The number of compounds from the following which can undergo reaction with Br₂/KOH/(alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is:
(see structures)ABCDEF
(A)5
(B)4
(C)3✓ Correct
(D)6
Answer: (C)
Solution
The Hofmann bromamide reaction with Br2/KOH needs an unsubstituted amide −CONH2, so the two N-substituted amides (the N-ethyl and the N-methyl amide) do not react at all. Of the unsubstituted amides, benzamide gives aniline, an aromatic amine that the Gabriel phthalimide synthesis cannot make because aryl halides do not undergo the required substitution. Phenylacetamide gives benzylamine, acetamide gives methylamine and cyclohexanecarboxamide gives cyclohexylamine, each a primary amine that the Gabriel route also delivers. Three compounds meet both requirements.
JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option