Chemistry - Halogen Derivative Question with Solution | TestHub

$\mathrm{C}-\mathrm{Br}$ bond is weaker than $\mathrm{C}-\mathrm{Cl}$ bond, therefore, alkyl bromide (II) reacts faster than alkyl chlorides, (III) and (IV). Since ${\mathrm{CH}}_2=\mathrm{CH}-$ is electron withdrawing therefore, ${\mathrm{CH}}_2$ has more +ve charge on III than on IV.
In other words, nucleophilic attack occurs faster on III than on IV. Further, since Williamson synthesis occurs by ${\mathrm{S}}_{\mathrm{N}}2$ mechanism, therefore, due to steric hindrance alkyl bromide (I) is the least reactive. Thus, the decreasing order of reactivity is II $>$ III $>$ IV $>$ I.