Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : S_N 2 reaction of C₆ H₅ CH₂ Br occurs more readily than the S_N 2 reaction of CH₃ CH₂ Br.
Reason (R) : The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Next to a phenyl ring, the p-orbital of the S_N2 transition state is stabilised by conjugation.

1. The transition state. In S_N2, the nucleophile attacks from the back as the bromide leaves. At the halfway point the carbon is sp²-like and the arrangement is trigonal bipyramidal. Its leftover p-orbital is partly bonded to both the nucleophile and the leaving group.
2. Check Assertion (A). Both C₆H₅CH₂Br and CH₃CH₂Br are primary halides, so steric crowding at the carbon is about the same. Even so, benzyl bromide reacts faster by S_N2, so (A) is correct.
3. Check Reason (R). In benzyl bromide that p-orbital sits next to the π system of the phenyl ring and overlaps with it. This conjugation spreads out the electron density and lowers the energy of the transition state. A lower transition state means a lower activation energy and a faster reaction, so (R) is correct and it explains (A).
4. So. Both statements are true and (R) gives the reason for (A), which is option (b).

The idea to keep. Benzylic and allylic halides are fast in both SN1 and SN2. In SN1 the ring stabilises the carbocation, and in SN2 it stabilises the p-orbital of the transition state by conjugation.

Answer from the NTA final answer key. Solution by PrepWiser.