Amines can be separated using Hinsberg method. Amines undergo a wide range of reactions like alkylation, acylation, reaction with nitrous acid, carbylamine reaction and condensation reactions with aldehydes. Which of the following on reduction with LiAlH \(_{4}\) will give \(\mathrm{N}\) -methylaniline? (a) Methyl isocyanide (b) Acetamide (c) Phenyl carbylamine (d) Benzonitrile

The first step is to write down the structures of all given compounds so we can analyze them properly. (a) Methyl isocyanide - CH\(_3\)NC (b) Acetamide - CH\(_3\)CONH\(_2\) (c) Phenyl carbylamine - C\(_6\)H\(_5\)NC(-CH\(_3\)) (d) Benzonitrile - C\(_6\)H\(_5\)CN

We can now analyze the reactivity of the given compounds with LiAlH\(_4\). (a) Methyl isocyanide (CH\(_3\)NC) gets reduced by LiAlH\(_4\) to form N-methylformamide (CH\(_3\)NHCHO), which can further be reduced by LiAlH\(_4\). Still, it won't give N-methylaniline as the final product. (b) Acetamide (CH\(_3\)CONH\(_2\)) on reduction with LiAlH\(_4\) will give ethylamine as the final product and not N-methylaniline. (c) Phenyl carbylamine (C\(_6\)H\(_5\)NC(-CH\(_3\))) on reduction with LiAlH\(_4\) will form N-methylaniline. The reaction can be written as: C\(_6\)H\(_5\)NC(-CH\(_3\)) + 4 [LiAlH\(_4\)] → C\(_6\)H\(_5\)NHCH\(_3\) + 4 [LiAlH\(_3\)] (d) Benzonitrile (C\(_6\)H\(_5\)CN) on reduction with LiAlH\(_4\) will form aniline (C\(_6\)H\(_5\)NH\(_2\)) and not N-methylaniline.

After analyzing the reactivity of all given compounds with LiAlH\(_4\), it is clear that the compound that yields N-methylaniline upon reduction with LiAlH\(_4\) is Phenyl carbylamine (C\(_6\)H\(_5\)NC(-CH\(_3\))). So the correct answer is (c) Phenyl carbylamine.