JEE Main 2024 — Haloalkanes And Haloarenes Question with Solution

To determine the correct answer, we need to analyze both the Assertion (A) and the Reason (R) independently, as well as see if the Reason (R) provides the correct explanation for the Assertion (A).

Assertion (A) Analysis:

Assertion (A) states that $${\mathrm{S}}_{\mathrm{N}}2$$ reaction of $${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{Br}$$ occurs more readily than the $${\mathrm{S}}_{\mathrm{N}}2$$ reaction of $${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}$$. In an $${\mathrm{S}}_{\mathrm{N}}2$$ reaction, the rate depends on the nucleophile as well as the substrate. Specifically, $${\mathrm{S}}_{\mathrm{N}}2$$ reactions proceed through a backside attack mechanism, leading to the formation of a transition state where the nucleophile and the leaving group are simultaneously bonded to the carbon atom. The benzyl group $${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2-$$ is more stable for $${\mathrm{S}}_{\mathrm{N}}2$$ reactions compared to the ethyl group $${\mathrm{CH}}_3{\mathrm{CH}}_2-$$ because the benzyl carbon's partial positive charge in the transition state is stabilized by the resonance of the phenyl ring.

Reason (R) Analysis:

Reason (R) explains that the partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring. This means that the presence of the phenyl ring can delocalize and stabilize the charge in the transition state, facilitating the $${\mathrm{S}}_{\mathrm{N}}2$$ reaction.

Combining Assertion (A) and Reason (R):

Given that the phenyl ring in $${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{Br}$$ can indeed stabilize the transition state through conjugation more effectively than an ethyl group, both the Assertion (A) and Reason (R) are correct. Furthermore, Reason (R) provides a valid explanation for Assertion (A), reinforcing why the $${\mathrm{S}}_{\mathrm{N}}2$$ reaction occurs more readily for $${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{Br}$$.

Thus, the most appropriate answer is:

Option B: Both (A) and (R) are correct and (R) is the correct explanation of (A)