Organic Chemistry

Section 11.5 Stereochemistry of E2 and E1 Reactions 413

PROBLEM 8

For each of the following reactions, (1) indicate whether elimination will occur via an E2

or an E1 reaction, and (2) give the major elimination product of each reaction, ignoring

stereoisomers:

a. d.

b. e.

c. f.

PROBLEM 9

The rate law for the reaction of with tert-butyl bromide to form an elimination product

in 75% ethanol 25% water at 30°C is the sum of the rate laws for the E2 and E1 reactions:

What percentage of the reaction takes place by the E2 pathway when these conditions exist?

a. b.

11.5 Stereochemistry of E2 and E1 Reactions

Stereochemistry of the E2 Reaction

An E2 reaction involves the removal of two groups from adjacent carbons. It is a con-

certed reaction because the two groups are eliminated in the same step. The bonds to

the groups to be eliminated (H and X) must be in the same plane because the orbital

of the carbon bonded to H and the orbital of the carbon bonded to X become over-

lapping porbitals in the alkene product. Therefore, the orbitals must overlap in the tran-

sition state. This overlap is optimal if the orbitals are parallel (i.e., in the same plane).

There are two ways in which the and bonds can be in the same

plane: They can be parallel to one another either on the same side of the molecule

(syn-periplanar) or on opposite sides of the molecule (anti-periplanar).

If an elimination reaction removes two substituents from the same side of the

bond, the reaction is called a syn elimination. If the substituents are removed

from opposite sides of the bond, the reaction is called an anti elimination.

Both types of elimination can occur, but syn elimination is a much slower reaction, so

anti elimination is highly favored in an E2 reaction. One reason anti elimination is

favored is that syn elimination requires the molecule to be in an eclipsed conformation,

whereas anti elimination requires it to be in a more stable, staggered conformation.

C¬C

C¬C

H

X

HX

substituents are anti-periplanar

a staggered conformer

substituents are syn-periplanar

an eclipsed conformer

C¬H C¬X

sp^3

sp^3

[HO-]=5.0 M [HO-]=0.0025 M

rate7.1: 10 ^5 [tert-butyl bromide][HO]1.5: 10 ^5 [tert-butyl bromide]

>

HO-

CH 3 C CHCH 3

CH 3 Br

CH 3

CH 3 CH 2 O−

DMSO

CH 3 CCH 3

Cl

CH 3

H 2 O

CH 3 C CHCH 3

CH 3 Br

CH 3

CH 3 CH 2 OH

CH 3 CH 2 CHCH 3

Br

CH 3 OH

CH 3 CCH 3

Cl

CH 3

HO−

CH 3 CH 2 CHCH (^3) DMF
Br
CH 3 O−
DMSO
An E2 reaction involves anti elimination.
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