COMEDK 2025 Evening Solved Paper

Concept:The compound is an optically active primary aliphatic amine with the formula $\mathrm{C}_8\mathrm{H}_{11}\mathrm{N}$.Explanation:The compound reacts with $\mathrm{CHCl}_3$ / Ethanolic KOH to form an isocyanide. This is the carbylamine reaction, which is given only by primary amines. The compound reacts with $\mathrm{NaNO}_2 / \mathrm{HCl}$ to yield an alcohol with liberation of $\mathrm{N}_2$. This is a deamination reaction characteristic of primary aliphatic amines. Aromatic primary amines form stable diazonium salts under these conditions, they do not liberate $\mathrm{N}_2$ readily. Thus, the compound must be a primary aliphatic amine, not an aromatic amine. The compound is optically active, so it must contain a chiral carbon atom. Among the options:Option A (2-ethylaniline) is an aromatic primary amine, not aliphatic, so it would form a diazonium salt and not liberate $\mathrm{N}_2$. Options C and D are also aromatic amines. Option B (1-phenylethanamine, $\mathrm{C}_6\mathrm{H}_5-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3$) is a primary aliphatic amine with a chiral carbon (the carbon attached to $\mathrm{NH}_2$). It will give the carbylamine test and liberate $\mathrm{N}_2$ with $\mathrm{NaNO}_2/\mathrm{HCl}$ forming 1-phenylethanol.Answer:1-Phenylethanamine (Option B).