Organic Chemistry

416 CHAPTER 11 Elimination Reactions of Alkyl Halides • Competition Between Substitution and Elimination

When only one hydrogen is bonded to

the -carbon, the major product of an

E2 reaction depends on the structure of

the alkene.

B

Molecular models can be helpful

whenever complex stereochemistry

is involved.

understanding how to decide which product is formed, see Problem 44 in the Study

Guide and Solutions Manual.)

PROBLEM 10

a. Determine the major product that would be obtained from an E2 reaction of each of the

following alkyl halides (in each case, indicate the configuration of the product):

1. 
   
   
   
   3. 
      
   
   
   
   

2.

b. Does the product obtained depend on whether you started with the Ror Senantiomer of

the reactant?

Stereochemistry of the E1 Reaction

We have seen that an E1 elimination reaction takes place in two steps. The leaving

group leaves in the first step, and a proton is lost from an adjacent carbon in the second

step, following Zaitsev’s rule in order to form the more stable alkene. The carbocation

formed in the first step is planar, so the electrons from a departing proton can move to-

ward the positively charged carbon from either side. Therefore, both syn and anti elim-

ination can occur.

Because both syn and anti elimination can occur in an E1 reaction, both the Eand Z

products are formed, regardless of whether the -carbon from which the proton is re-

moved is bonded to one or two hydrogens. The major product is the one with the bulkiest

groups on opposite sides of the double bond, because it is the more stable alkene.

b

CH 3 CH 2 CHCH 2

Cl

CH 3 CH 2 CHCH 2 CH

Cl

CH 3 CH 2 CHCHCH 3 CH 2

BrCH 3

(2S,3S)-2-bromo-3-phenylbutane

CC

H CH 3

H

H 3 C

C 6 H 5 Br

CH 3 O−

CH 3 O−

(E)-2-phenyl-2-butene

H 3 C CH 3

C 6 H 5 H Br−

CH 3 OH

(2S,3R)-2-bromo-3-phenylbutane

CC

H CH 3

H

H 3 C

C 6 H 5

Br

(Z)-2-phenyl-2-butene

CH 3

H 3 CH

C 6 H 5

Br−

CH 3 OH

CC

CC

The major product of an E1 reaction

is the alkene with the bulkiest

substituents on opposite sides

of the double bond.

CH 3 CH 2 CH C CH 2 CH 3 CC CC

Cl

CH 3 CH 3 CH 3

CH 3 CH 2 CH C

+

CH 2 CH 3

Cl−

H+

(E)-3,4-dimethyl-

3-hexene

major product

+

+

+

CH 3 H^3 C CH^2 CH^3

CH 3 CH 2 CH 3

(Z)-3,4-dimethyl-

3-hexene

minor product

H 3 C CH 3

CH 3 CH 2 CH 2 CH 3

-carbon has

one hydrogen

CC CC

CH 3

CH 3 CH 2 CHBr

CH 3 H

H

CH 3 CC

+

CH 3

-carbon has Br−

two hydrogens

H+

(E)-2-butene

major product

++

+

H H 3 C H

H

(Z)-2-butene

minor product

H 3 C CH 3

H H

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