JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

Statement A: In benzylamine (C_6H_5CH_2NH_2), the lone pair of electrons on the nitrogen atom is localized and not involved in resonance. In aniline (C_6H_5NH_2), the lone pair is delocalized over the benzene ring. Hence, benzylamine is a stronger base than aniline. Statement A is correct. Statement B: Gabriel phthalimide synthesis involves the nucleophilic substitution (S_N2) of an alkyl halide by the phthalimide anion. Aryl halides do not undergo nucleophilic substitution easily due to the partial double bond character of the C-X bond. Therefore, primary aromatic amines (like p-methoxyaniline) cannot be synthesized by this method. Statement B is incorrect. Statement C: The reaction of 2-phenylacetamide (C_6H_5CH_2CONH_2) with Br_2 and NaOH is the Hoffmann bromamide degradation, which yields benzylamine (C_6H_5CH_2NH_2). Benzylamine is a primary aliphatic (aralkyl) amine, not a primary aromatic amine (where the -NH_2 group must be directly attached to the benzene ring). Statement C is incorrect. Statement D: p-Nitroaniline undergoes diazotization with NaNO_2 and HCl at 0^ C to form p-nitrobenzenediazonium chloride. Heating this aqueous solution ( ) yields p-nitrophenol. Since p-nitrophenol is acidic, it dissolves in aqueous NaOH to form a soluble sodium salt. Statement D is correct. The incorrect statements are B and C. Answer: B and C Only