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.
Compound (B) of the passage is heated with Br2 and KOH to give compound (C). Which of the following inorganic products are formed along with compound (C) in this reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → K2CO3 and KBr
Heating an amide with bromine and an alkali is the Hoffmann bromamide degradation. The alkyl or aryl group moves from the carbonyl carbon to the nitrogen atom, so the amine formed has one carbon atom less than the amide.
The carbonyl carbon that is lost leaves as carbonate, and the bromine ends up as bromide ions. With KOH as the alkali, these are potassium salts (with NaOH as the alkali, they are Na2CO3 and NaBr):
\( \displaystyle \ce{R-CO-NH2 + Br2 + 4KOH -> R-NH2 + K2CO3 + 2KBr + 2H2O} \)
So the inorganic products formed along with the amine are potassium carbonate and potassium bromide (and water).
Why the other options are incorrect
Option (2) KBrO3 and CO — Incorrect. Bromine (oxidation state 0) is reduced to bromide (−1), not oxidised to bromate. In the strongly alkaline mixture the lost carbon appears as carbonate, not carbon monoxide.
Option (3) KBr and NH3 — Incorrect. The nitrogen of the amide is kept in the amine formed; it is not given off as ammonia.
Option (4) KCN and K2CO3 — Incorrect. No cyanide is formed. The carbon and the nitrogen of the –CONH2 group separate: the nitrogen stays with the alkyl or aryl group, and the carbon leaves as carbonate.
NCERT (2026-27 edition):
9.4 Preparation of Amines – “6. Hoffmann bromamide degradation reaction: Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.”
\( \displaystyle \ce{R-CO-NH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)