Three isomers A, B, and C (mol. formula C8H11N) give the following results:

• Diazotization (i) Hydrolysis: A and C → P + Q (ii) Oxidation: KMnO₄ + H⁺ → R (product of A) + S (product of C)
• R has a lower boiling point than S
• C₆H₅SO₂Cl
• B → alkali-insoluble product

A, B, and C respectively are:

To solve this problem, we need to analyze the chemical reactions and the properties given for the three isomers A, B, and C with the molecular formula C₈H₁₁N. Let's break down the problem step by step:

Key Information Given:

1. Molecular formula: C₈H₁₁N
   • This suggests that all three compounds are amines with a molecular formula that fits C₈H₁₁N.
2. Diazotization and Hydrolysis:
   • A and C undergo hydrolysis to give products P and Q.
   • Hydrolysis typically involves the reaction of amines with water to produce primary or secondary amines.
3. Oxidation with KMnO₄:
   • KMnO₄ is a strong oxidizing agent and will oxidize primary amines to carboxylic acids.
   • The products of oxidation are R (from A) and S (from C).
   • R has a lower boiling point than S.
4. C₆H₅SO₂Cl:
   • This is benzenesulfonyl chloride, which is commonly used in electrophilic substitution reactions with amines, suggesting that one of the isomers is undergoing a reaction with this group.
5. Alkali-insoluble product:
   • B forms an alkali-insoluble product, which likely means that B is a tertiary amine or a compound that doesn't react with alkaline solutions to form soluble salts.

Step-by-Step Analysis:

1. Diazotization and Hydrolysis:
   • A and C undergo hydrolysis to give two products (P and Q). Given that these are amines, they likely produce corresponding alcohols or acids.
   • This suggests A and C are likely primary or secondary amines.
2. Oxidation with KMnO₄:
   • KMnO₄ + H⁺ oxidizes amines to carboxylic acids. So:
     • R (product of A) could be a carboxylic acid derived from A.
     • S (product of C) could be a carboxylic acid derived from C.
   • Since R has a lower boiling point than S, R is likely a smaller molecule or a less-substituted carboxylic acid compared to S.
3. C₆H₅SO₂Cl Reaction:
   • This group suggests that one of the amines undergoes sulfonation to form an alkyl-substituted sulfonamide. This could be A or C reacting with benzenesulfonyl chloride.
4. Alkali-insoluble Product:
   • B is described as forming an alkali-insoluble product, indicating that B could be a tertiary amine or a bulky amine that does not form soluble salts in alkali solutions.

Deduction of the Isomers:

• A and C are likely primary or secondary amines.
• B is likely a tertiary amine because it forms an alkali-insoluble product.

Final Conclusion:

Based on the analysis of the reactions and the given information, the isomers A, B, and C can be identified as:

• A: Aniline (C₆H₅NH₂) — Reacts with KMnO₄ to form a carboxylic acid and is likely involved in electrophilic substitution with C₆H₅SO₂Cl.
• B: N,N-Dimethylaniline — A tertiary amine that forms an alkali-insoluble product.
• C: N-ethyl aniline — Reacts with KMnO₄ to give a carboxylic acid, with a higher boiling point than R.

Therefore, A, B, and C respectively are:

• A: Aniline
• B: N,N-Dimethylaniline
• C: N-ethyl aniline