Hinsberg's reagent is benzenesulfonyl chloride ($C_6H_5SO_2Cl$). It's used to distinguish between primary, secondary, and tertiary amines.
1. Reaction with Different Amines:
- Primary amines react with Hinsberg's reagent to form an N-alkyl or N-aryl benzenesulfonamide. The product is soluble in alkali due to the presence of acidic hydrogen.
- Secondary amines react to form N,N-dialkyl or N,N-diaryl benzenesulfonamides. The product is insoluble in alkali because it has no acidic hydrogen.
- Tertiary amines do not react with Hinsberg's reagent because they do not have a replaceable hydrogen atom on the nitrogen.
2. Analyzing the Given Options:
- (A) Aniline ($C_6H_5NH_2$) is a primary amine.
- (2) N,N-Dimethylaniline ($C_6H_5N(CH_3)_2$) is a tertiary amine.
- (C) Methylamine ($CH_3NH_2$) is a primary amine.
- (4) Trimethylamine ($N(CH_3)_3$) is a tertiary amine.
- (E) Diphenylamine ($(C_6H_5)_2NH$) is a secondary amine.
3. Conclusion:
Therefore, options A, C, and E will react with Hinsberg's reagent.
Final Answer:
The final answer is $A, C, E$.