An organic compound (A) on treatment with NH3 gives (B), which on heating gives (C). (C) on treatment with Br2 and KOH gives ethylamine. Compound (A) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → CH3CH2COOH
The last step, an amide with \( \ce{Br2} \) and KOH, is the Hoffmann bromamide degradation, which gives a primary amine with one carbon less than the amide. Ethylamine has 2 carbon atoms, so (C) is the 3-carbon amide propanamide, and (A) is the 3-carbon acid, propanoic acid.
Step 1 – (A) → (B): the acid reacts with ammonia to give the ammonium salt.
\( \displaystyle \ce{CH3CH2COOH + NH3 -> CH3CH2COONH4} \)
Step 2 – (B) → (C): heating the ammonium salt removes water and gives the amide.
\( \displaystyle \ce{CH3CH2COONH4} \xrightarrow{\Delta} \ce{CH3CH2CONH2 + H2O} \)
Step 3 – (C) → ethylamine: the ethyl group moves from the carbonyl carbon to nitrogen, and the carbonyl carbon leaves as carbonate.
\( \displaystyle \ce{CH3CH2CONH2 + Br2 + 4KOH -> CH3CH2NH2 + K2CO3 + 2KBr + 2H2O} \)
NCERT (2026-27 edition):
8.9.2 Reactions Involving Cleavage of C–OH Bond – “4. Reaction with ammonia
Carboxylic acids react with ammonia to give ammonium salt which on further heating at high temperature give amides.”
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.”