NEETChemistryHydrocarbons
Match List I with List II. List I (Reaction Sequence) List II (Major Product) A. Aniline (i) Br ₂ (aq) (ii) NaNO ₂ /HCl, 273 K (iii) H ₃ PO ₂ I. 1,3,5-Tribromobenzene B. Aniline (i) Ac ₂ O, pyridine (ii) Br ₂ /CH ₃ COOH (iii) H ₃ O ^+ (iv) NaNO ₂ /HCl, 273 K (v) H ₃ PO ₂ II. Bromobenzene C. p -Toluidine (i) NaNO ₂ /HCl, 273 K (ii) CuCN III. 4-Methylbenzonitrile D. p -Nitroaniline (i) NaNO ₂ /HCl, 273 K (ii) HBF ₄ , I
Options
- AA-I, B-II, C-III, D-IV
- BA-II, B-I, C-III, D-IV
- CA-I, B-II, C-IV, D-III
- DA-II, B-I, C-IV, D-III
Correct answer
A. A-I, B-II, C-III, D-IV
Step-by-step solution
A. Aniline reacts with bromine water to form 2,4,6-tribromoaniline. Diazotization followed by reduction with H ₃ PO ₂ replaces the diazonium group with hydrogen, yielding 1,3,5-tribromobenzene. (A I) B. Protection of aniline with acetic anhydride yields acetanilide, which on bromination gives p -bromoacetanilide as the major product. Hydrolysis yields p -bromoaniline. Diazotization and subsequent reduction with H ₃ PO ₂ removes the amino group, yielding bromobenzene. (B II) C. p -Toluidine is diazotized and treated