Question

Explain how the following changes would affect the rate of the reaction of 2-bromo-3-methylbutane with methanol: Part A The alkyl halide is changed to 2-chloro-3-methylbutane. The alkyl halide is changed to 2-chloro-3-methylbutane. The reaction will be slower because the leaving group will be poorer. The reaction will be faster because the leaving group will be better. The reaction will be slower because the leaving group will be better. The rate of the reaction does not change.

Answer 1

The reaction will be slower because the leaving group will be poorer.

Step-by-step explanation:

The given reaction is an example of
`S_(N)1` reaction.

In the first step, bromide ion leaves the molecule to produce a carbocation. This is the rate determining step.

In the second step, methanol attacks the carbocation to form product of the reaction. This is a fast step and does not have influence on rate of reaction.

Therefore, if we replace Br with Cl then formation of carbocation will require more activation energy as
`Br^(-)` is a better leaving group than
`Cl^(-)` due to higher polarizability of

Hence, the reaction will be slower when the alkyl halide is changed to 2-chloro-3-methylbutane.