Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : $$S_N2$$ reaction of $$C_6H_5CH_2Br$$ occurs more readily than the $$S_N2$$ reaction of $$CH_3CH_2Br$$. 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 :

We need to evaluate the Assertion (A) and Reason (R) about $$S_N2$$ reactions of benzyl bromide versus ethyl bromide.

Considering Assertion (A): "$$S_N2$$ reaction of $$C_6H_5CH_2Br$$ (benzyl bromide) occurs more readily than the $$S_N2$$ reaction of $$CH_3CH_2Br$$ (ethyl bromide)."

This assertion is correct. Benzyl bromide undergoes $$S_N2$$ reactions faster than ethyl bromide. The relative reactivity order for $$S_N2$$ reactions is: benzyl > allyl > primary > secondary. The enhanced reactivity of benzyl halides in $$S_N2$$ is due to stabilization of the transition state (explained below).

Considering Reason (R): "The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring."

This reason is correct and provides the correct explanation for Assertion (A). Here is why:

In an $$S_N2$$ mechanism, the nucleophile attacks the substrate carbon from the backside while the leaving group departs simultaneously. The transition state has a trigonal bipyramidal geometry around the central carbon.

In this transition state, the central carbon changes from $$sp^3$$ hybridization (in the reactant) toward $$sp^2$$-like character. The carbon develops a partially bonded, unhybridized p-orbital that is perpendicular to the plane formed by the three remaining substituents.

In benzyl bromide, this developing p-orbital at the transition state can overlap (conjugate) with the $$\pi$$-electron system of the adjacent phenyl ring. This conjugation delocalizes electron density and stabilizes the transition state, thereby lowering the activation energy.

In ethyl bromide, there is no such conjugation possible with the methyl group, so the transition state is not stabilized to the same extent.

Since both (A) and (R) are correct, and (R) is the correct explanation for (A), the correct answer is Option (4): Both (A) and (R) are correct and (R) is the correct explanation of (A).