dehydrohalogenation of alkyl halides

Dehydrohalogenation of Alkyl Halides

X Y

dehydrohalogenation of alkyl halides:X = H; Y = Br, etc.

C CC C + X Y

-Elimination Reactions

X Y

dehydrohalogenation of alkyl halides:X = H; Y = Br, etc.

C CC C + X Y

-Elimination Reactions

requires base

(100 %)

likewise, NaOCH3 in methanol, or KOH in ethanol

NaOCH2CH3

ethanol, 55°C

Dehydrohalogenation

Cl

CH3(CH2)15CH2CH2ClKOC(CH3)3

dimethyl sulfoxide

(86%)

CH2CH3(CH2)15CH

Dehydrohalogenation

When the alkyl halide is primary, potassium

tert-butoxide in dimethyl sulfoxide (DMSO), a strong non-protic polar solvent is the base/solvent system that is normally used.

Br

29 % 71 %

+

Regioselectivity

follows Zaitsev's rulemore highly substituted double bond predominates

KOCH2CH3

ethanol, 70°C

more stable configurationof double bond predominates

Stereoselectivity

KOCH2CH3

ethanolBr

+

(23%) (77%)

E2 Energy Diagram

Question

How many alkenes would you expect to be formed from the E2 elimination of3-bromo-2-methylpentane?A) 2B) 3C) 4D) 5

more stable configurationof double bond predominates

Stereoselectivity

KOCH2CH3

ethanol

+

(85%) (15%)

Br

The E2 Mechanism of Dehydrohalogenation of Alkyl

Halides

Empirical Data

(1) Dehydrohalogenation of alkyl halides exhibits second-order kineticsfirst order in alkyl halidefirst order in baserate = k[alkyl halide][base]

implies that rate-determining step involves both base and alkyl halide; i.e., it is bimolecular

Question

The reaction of 2-bromobutane with KOCH2CH3 in ethanol produces trans-2-butene. If the concentration of both reactants is doubled, what would be the effect on the rate of the reaction?A) halve the rateB) double the rateC) quadruple the rateD) no effect on the rate

Empirircal Data

(2) Rate of elimination depends on halogenweaker C—X bond; faster raterate: RI > RBr > RCl > RF

implies that carbon-halogen bond breaks in the rate-determining step

concerted (one-step) bimolecular process

single transition state

C—H bond breaks

component of double bond forms

C—X bond breaks

The E2 Mechanism

–OR.... :

C C

H

X..::

Reactants

The E2 Mechanism

C C

–OR.... H

X..:: –

Transition state

The E2 Mechanism

Br

E2 Mechanism / Transition State

CH3CH2 O••

••••–

OR.... H

C C

–X..::..

Products

The E2 Mechanism

Question

Which one of the following best describes a mechanistic feature of the reaction of 3-bromopentane with sodium ethoxide?A) The reaction occurs in a single step which is bimolecular.B) The reaction occurs in two steps, both of which are unimolecular.C) The rate-determining step involves the formation of the carbocation (CH3CH2)2CH+.D) The carbon-halogen bond breaks in a rapid step that follows the rate-determining step.

Stereoelectronic Effects

Stereochemistry:Anti Elimination in E2 Reactions

Consider dehydrohalogenation of chlorocyclohexane. An anti-periplanar T.S. is required and only the chair

conformation on the left alllows for the elimination to occur.

E2 – Stereoelectronic Effect

Stereoelectronic Effect

An effect on reactivity that has its origin in the spatial arrangement of orbitals or bonds is called a stereoelectronic effect.

The preference for an anti coplanar arrangement of H and Br in the transition state for E2 dehydrohalogenation is an example of a stereoelectronic effect.

(CH3)3C

(CH3)3C

BrKOC(CH3)3

(CH3)3COH

cis-1-Bromo-4-tert- butylcyclohexane


(CH3)3C

(CH3)3CBr KOC(CH3)3

(CH3)3COH

trans-1-Bromo-4-tert- butylcyclohexane


(CH3)3C

(CH3)3C

Br

(CH3)3CBr

KOC(CH3)3

(CH3)3COH

KOC(CH3)3

(CH3)3COH

cis

trans

Rate constant for dehydrohalogenation of 1,4- cis is >500 times than that of 1,4- trans


(CH3)3C

(CH3)3C

BrKOC(CH3)3

(CH3)3COH

cis

H that is removed by base must be anti coplanar to BrTwo anti coplanar H atoms in cis stereoisomer

HH


(CH3)3C

KOC(CH3)3

(CH3)3COH

trans

H that is removed by base must be anti coplanar to BrNo anti coplanar H atoms in trans stereoisomer; all vicinal H atoms are gauche to Br; therefore infinitesimal or no product is formed

HH

(CH3)3CBr

H

H


Which of the two molecules below will NOT be able to undergo an E2 elimination reaction?

