(a) Identify X and Y in the following reaction sequence and also write the names of the two reactions involved:
$R - C O N H_{2} B r_{2} / N a O H - ------ \to X C H C l_{3} / K O H (a l c .) - ---------- \to Y$
(b) Alkanamine is more basic than ammonia. Give reasons.

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Solution

(a) The two reactions involved are (i) Hoffmann bromide degradation which converts an amide to amine (with one C atom less) and (ii) Hoffmann's carbylamine test/ isocyanide test which converts amine to isocyanide.
$R - C O N H_{2} B r_{2} / N a O H - -------------- \to N a_{2} C O_{3} + 2 N a B r + 2 H_{2} O R - N H_{2} X C H C l_{3} / K O H (a l c .) - ---------- \to - K C l + H_{2} O R - N C Y$
(b) Alkanamine is more basic than ammonia due to $+ I$ effect of alkyl group. Due to this, the electron density on nitrogen increases and the lone pair of electrons becomes readily available for protonation. The alkyl ammonium ion is stabilized due to dispersal of positive charge by $+ I$ effect of alkyl group.

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