Comprehension:
Physical and Chemical properties of Amines
Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp3 hybridised. Each of the three sp3 hybridised orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines. The fourth orbital of nitrogen in all amines contains an unshared pair of electrons. Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base as well as act as a nucleophile. Lower aliphatic amines are soluble in water because they form hydrogen bond with water molecules and also form strong intermolecular hydrogen bonding. Solvation, +I and +R groups affect the electron density over nitrogen atom. Their pKb value lies in the range of 3 to 4.22. The most widely used method for preparing amines involves the reduction of nitro group, nitriles, amides, etc. using different reagents like H2/Pd, Sn+HCl, H2/Ni, LiAlH4+H2O, Br2+NaOH etc. Primary amines can be easily detected by the carbylamine test and Hinsberg's reagent. Diazonium salt can be synthesized from amines easily for synthesis of variety of new compounds.
The product (A) for the given reaction will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → CH3NH2
When a primary amide is warmed with bromine and an aqueous or ethanolic solution of sodium hydroxide, it gives a primary amine with one carbon atom less than the amide. This is the Hofmann bromamide degradation. The carbonyl carbon is removed as carbonate, and the alkyl group moves from the carbonyl carbon to the nitrogen:
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
The given reaction is this degradation, with R = \( \ce{CH3} \). Acetamide (ethanamide, two carbon atoms) therefore gives methylamine (methanamine, one carbon atom):
\( \displaystyle \ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Two other routes from amides and nitriles are easy to confuse with it: reduction of an amide with LiAlH4 gives an amine with the same number of carbon atoms (\( \ce{CH3CONH2} \) → \( \ce{CH3CH2NH2} \)), and hydrolysis of an amide gives the carboxylic acid (\( \ce{CH3CONH2} \) → \( \ce{CH3COOH} \)).
So the product A is CH3NH2.
Why the other options are incorrect
Option (2) CH3CH2NO2 — Incorrect. CH3CH2NO2 is a nitro compound with two carbon atoms, as many as acetamide; Hofmann degradation removes one carbon and gives an amine, not a nitro compound.
Option (3) CH3COOH — Incorrect. Acetic acid is the product of hydrolysis of the amide, which does not need bromine and keeps the carbonyl carbon.
Option (4) CH2CH3NH2 — Incorrect. An ethylamine (two carbons) would need LiAlH4 reduction, which keeps all the carbon atoms.
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction
