Organic Chemistry

Section 11.2 The Regioselectivity of the E2 Reaction 403

Progress of the reaction

Free energy

1- butene + CH 3 OH + Br–

2- butene + CH 3 OH + Br–

2- bromobutane + CH 3 O−

>Figure 11.1

Reaction coordinate diagram for

the E2 reaction of 2-bromobutane

and methoxide ion.

To answer this question, we must determine which of the alkenes is formed more

easily—that is, which is formed faster. The reaction coordinate diagram for the E2

reaction of 2-bromobutane is shown in Figure 11.1.

In the transition state leading to an alkene, the and bonds are par-

tially broken and the double bond is partially formed (partially broken and partially

formed bonds are indicated by dashed lines), giving the transition state an alkene-like

structure. Because the transition state has an alkene-like structure, any factors that

stabilize the alkene will also stabilize the transition state leading to its formation,

allowing the alkene to be formed faster. The difference in the rate of formation of the

two alkenes is not very great. Consequently, both products are formed, but the more

stableof the two alkenes will be the major product of the reaction.

We know that the stability of an alkene depends on the number of alkyl substituents

bonded to its carbons: The greater the number of substituents, the more stable is

the alkene (Section 4.11). Therefore, 2-butene, with a total of two methyl substituents

bonded to its carbons, is more stable than 1-butene, with one ethyl substituent.

The reaction of 2-bromo-2-methylbutane with hydroxide ion produces both

2-methyl-2-butene and 2-methyl-1-butene. Because 2-methyl-2-butene is the more

substituted alkene (it has a greater number of alkyl substituents bonded to its car-

bons), it is the more stable of the two alkenes and, therefore, is the major product of

the elimination reaction.

Alexander M. Zaitsev, a nineteenth-century Russian chemist, devised a shortcut to

predict the more substituted alkene product. He pointed out that the more substituted

alkene product is obtained when a proton is removed from the -carbon that is bond-

ed to the fewest hydrogens. This is called Zaitsev’s rule. In 2-chloropentane, for ex-

ample, one -carbon is bonded to three hydrogens and the other -carbon is bondedb b

b

sp^2

sp^2

sp^2

CH 3 CH CHCH 3

H

OCH 3

Br

transition state leading

to 2-butene

δ− OCH

3

δ−

δ−

CH 2 CHCH 2 CH 3

H

Br

transition state leading

to 1-butene

δ−

more stable less stable

C¬H C¬Br

The major product of an E2 reaction is

the most stable alkene.

CH 3 CCH 2 CH 3 HO− H

2 O

+ CH 3 C CHCH 3 + CH 2 CCH 2 CH 3 + H 2 O + Br−

Br

CH 3 CH 3 CH 3

2-bromo-2-methylbutane

2-methyl-2-butene

70%

2-methyl-1-butene

30%

3-D Molecules:

2-Methyl-2-butene;

2-Methyl-1-butene

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