Organic Chemistry

348 CHAPTER 9 Reactions of Alkanes • Radicals

PROBLEM 11

Two products are formed when methylenecyclohexane reacts with NBS. Explain how each

is formed.

PROBLEM 12 SOLVED

How many allylic substituted bromoalkenes are formed from the reaction of 2-pentene

with NBS? Ignore stereoisomers.

SOLUTION The bromine radical will abstract a secondary allylic hydrogen from C-4 of

2-pentene in preference to a primary allylic hydrogen from C-1. The resonance contribu-

tors of the resulting radical intermediate have the same groups attached to the carbons,

so only one bromoalkene is formed. Because of the high selectivity of the bromine radical,

an insignificant amount of radical will be formed by abstraction of a hydrogen from the

less reactive primary allylic position.

9.6 Stereochemistry of Radical

Substitution Reactions

If a reactant does not have an asymmetric carbon and a radical substitution reaction

forms a product with an asymmetric carbon, a racemic mixture will be obtained.

To understand why both enantiomers are formed, we must look at the propagation

steps of the radical substitution reaction. In the first propagation step, the bromine rad-

ical removes a hydrogen atom from the alkane, creating a radical intermediate. The

sp^2

NBS, ∆

peroxide

Br

CH 2 CH 2 Br

+

CH 2

CH 3 CH 2 CH 2 CH 3 ++Br 2 CH 3 CH 2 CHCH 3 HBr

Br

h

an asymmetric carbon

Br C

H

CH 2 CH 3

CH 3

C Br

H

CH 3 CH (^2) CH
3
H
Br
CH 3 CH 2 CH 3 +
Br
H
a pair of enantiomers
Fischer projections
a pair of enantiomers
perspective formulas
CH 3 CH 2 CH 3
configuration of the products
CH 3 CH CHCHCH 3 CH 3 CHCH CHCH 3 + HBr
CH 3 CH CHCHCH 3
CH 3 CH CHCH 2 CH 3
Br
NBS, ∆
peroxide