Decreasing order of reactivity in Williamson synthesis of the following : I. $\mathrm{Me}_3 \mathrm{CCH}_2…
Decreasing order of reactivity in Williamson synthesis of the following :
I. $\mathrm{Me}_3 \mathrm{CCH}_2 \mathrm{Br}$
II. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
III. $\mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{Cl}$
IV. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$
III $>$ II $>$ IV $>$ I
$\mathrm{I}>\mathrm{II}>\mathrm{IV}>\mathrm{III}$
II $>$ III $>$ IV $>$ I
I $>$ III $>$ II $>$ IV
Solution
$\mathrm{C}-\mathrm{Br}$ 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.
