Question

Arrange the following alkyl halides in decreasing order of the rate of $\beta$- elimination reaction with alcoholic KOH.
${CH}_3{CH}_2{CH}_{2}Br{CH}_3-\underset{\underset{{CH}_3}{|}}{CH}-{CH}_{2}Br{CH}_3-{CH}_2-\underset{\underset{Br}{|}}{CH}-{CH}_3$
(i) (ii) (iii)

• A. $\left(ii\right)>\left(iii\right)>\left(i\right)$
• B. $\left(iii\right)>\left(ii\right)>\left(i\right)$
• C. $\left(i\right)>\left(ii\right)>\left(iii\right)$
• D. $\left(ii\right)>\left(i\right)>\left(iii\right)$

Answer 1

Answer: (B)

$\left(iii\right)>\left(ii\right)>\left(i\right)$

Rate of dehydrohalogenation in alkyl halide is in the order $\Rightarrow 3°>2°>1°$.

Thus, correct order is $\left(iii\right)>\left(ii\right)>\left(i\right)$.