Identify the reactions which give aniline as the major product. (A) C₆H₅CN →[LiAlH₄]; (B) C₆H₅CONH₂ →[KOH, Br₂]; (C) C₆H₅NO₂ →[NaBH₄]; (D) C₆H₅NHCOCH₃ →[HCl, H₂O, Δ]. Choose the correct answer:

Solution & Explanation

Concept: identify which routes convert the benzene-attached group into a free –NH₂ on the ring (aniline, C₆H₅NH₂), without adding an extra carbon. A. C₆H₅CN + LiAlH₄ → C₆H₅CH₂NH₂ (benzylamine). The nitrile is reduced to a –CH₂NH₂, giving a one-carbon-longer benzylamine, NOT aniline. ✗ B. C₆H₅CONH₂ + Br₂/KOH → C₆H₅NH₂. This is the Hofmann bromamide degradation: the amide loses its carbonyl carbon (as carbonate) and the nitrogen migrates onto the ring → aniline. ✓ C. C₆H₅NO₂ + NaBH₄ → no reaction. NaBH₄ is too mild to reduce an aromatic –NO₂ group (needs Sn/HCl, Fe/HCl or H₂/catalyst). ✗ D. C₆H₅NHCOCH₃ + HCl/H₂O, Δ → C₆H₅NH₂ + CH₃COOH. Acid hydrolysis of acetanilide cleaves the amide to regenerate aniline. ✓ Trap: LiAlH₄ on a nitrile adds a CH₂ (benzylamine, not aniline), and NaBH₄ cannot touch the aromatic nitro group. So aniline is the major product in B and D only — the answer is B.