Organic Chemistry

Section 9.6 Stereochemistry of Radical Substitution Reactions 349

carbon bearing the unpaired electron is hybridized; therefore, the three atoms to
which it is bonded lie in a plane. In the second propagation step, the incoming halogen
has equal access to both sides of the plane. As a result, both the Rand Senantiomers
are formed. Identical amounts of the Renantiomer and the Senantiomer are obtained,
so the reaction is not stereoselective.
Notice that the stereochemical outcome of a radical substitution reactionis identi-
cal to the stereochemical outcome of a radical addition reaction(Section 5.19). This is
because both reactions form a radical intermediate, and it is the reaction of the inter-
mediate that determines the configuration of the products.
Identical amounts of the Rand Senantiomers are also obtained if a hydrogen bond-
ed to an asymmetric carbon is substituted by a halogen. Breaking the bond to the
asymmetric carbon destroys the configuration at the asymmetric carbon and forms a
planar radical intermediate. The incoming halogen has equal access to both sides of
the plane, so identical amounts of the two enantiomers are obtained.

What happens if the reactant already has an asymmetric carbon and the radical sub-
stitution reaction creates a second asymmetric carbon? In this case, a pair of diastere-
omers will be formed in unequal amounts.

Diastereomers are formed because the new asymmetric carbon created in the product
can have either the Ror the Sconfiguration, but the configuration of the asymmetric
carbon in the reactant will be unchanged in the product because none of the bonds to
that carbon are broken during the course of the reaction.

More of one diastereomer will be formed than the other because the transition states
leading to their formation are diastereomeric and, therefore, do not have the same
energy.

PROBLEM 13

a. What hydrocarbon with molecular formula forms only two monochlorinated

products? Both products are achiral.

b. What hydrocarbon with the same molecular formula as in part a forms only three mono-

chlorinated products? One is achiral and two are chiral.

C 4 H 10

CH 2 CH 3 CH 2 CH 3

CH 3

CH 2

H Cl +

CH 3

new CH Br

asymmetric

carbon

original

asymmetric

carbon

Br 2 H Cl + HBr

h

sp^2

H

CH 3

Br +

CH 2 CH 3

C

CH 3

CH 2 CH 3

C Br 2

Br

CH 3

CH 2 CH 3

C

Br

H 3 C

CH 2 CH 3

C

+ HBr + Br

C Br Br

CH 3

H

a radical intermediate

CH 3 CH 2

CH 3

CH 2 CH 3

+ H Cl +

H Br

CH 3

CH 2 CH 3

a pair of diastereomers

Fischer projections

a pair of diastereomers

perspective formulas

H Cl

Br H

configuration of the products

Br

C C H

CH 2 CH 3

H 3 C

Cl

H

configuration does

configuration does not change

not change

Br

C C H

CH 2 CH 3

H 3 C

Cl

H

3-D Molecule:

sec-Butyl radical