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JEE Main 2026 · 6 Apr · Shift 2, Chemistry question 68

The number of compounds from the following which can undergo reaction with BrX2\ce{Br2}/KOH (alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is :

Six amides: benzamide C₆H₅CONH₂; 2-phenylacetamide C₆H₅CH₂CONH₂; acetamide CH₃CONH₂; N-ethylcyclohexanecarboxamide C₆H₁₁CONHCH₂CH₃; N-methyl-2,2-dimethylpropanamide (H₃C)₃CCONHCH₃; cyclohexanecarboxamide C₆H₁₁CONH₂.
Choose one option
Answer and worked solution

Official answer: C

Hofmann products and Gabriel. Only primary amides (−CONHX2\ce{-CONH2}) undergo Hofmann degradation, giving an amine with one carbon fewer.

  • Benzamide gives aniline. Gabriel cannot make aryl amines.
  • 2-Phenylacetamide gives benzylamine. Gabriel works.
  • Acetamide gives methylamine. Gabriel works.
  • Cyclohexanecarboxamide gives cyclohexylamine. Counted as obtainable by Gabriel.
  • The two N-substituted amides do not undergo Hofmann degradation.

Count: 3.

(C)

Topics:AminesAcid derivatives
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