An aromatic compound 'A' on heating with Br₂ and KOH forms compound 'B' of molecular formula C₆H₇N, which on reacting with CHCl₃ and alcoholic KOH produces a foul-smelling compound 'C'. Write the structures and IUPAC names of compounds A, B and C.
Answer: A = benzamide C₆H₅CONH₂; B = aniline C₆H₅NH₂ (benzenamine); C = phenyl isocyanide C₆H₅NC (phenyl carbylamine).
Solution
Br/KOH on A giving CHN (one C fewer) ⇒ Hofmann bromamide ⇒ A is the amide:
A = \textbf{benzamide}, CHCONH.
B = \textbf{aniline (benzenamine)}, CHNH.
Carbylamine on B:
C = \textbf{phenyl isocyanide (isocyanobenzene)}, CHNC — the foul-smelling product.
A = \textbf{benzamide}, CHCONH.
B = \textbf{aniline (benzenamine)}, CHNH.
Carbylamine on B:
C = \textbf{phenyl isocyanide (isocyanobenzene)}, CHNC — the foul-smelling product.
CBSE Class 12 Board · 2019 · Subjective