NEETChemistryAmines
Match List I with List II: List I (Reactions) List II (Products) A. CH₃CONH₂ + Br₂/NaOH I. Benzylamine B. C₆H₅CN + LiAlH₄ II. Aniline C. C₆H₅NC + LiAlH₄ III. Methanamine D. C₆H₅NO₂ + Sn/HCl IV. N-Methylaniline Choose the correct answer from the options given below:
Options
- AA-III, B-I, C-IV, D-II
- BA-III, B-IV, C-I, D-II
- CA-III, B-I, C-II, D-IV
- DA-IV, B-I, C-III, D-II
Correct answer
A. A-III, B-I, C-IV, D-II
Step-by-step solution
A. Acetamide ( CH₃CONH₂ ) undergoes Hoffmann bromamide degradation with Br₂/NaOH to form methanamine ( CH₃NH₂ ), a primary amine with one carbon less. (A-III) B. Benzonitrile ( C₆H₅CN ) is reduced by LiAlH₄ to form benzylamine ( C₆H₅CH₂NH₂ ). (B-I) C. Phenyl isocyanide ( C₆H₅NC ) is reduced by LiAlH₄ to form a secondary amine, N-methylaniline ( C₆H₅NHCH₃ ). (C-IV) D. Nitrobenzene ( C₆H₅NO₂ ) is reduced by Sn/HCl to form aniline ( C₆H₅NH₂ ). (D-II) Therefore, the correct matching is A-III, B-I, C-IV, D-II. Answer: A