JEE Main 2026 — Amines Question with Solution

Statement A: In benzylamine ($C_6{H}_{5}C{H}_{2}N{H}_2$), the lone pair of electrons on the nitrogen atom is localized and not involved in resonance. In aniline ($C_6{H}_{5}N{H}_2$), 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 ($S_{N}2$) 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 ($C_6{H}_{5}C{H}_{2}CON{H}_2$) with $B{r}_2$ and $NaOH$ is the Hoffmann bromamide degradation, which yields benzylamine ($C_6{H}_{5}C{H}_{2}N{H}_2$). Benzylamine is a primary aliphatic (aralkyl) amine, not a primary aromatic amine (where the $-N{H}_2$ group must be directly attached to the benzene ring). Statement C is incorrect.

Statement D: p-Nitroaniline undergoes diazotization with $NaN{O}_2$ and $HCl$ at $0^{\circ }C$ to form p-nitrobenzenediazonium chloride. Heating this aqueous solution ($\Delta$) 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