The correct order of reactivity of the follow | vedclass

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(B) The reaction of benzyl halides with $KCN$ typically proceeds via an $S_{N}1$ mechanism in polar protic solvents or $S_{N}2$ depending on condition

Vedclass · Accepted Answer

$b > a > d > c$

Vedclass · Answer

(B) The reaction of benzyl halides with $KCN$ typically proceeds via an $S_{N}1$ mechanism in polar protic solvents or $S_{N}2$ depending on conditions. However, for benzyl halides, the rate is governed by the stability of the intermediate carbocation in $S_{N}1$ or steric/electronic factors in $S_{N}2$. In both cases, electron-donating groups $(EDG)$ increase reactivity, while electron-withdrawing groups $(EWG)$ decrease it. $a$ has $-OH$ (strong $EDG$ via resonance), $b$ has $-NH_2$ (strong $EDG$ via resonance), $c$ has $-NO_2$ (strong $EWG$ at para position), and $d$ has $-NO_2$ (strong $EWG$ at meta position). Comparing $a$ and $b$: $-NH_2$ is a stronger $EDG$ than $-OH$ due to lower electronegativity of $N$ compared to $O$, making $b$ more reactive than $a$. Comparing $c$ and $d$: $-NO_2$ at the para position $(c)$ exerts a stronger destabilizing effect on the carbocation than at the meta position $(d)$ due to direct resonance destabilization. Thus, the order is $b > a > d > c$.

The correct order of reactivity of the following benzyl halides towards reaction with $KCN$ is:

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