Question

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

(I) $CH_{3} - \underset{CH_{3}}{\overset{H}{\overset{|}{\underset{|}{C}}} -}CH_{2}Br$

(II) $CH_{3} - CH_{2} - Br$

(III) $CH_{3} - CH_{2} - CH_{2} - Br$

• A. I > II > III
• B. III > II > I
• C. II > III > I
• D. I > III > II

Answer 1

Answer: (D)

I > III > II

Answer 2

:

More the number of $\beta -$substituents (alkyl groups), more stable alkene it will form on $\beta -$elimination and more will be the reactivity. Thus the decreasing order of the rate of $\beta -$elimination reaction with alcoholic KOH is. I > III > II