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