An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br2 and KOH forms compound (R) having molecular formula C6H7N. Names of P, Q and R respectively are.
- Toluic acid, methylbenzamide, 2-methylaniline
- Benzoic acid, benzamide, aniline
- Benzoic acid, 4-methylbenzamide, 4-methylaniline
- Phenylethanoic acid, phenylethanamide, benzamine
Answer
(2) Benzoic acid, benzamide, aniline
Solution
R = C6H7N = C6H5NH2 (aniline).
Hoffmann bromamide degradation: RCONH2 + Br2 + 4KOH RNH2 + K2CO3 + 2KBr + 2H2O
The amine has one carbon atom less than the amide Q = C6H5CONH2 (benzamide).
An acid on heating with ammonia gives the ammonium salt, which loses water to give the amide:
C6H5COOH + NH3 C6H5COO-NH4+ C6H5CONH2 + H2O
P = benzoic acid, Q = benzamide, R = aniline.
(Option (3) would give 4-methylaniline, C7H9N; option (4) would give benzylamine, C7H9N.)
Answer: (2)