An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with Br₂ and KOH forms a compound 'C' of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Compound A: Structure: C₆H₅COOH IUPAC Name: Benzoic acid
Compound B: Structure: C₆H₅CONH₂ IUPAC Name: Benzamide
Compound C: Structure: C₆H₅NH₂ IUPAC Name: Benzenamine (or Aniline)
Reaction Sequence: C₆H₅COOH (A) →[NH₃, Δ] C₆H₅CONH₂ (B) →[Br₂, KOH] C₆H₅NH₂ (C)
Explanation
This question tests the student's understanding of Hofmann bromamide reaction. The key insight is that heating an amide with Br₂ and KOH (Hofmann bromamide reaction) produces a primary amine with one less carbon atom. Since compound C has molecular formula C₆H₇N (aniline/benzenamine), compound B must be benzamide (C₆H₅CONH₂) which has 7 carbon atoms. The reaction of benzoic acid with aqueous ammonia followed by heating produces benzamide through the formation of ammonium benzoate intermediate, which undergoes dehydration on heating.
Solution Steps
Step 1: Identify that Br₂ and KOH with heating indicates Hofmann bromamide reaction, which converts an amide to a primary amine with one less carbon.
Step 2: Compound C (C₆H₇N) is an aromatic primary amine - aniline (benzenamine).
Step 3: Since Hofmann reaction removes one carbon, compound B must be benzamide (C₆H₅CONH₂) with 7 carbons.
Step 4: Compound A must be benzoic acid (C₆H₅COOH), which reacts with aqueous ammonia and heating to form benzamide.