A) Aniline
B) Methylamine
C) Diphenylamine
D) Ethylamine
Correct Answer: A
Solution :
Key Idea: Only \[{{1}^{o}}\] aromatic amines on reaction with \[\text{HN}{{\text{O}}_{\text{2}}}\]and phenol produce dye. Aliphatic amines liberate \[{{\text{N}}_{\text{2}}}\]and form alcohol on reaction with\[HN{{O}_{2}}.\] [a] [ b] \[C{{H}_{3}}N{{H}_{2}}\] [c] \[\underset{{{C}_{6}}{{H}_{5}}}{\overset{H}{\mathop{\underset{|}{\overset{|}{\mathop{N}}}\,}}}\,-{{C}_{6}}{{H}_{5}}\] [d] \[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\] \[\because \] [a] is \[{{1}^{o}}\] aromatic amine (aniline) \[\therefore \]It gives dye test.You need to login to perform this action.
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