The major product $Z$ formed in the following sequence of reactions is: $\mathrm{C}{}_2\mathrm{H}{}_6\stackrel{\mathrm{Cl}{}_2/\mathrm{UV}}{\to }\mathrm{X}\stackrel{\mathrm{NH}{}_3}{\to }\mathrm{Y}\underset{(\mathrm{ii})\mathrm{H}{}_2\mathrm{O}}{\overset{(\mathrm{i})\mathrm{NaNO}{}_2/\mathrm{HCl}}{\to }}\mathrm{Z}$

Solution & Explanation

\textbf{Answer:} (C) A primary aliphatic amine reacts with nitrous acid to give an unstable diazonium salt that instantly loses $\mathrm{N}{}_2$ and is converted to the alcohol. \textbf{Solution:} Free-radical monochlorination of ethane gives chloroethane: $\mathrm{C}{}_2\mathrm{H}{}_6\stackrel{\mathrm{Cl}{}_2/\mathrm{UV}}{\to }\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{Cl}$ ($X$). Reaction with ammonia (ammonolysis) gives the primary amine ethylamine: $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{Cl}\stackrel{\mathrm{NH}{}_3}{\to }\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NH}{}_2$ ($Y$). Treatment with nitrous acid ($\mathrm{NaNO}{}_2/\mathrm{HCl}$) diazotises the primary aliphatic amine, but unlike aromatic diazonium salts the aliphatic diazonium ion $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}$ has no ring resonance to stabilise it and decomposes instantly, evolving $\mathrm{N}{}_2$; with water present it yields the alcohol: $\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{NH}{}_2\underset{\mathrm{H}{}_2\mathrm{O}}{\overset{\mathrm{NaNO}{}_2/\mathrm{HCl}}{\to }}\mathrm{C}{}_2\mathrm{H}{}_5\mathrm{OH}+\mathrm{N}{}_2\uparrow$. Thus $Z$ is ethanol (C). The nitro compound (A), the diazoate (B) and the unreacted amine (D) are not the major products.