Question

Arrange the following alkyl halides in decreasing
order of the rate of β–elimination reaction with
alcoholic KOH.
H
丨
(A) CH₃—C—CH₂Br
丨
CH₃
(B) CH₃—CH₂—Br
(C) CH₃—CH₂—CH₂—Br
(a) A > B > C (b) C > B > A
(c) B > C > A (d) A > C > B

Answer 1

d) Option is correct.....

Answer 2

(d) A > C > B

Explanation:

More the number of β- substituents (alkyl groups), more stable alkene it will form on β- elimination and more will be the reactivity. Hence, the decreasing order of the rate of β-elimination reaction with alcoholic KOH is will be: 2 β-substituents > 1 β-substituent > 0 β-substituent.

So A > C > B

Option D is the answer.