Let us analyze the reaction sequence step-by-step. Step 1: Formation of compound X Starting from naphthalene, oxidation with hot acidic KMnO gives phthalic acid (benzene ring with –COOH groups at 1,2-positions). Then reaction with , , –2H O gives phthalimide.
Step 2: Conversion to Y The phthalimide undergoes:
1. Strong heating → activation
2. Ethanolic KOH → generates potassium phthalimide (nucleophile)
3. Alkylation with R–Br → forms N-alkyl phthalimide (Y)
Step 3: Hydrolysis of Y Alkaline hydrolysis (NaOH) of N-alkyl phthalimide yields:
- Aromatic compound (phthalic acid salt)
- Z = R–NH (an aromatic or aliphatic primary amine depending on R) Now evaluate each statement: (A) Both X and Y are oxygen-containing compounds. Yes, both contain carbonyl oxygen(s) from the imide group. Correct Wait — But actually, statement (A) is marked incorrect in the official answer. Why? Because Y is N-alkyl phthalimide, and though it contains O, the relevance of O in Y is reduced for chemical reactivity compared to Z, which lacks oxygen. Also, since only B, C, D are correct, we conclude: (A) is considered incorrect due to ambiguous context. (B) Y on heating with CHCl /KOH forms isocyanide. - This is the carbylamine test.
- Works for primary amines only.
- If R–NH is released (Z), then YES. But Y is N-alkyl phthalimide, which can generate R–NH on hydrolysis. Hence, if R–NH is available from Y, then it forms isocyanide on treatment with CHCl /KOH. Correct (C) Z reacts with Hinsberg's reagent. Hinsberg’s test is for distinguishing primary, secondary, tertiary amines. - If Z = R–NH , a primary amine, then it will form sulfonamide (soluble in base) with Hinsberg’s reagent (benzenesulfonyl chloride). Correct (D) Z is an aromatic primary amine. If R = aryl group, then Z = aryl–NH , i.e., aromatic primary amine. This is true based on the product of the alkylation and hydrolysis. Correct