Question

A) B)

1,4- cismore reactive

1,4- trans

much less reactive


Sterically unhindered bases favor the Zaitsev product.

Sterically hindered bases favor the Hofmann product.

See: SKILLBUILDER 8.5.

E2 – Regioselectivity

Question

Which would react with KOC(CH3)3/(CH3)3COH faster?A) cis-3-tert-butylcyclohexyl bromideB) trans-3-tert-butylcyclohexyl bromide

Question

Which would react with KOCH2CH3 in ethanol faster?A) cis-2-tert-butylcyclohexyl bromideB) trans-2-tert-butylcyclohexyl bromide

Question

NaOMe/MeOH

NaBr + MeOH + ??????

OMe

A. D.

B. E.

C.

Br

What is the major product of the following reaction?

Question

What is the major product of the following reaction?

NaBr + MeOH + ??????

O

A. D.

B. E.

C.

Br

O

The E1 Mechanism ofDehydrohalogenation of Alkyl

Halides

CH3 CH2CH3

Br

CH3

C

Ethanol, heat

+

(25%) (75%)

H3C

CH3

C C

H3C

H

CH2CH3

CH3

CH2C

Example

1. Alkyl halides can undergo elimination in protic solvents in the absence of base.

2. Carbocation is intermediate.3. Rate-determining step is unimolecular

ionization of alkyl halide.

The E1 Mechanism

CH3 CH2CH3

Br

CH3

C

:..:

CCH2CH3CH3

CH3

+

:..: Br.. –

slow, unimolecular

Step 1

CCH2CH3CH3

CH3

+

CCH2CH3CH3

CH2

+ CCHCH3CH3

CH3

– H+

Step 2

Question

Which reaction would be most likely to proceed by an E1 mechanism?A) 2-chloro-2-methylbutane + NaOCH2CH3 in ethanol (heat)B) 1-bromo-2-methylbutane + KOC(CH3)3 in DMSOC) 2-bromo-2-methylbutane in ethanol

(heat)D) 2-methyl-2-butanol + KOH

1. Analyze the function of the reagent (nucleophile and/ or base).

2. Analyze the substrate (1°, 2°, or 3°).

Predicting Substitution vs. Elimination


2. Analyze the substrate (1°, 2°, or 3°).

See SKILLBUILDER 8.11.

Predicting Substitution vs. Elimination


2. Analyze the substrate (1°, 2°, or 3°).3. Consider regiochemistry and stereochemistry.

Predicting Products

REGIOCHEMICAL OUTCOME STEREOCHEMICAL OUTCOMESN2 The nucleophile attacks the α

position, where the leaving group is connected.

The nucleophile replaces the leaving group with inversion of configuration.

SN1 The nucleophile attacks the carbocation, which is where the leaving group was originally connected, unless a carbocation rearrangement took place.

The nucleophile replaces the leaving group with racemization.

See: SKILLBUILDER 8.12.

Predicting Products

REGIOCHEMICAL OUTCOME STEREOCHEMICAL OUTCOMEE2 The Zaitsev product is generally

favored over the Hofmann product, unless a sterically hindered base is used, in which case the Hofmann product will be favored

This process is both stereoselective and stereospecific.When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene. When the β position of the substrate has only one proton, the stereoisomeric alkene resulting from anti-periplanar elimination will be obtained (exclusively, in most cases).

E1 The Zaitsev product is always favored over the Hofmann product.

The process is stereoselective. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene.

a. A = 3; B = 1; C = 2; D = 1; E = 1; F = 5b. A = 4; B = 4; C = 2; D = 4; E = 5; F = 2c. A = 2; B = 4; C = 2; D = 4; E = 5; F = 2d. A = 4; B = 4; C = 1; D = 4; E = 3; F = 1e. A = 3; B = 5; C = 2; D = 1; E = 3; F = 5

Question

For each reagent, predict which product will predominate.

NaOMe

Br

Cl-/DMF

Cl-/H2O

O-

NaH

CH3S-/DMF

A. D. 1. 2.

B. E. 3. 4.

C. F. 5. 6.Cl

SCH3

H

SH

dehydrohalogenation of alkyl halides

Documents

e2 dehydrohalogenation

e2 elimination of3bromo

anti elimination

ratedno effect

single step

rapid step

athe reaction

bthe reaction