Major Product in Phthalimide Alkylation Reaction

To determine the major product of the reaction, let us analyze the reaction step-by-step:

Step 1: Deprotonation of Phthalimide
Phthalimide has an acidic proton on the nitrogen atom ($pK_a \approx 8.3$) due to resonance stabilization of the conjugate base across the two adjacent carbonyl groups:
$\text{Phthalimide} + \text{KOH} \longrightarrow \text{Potassium phthalimide} + \text{H}_2\text{O}$
The resulting phthalimide anion is an effective nitrogen nucleophile.

Step 2: Nucleophilic Substitution Reaction
The nucleophile reacts with 4-bromobenzyl chloride ($\text{Br}-\text{C}_6\text{H}_4-\text{CH}_2\text{Cl}$), which contains two potential reaction sites:

1. The aryl bromide site ($\text{C}_{sp^2}-\text{Br}$): The aryl $\text{C}-\text{Br}$ bond possesses partial double-bond character due to resonance with the aromatic $\pi$-system. Additionally, the geometry of the benzene ring prevents backside attack, making $\text{S}_\text{N}2$ displacement virtually impossible under standard conditions.
2. The benzylic chloride site ($\text{C}_{sp^3}-\text{Cl}$): The primary benzylic carbon is sterically accessible and the transition state for an $\text{S}_\text{N}2$ displacement is strongly stabilized by the $\pi$-electron system of the aromatic ring.

Therefore, the reaction proceeds exclusively at the benzylic carbon via an $\text{S}_\text{N}2$ pathway, where the nitrogen atom of the phthalimide anion attacks the benzylic carbon, displacing the chloride ion ($\text{Cl}^-$) as the leaving group:
$\text{Phthalimide}^- + \text{Cl}-\text{CH}_2-\text{C}_6\text{H}_4-\text{Br} \longrightarrow \text{Phthaloyl}-\text{N}-\text{CH}_2-\text{C}_6\text{H}_4-\text{Br} + \text{Cl}^-$

The $\text{N}$-alkylated product formed is $N$-(4-bromobenzyl)phthalimide.

This product corresponds to structure (A).

Correct Option: A