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Question
Arrange the following alkyl halides in decreasing order of the rate of β– elimination reaction with alcoholic KOH.
| (A) | \[\begin{array}{cc} \ce{H}\phantom{...}\\ |\phantom{...}\\ \ce{CH3 - C - CH2Br}\\ |\phantom{...}\\ \ce{CH3}\phantom{} \end{array}\] |
| (B) | \[\ce{CH3 - CH2 - Br}\] |
| (C) | \[\ce{CH3 - CH2 - CH2 - Br}\] |
Options
A > B > C
C > B > A
B > C > A
A > C > B
Solution
A > C > B
Explanation:
\[\begin{array}{cc}
\ce{H}\phantom{................................................}\\
|\phantom{................................................}\\
\ce{CH3 - \overset{α}{\underset{|}{C}} - \overset{α}{C}H2Br > \underset{\underset{(has no β-substitutent)}{(II)}}{\overset{α}{C}H3 - \overset{α}{C}H2 - Br} > \underset{\underset{(has 1 β-substitutent)}{(III)}}{CH3 - \overset{β}{C}H2 - \overset{α}{C}H2 - Br}}\\
\ce{\underset{(has 2 β-substitutents)}{\underset{(I)}{CH3}}}\phantom{.............................................}
\end{array}\]
More the number of β-substituents (alkyl groups), more stable alkene will be formed on β -elimination and more will be the reactivity. Thus, the decreasing order of the rate of β -elimination reaction with alcoholic KOH is: A > C > B.
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