Amines, on the exam screen.

Nitriles $\mathrm{R}-\mathrm{CN}$ are reduced to primary amines $\mathrm{R}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$ by $\mathrm{LiAlH}_{4}$ followed by hydrolysis, by catalytic hydrogenation $\mathrm{H}_{2}/\mathrm{Ni}$, and by $\mathrm{Na}(\mathrm{Hg})/\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$ (sodium amalgam in ethanol). So Statements I, III and IV are correct. $\mathrm{Sn} + \mathrm{HCl}$ (Stephen reduction) stops at the aldehyde, and $\mathrm{Br}_{2}$/aq. NaOH is the Hoffmann bromamide reaction on amides, not a nitrile reduction.

With silver cyanide ($\mathrm{AgCN}$) the alkyl halide gives the isocyanide, so ethyl chloride yields foul-smelling ethyl isocyanide, $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NC}$ (X = $\mathrm{AgCN}$). In the second route, propanamide undergoes Hoffmann bromamide degradation ($\mathrm{Br}_{2}/\mathrm{NaOH}$) to ethylamine, which then gives ethyl isocyanide via the carbylamine reaction ($\mathrm{CHCl}_{3}$ + ethanolic $\mathrm{KOH}$). Both paths converge on Z = $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NC}$. $\mathrm{KCN}$ would give the nitrile ($\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{CN}$), which is not foul-smelling.

Why the other options are wrong:

- (2) Wrong: $\mathrm{KCN}$ gives nitrile, and nitrile is not the foul-smelling carbylamine product.
- (3) Wrong: $\mathrm{AgCN}$ does not give the nitrile.
- (4) Wrong: $\mathrm{KCN}$ gives nitrile, not isocyanide.

Monochlorination of ethane gives chloroethane $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{Cl}$ (X). Reaction with $\mathrm{NH}_{3}$ gives ethylamine $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2}$ (Y). Treating this primary aliphatic amine with $\mathrm{NaNO}_{2}/\mathrm{HCl}$ forms a highly unstable diazonium salt that reacts with water to release $\mathrm{N}_{2}$ and yield ethanol $\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}$ (Z). Hence (3) is correct.

Reaction B is the Hofmann bromamide degradation, which removes one carbon from an amide and delivers the amine with the nitrogen now bonded directly to the ring - aniline. Reaction D is simple acid hydrolysis of an anilide, cleaving the amide bond to regenerate aniline and acetic acid. Reaction A gives benzylamine, since $\mathrm{LiAlH}_{4}$ reduces the nitrile carbon to $\mathrm{CH}_{2}$ and leaves an extra carbon in the chain, and reaction C fails because $\mathrm{NaBH}_{4}$ is too mild to reduce an aromatic nitro group. NCERT Class 12, Amines (methods of preparation).

Why the other options are wrong:

- (1) includes the nitrile reduction, forgetting that the nitrile carbon is retained and gives benzylamine
- (3) includes both the nitrile reduction and the $\mathrm{NaBH}_{4}$ attempt, neither of which yields aniline
- (4) includes the $\mathrm{NaBH}_{4}$ reduction, confusing it with $\mathrm{Sn}/\mathrm{HCl}$ or catalytic hydrogenation

Each reagent attacks a different group on the same starting material. Acidic permanganate oxidises the methyl group to $-\mathrm{COOH}$, leaving the nitro group intact; the resulting p-nitrobenzoic acid is stronger than benzoic acid because the electron-withdrawing nitro group stabilises the carboxylate anion. $\mathrm{Sn}/\mathrm{HCl}$ instead reduces the nitro group to $-\mathrm{NH}_{2}$, leaving the methyl in place; the resulting p-toluidine is more basic than aniline because the electron-releasing methyl group raises the electron density on nitrogen. Both statements are therefore correct. NCERT Class 12, Amines; Aldehydes, Ketones and Carboxylic Acids.

Only primary amines give the carbylamine test. Aniline is primary, whereas N-methyl-, N-ethyl- and N,N-dimethylaniline are secondary or tertiary. Chapter: Amines.

Why the other options are wrong:

- (1) N-methylaniline is a secondary amine with only one N-H, and an isocyanide needs two N-H hydrogens to form with $\mathrm{CHCl}_{3}$ and alcoholic $\mathrm{KOH}$.
- (2) N,N-dimethylaniline is a tertiary amine with no N-H at all, so no isocyanide can form.
- (3) N-ethylaniline is a secondary amine; its single N-H is not enough to give the foul-smelling isocyanide.