The product A is
\(C_6H_5-NH_2+CHCl_3 + KOH (alc.)\ \rightarrow A\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(C_6H_5NC\)
Aniline is a primary aromatic amine. On heating with chloroform and ethanolic (alcoholic) KOH it gives phenyl isocyanide (phenyl carbylamine), a foul-smelling compound. This is the carbylamine reaction, the test for primary amines:
\( \ce{C6H5NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{C6H5NC + 3KCl + 3H2O} \)
Secondary and tertiary amines do not give this reaction.
Why the other options are incorrect
Option (2) \(C_6H_5NH_2\) — Incorrect. This is aniline itself, the starting material.
Option (3) \(C_6H_5CN\) — Incorrect. Benzonitrile (phenyl cyanide) is the isomer bonded through C; in the carbylamine reaction the new carbon is attached to nitrogen, giving the isocyanide.
Option (4) \(C_6H_5OH\) — Incorrect. Phenol is made from aniline through the diazonium salt and warm water, not with \( \ce{CHCl3} \)/KOH.
NCERT (2026-27 edition):
Carbylamine Reaction – “Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction.”