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Question: Arrange the following alkyl halides in decreasing order of the rate of \(\beta\) -elimination reacti...

Arrange the following alkyl halides in decreasing order of the rate of β\beta -elimination reaction with alcoholic KOH.

(i) CH3−C∣H∣CH3−CH2BrCH_{3} - \underset{}{\underset{\underset{\text{C}\text{H}_{3}}{|}}{\overset{H}{\overset{|}{C}}} - CH_{2}Br}

(ii) CH3−CH2−BrCH_{3} - CH_{2} - Br

(iii) CH3−CH2−CH2−BrCH_{3} - CH_{2} - CH_{2} - Br

A

(i) > (ii) > (iii)

B

(iii) > (ii) > (i)

C

(ii) > (iii) > (i)

D

(i) > (iii) > (ii)

Answer

(i) > (iii) > (ii)

Explanation

Solution

: More the number of β−\beta -substituent more stable alkene it will give on β−\beta -elimination. Since (i) has two, (iii) has one β−\beta -methyl substituent while (ii) has no β−\beta -methyl substituent, therefore reactivity towards β−\beta -elimination decreases in the order : (i) > (iii) > (ii)