JEE Main202621 January 2026Morning ShiftChemistryAminesActual
An organic compound ( P ) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br ₂ and KOH forms compound (R) having molecular formula C ₆ H ₇ ~N . Names of P, Q and R respectively are.
Options
- APhenylethanoic acid, phenylethanamide, benzamine
- BBenzoic acid, 4-methylbenzamide, 4-methylaniline
- CToluic acid, methylbenzamide, 2-methylaniline
- DBenzoic acid, benzamide, aniline
Correct answer
D. Benzoic acid, benzamide, aniline
Step-by-step solution
Q + Br₂ /KOH → R is Hofmann bromamide degradation: Amide → Amine (with one less carbon). R = C₆H₇N = C₆H₅NH₂ = Aniline For Q to give aniline, Q = Benzamide ( C₆H₅CONH₂ ) C₆H₅CONH₂ Br₂/KOH C₆H₅NH₂ P + hot NH₃ → Q: Carboxylic acid + NH₃ → Amide P = Benzoic acid ( C₆H₅COOH ) C₆H₅COOH NH₃, C₆H₅CONH₂ P = Benzoic acid, Q = Benzamide, R = Aniline