ce to n e o r C r O 3 /H 2 S O 4 / a c e t o n e

R

3

+

+ 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 reduction since you’re adding an O and an H

( w e a k o x i d iz e r )

N

no reaction

2

1

PCC or  chromic acid

OH

o

PCC or  chromic acid

OH

o

PCC

o

( str on g o xid iz e rs )

RCH

O

O

2 OH

H

1

chromic acid

o

RCH

O

2 OH

OH

 Al co hol s  Alkyl Halides from Alcohols

Reductions :

O H

Cl H

H

O H

R

H

O

R'

R

RCH 2OH

NaBH 4 not strong enough to do this

RCH 2OH

NaBH 4 not strong enough to do this

RCH 2NR

NaBH 4 not strong enough to do this

LiAlH 4

R

H

heat

(reducing agents)

strong

OH

R'

HO

LiAlH 4

R

H+

-

OR O

O R

R LiAlH 4

R

Pinacol rearrangement 

NaBH 4 and LiAlH 4 = H mild

O

H

OH

R

B r 

P B r 3 Ether 

NaBH 4 or  LiAlH 4

O

S O C l2 Pyridine

NR

OH

CARBONYLS AND AMINES

R

Tosylation CH3

CH3

 pyr  R

OH

Tos

Cl

R

CH3

 NuSn2 OTos

Nu

Carbonyl reactions: R

O

Epoxide Formation R

R'

H

These reactions produce an alcohol. Nucleophiles include H 2O, CN- or RMgX H R

R'

R

R'

Wittig Reaction R O

 Alkyl halide formation from ether 

+ HBr

δ+

nucleophiles react here

H

H

δ−

O

RCO3H

O

Nucleophilic addition reactions

CH

-

+ Ph

2

P

Ph

ROH + RBr  

R

Ph

Conjugate addition H N

O

O

N

ethanol

Oxidations: O

(

) Cu Li 2

O

 Ami nes Basicity Review  A base is an atom with a lone pair of electrons. The best bases are negatively charged.

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

O

OH

O

mild base

NH2

H

R

R

H

 _ 

H

> R 2 N

RCH2

H

 _ 

 _  > OH > R 3 N > RNH2 >

O

N

>

+ > ROH > NR 4

> R

OH

O H

R

H3C

3

3

2 5 4

aldol products can dehydrate under acidic or basic conditions to give the conjugated product

H O

O 1

Reductive Aminolysis

O

R

2

4

N

R

a secondary a m in e

H HO

H N

O

N

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

O

O

(know this forward and backwards)

+ H2O

an imine or Schiff base

5

O

H+

O + H2 NR  1o amine

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

1

H

NR2

H

O

+

mild base OEt

R

R

OEt

a n e n a m in e

OEt

Hoffmann Elimination: CH3I CH3(CH2)5 N(CH3)3I CH3(CH2)5 NH2 (excess) Hexylamine Hexyltrimethylammonium iodide

Michael Addition O -

O

O

LDA THF o -78 C

-

Diazotization Reaction :  N

O

HSO4

+ 2 H2O

2

W o l f f- K i s c h n e r re d u c t i o n

O

O

Y

OH

CrO 3

H + /heat

O

CrO 3

R O

aldyhydes carboxylic acids

R'

H / H 2O

R'NH 2 H+

O

R' N H

SOCl 2

+

H2O

+

H2O

amides

O

O Nucleophilic acyl substitutions

 Nu

Cl

acid chlorides

R

 N

N HSO4

Nu

Y

 N

(E)

 N

O

R

R

H2O

esters

R nitriles

+

anhydrides

OH

R +

R

O

R'OH H+

R

H

N

H2SO4

Diazonium Coupling Reaction: LiAlH 4

C

+

KOH

Carboxylic Acid

R

NH 2 NH

OH

O

 N + HNO2

O

O

O

CH2  + N(CH3)3

1-Hexene (60%)

+

H2O

alcohols

CH3(CH2)3CH

O Li

O

 NH2

R

Ag2O H2O, heat

where Y =

OH or

NR 2

An azo compound

- Then look for the number of CH3 peaks there are (integrate for 3 protons) - Then use all the data you’ve learned from the molecular formula, IR and NMR to draw possible structures. Then look at each structure and compare them to the number of chemical shifts in the NMR and the splitting and integration observed in the NMR. The structure should match the observed NMR data perfectly. If it doesn’t it is not the correct structure. Eliminate it and look at the next possible structure.

BIOCHEMISTRY AND LAB TECHNIQUES Lipids: hydrolysis O O O

C

O

C

O

C

O O

R

1) OH R

1

R

2

 -

2) H 3 O

HO

C O

R

HO

C

R

OH

in H 2 O, heat +

OH

+

1

O

OH

Glycerol

HO

C

R

2

Fatty acids

 Ami no Aci ds : O

O NH2

O

+

HO

NH2

HN

HO

NH2 R'

R

O

HO

amino acids

R

+ H2O

R'

dipeptide

Carbohydrates H HO O H

H H

OH OH

HO

H

O H

H OH

OH H

OH

H

OH

H

C H2O H

OH

D-glucose

The human body can assimilate only D-fructose and D- glucose and cannot assimilate L-fructose and L-glucose.  An App ro ach To St ru ct ur e Deter mi nat io n: 1. 2.

Determine the units of unsaturation 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 there are heteroatoms in the molecular formula - ex. If there is an O in the molecular formula the IR can tell you if it is a ketone or aldehyde, carboxylic acid or alcohol. If none of these peaks 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 are observed the N may be a tertiary amine or amide. If it was 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 if there is an aromatic ring in the structure (peak in the NMR spectrum between 6.8-8 ppm)