Q15 / 150 · MHT CET 2026 (17 April Shift 2) · Chemistry
Which from following amines when heated with ethanolic KOH and chloroform forms aryl isocyanide ?
- Correct answerABenzenamine
- BN -methylaniline
- CN, N - Dimethylaniline
- DEthylethanamine
Solution
The reaction of an amine with chloroform and ethanolic KOH to form an isocyanide is known as the carbylamine reaction.
This reaction is a characteristic test given only by primary aliphatic and primary aromatic amines.
Among the given options, benzenamine (aniline) is a primary aromatic amine, while N-methylaniline and ethylethanamine are secondary amines, and N,N-dimethylaniline is a tertiary amine.
Therefore, benzenamine reacts with chloroform and ethanolic KOH to form phenyl isocyanide (an aryl isocyanide).
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
Answer: Benzenamine
This reaction is a characteristic test given only by primary aliphatic and primary aromatic amines.
Among the given options, benzenamine (aniline) is a primary aromatic amine, while N-methylaniline and ethylethanamine are secondary amines, and N,N-dimethylaniline is a tertiary amine.
Therefore, benzenamine reacts with chloroform and ethanolic KOH to form phenyl isocyanide (an aryl isocyanide).
$C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
Answer: Benzenamine
Hence, the correct answer is option: 'A'.