JEE Chemistry 2026 Question with Solution

Given: The reactivity order of substituted benzyl halides towards reaction with KCN is to be determined.

Find: The correct sequence among a, b, c, d.

For benzylic substitution with KCN in a polar aprotic medium, the reaction generally proceeds through an $S_N2$ pathway, but the benzylic transition state is influenced by substituents through resonance and inductive effects.

The solution states that the developing positive character in the transition state is stabilized by electron-donating groups.

1. p-amino (b): $-NH_2$ is a strong electron-donating group by the $+M$ effect, so it gives maximum stabilization.
2. p-hydroxy (a): $-OH$ also donates by the $+M$ effect, but less strongly than $-NH_2$.
3. m-nitro (d): $-NO_2$ at meta position mainly shows the $-I$ effect.
4. p-nitro (c): $-NO_2$ at para position shows both $-I$ and strong $-M$ effects, so it destabilizes the cation-like transition state the most.

Therefore, the reactivity order is b $>$ a $>$ d $>$ c.

The correct option is D.

Given: Four benzyl bromides differ by ring substituents: para hydroxy, para amino, para nitro, and meta nitro.

Find: Which one reacts fastest with KCN, and hence the overall order.

The key idea from the solution is stabilization of the benzylic substitution transition state.

• Groups with strong resonance donation increase reactivity.
• Groups with strong withdrawal decrease reactivity.
• The position of the substituent matters because resonance interaction is stronger at para than at meta.

Thus:

• b with $-NH_2$ is most reactive.
• a with $-OH$ is next.
• d with meta $-NO_2$ is less reactive than donor-substituted compounds but more reactive than para nitro.
• c with para $-NO_2$ is least reactive.

Hence the final order is b $>$ a $>$ d $>$ c, so the correct option is D.