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Organic Chemistry
Found in: Page 306

Short Answer

Consider an E2 reaction between and . What effect does each of the following changes have on the rate of elimination? (a) The base is changed to KOH. (b) The alkyl halide is changed to .

(a) The rate decreases.

(b) The chlorine atom does not depart from the compound readily compared to the bromine atom.

See the step by step solution

Step by Step Solution

Identification of the base

KOH is considered to be a weaker base because of the lack of a bulky group. The more the bulky group, the more it favors elimination reaction.

In the case of the tert-butoxide base, the bulky methyl groups are present, which through the +I effect donate electrons and enriches the electron density of the oxygen atom. Hence, they are more favorable towards elimination reaction.

Identification of the leaving group

As the size of the leaving group increases, the rate of elimination simultaneously increases. Besides this, the rate also increases when the leaving group is a strong conjugate base.

Since the size of Br and conjugate basicity of Br is more than Cl, the rate of elimination increases when Br is present as a leaving group.


(a) KOH is a weaker base than the tert-butoxide base because of the lack of +I effect group, which increases the electron density over the oxygen atom. It doesn’t abstract protons readily; hence the elimination reaction occurs slowly.

By observation, it is considered that when KOH is used, the rate of the elimination reaction decreases. Whereas when is used, the rate of the elimination reaction decreases in this case.

(b) When the alkyl halide is changed to chloroethane from bromoethane, the rate of the E2 reaction decreases since the Br is a more powerful leaving group due to its large size and strong conjugate basicity as compared to Cl.

So, chlorine doesn’t leave the compound as readily as compared to bromine.

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