A reduction B CHCl3/KOH C reduction N-methylaniline then A is
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- CH3NH2
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Answer
(2) [figure]
Solution
Working backwards:
∵ Reduction of an isocyanide gives a secondary amine containing a methyl group ⇒ C is phenyl isocyanide, C6H5NC, which gives N-methylaniline, C6H5NHCH3.
∵ As per NCERT, a primary amine heated with chloroform and ethanolic KOH forms an isocyanide (carbylamine reaction) ⇒ B is aniline, C6H5NH2.
∵ Aniline is obtained by reduction of nitrobenzene ⇒ A is nitrobenzene.
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Answer: (2)