Question:

Which among the following react with Hinsberg's reagent?


Choose the correct answer from the options given below:

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Hinsberg's reagent is a useful test for distinguishing primary and secondary amines from tertiary amines, as tertiary amines do not react due to the absence of a replaceable hydrogen atom on nitrogen.
Updated On: Nov 24, 2025
  • B and D only
  • C and D only
  • A, B and E only
  • A, C and E only
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The Correct Option is D

Approach Solution - 1

In order to determine which compounds react with Hinsberg's reagent (benzenesulfonyl chloride), we must understand the reaction mechanism:

  • Primary amines react with Hinsberg's reagent to form a sulfonamide, which is soluble in alkaline solutions but insoluble in acidic solutions.
  • Secondary amines react to form a sulfonamide, which is insoluble in both acidic and alkaline solutions, making them less useful for separation.
  • Tertiary amines do not react with Hinsberg's reagent because they lack a hydrogen atom on the nitrogen, which is necessary for the formation of a sulfonamide. 

Given these principles, let's analyze each option:

  1. (A) Aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)): This is a primary amine, which reacts with Hinsberg's reagent.
  2. (B) N,N-dimethylaniline (\( \text{C}_6\text{H}_5\text{N(CH}_3)_2 \)): This is a secondary amine, which reacts with Hinsberg's reagent but is insoluble in acids and bases.
  3. (C) Methylamine (\( \text{CH}_3\text{NH}_2 \)): This is a primary amine, thus it reacts with Hinsberg's reagent.
  4. (D) Trimethylamine (\( \text{N(CH}_3)_3 \)): This is a tertiary amine, so it does not react with Hinsberg's reagent.
  5. (E) Diphenylamine (\( \text{C}_6\text{H}_5\text{NH}\text{C}_6\text{H}_5 \)): This is a secondary amine, and it will also react with Hinsberg's reagent.

Based on the above analysis, the options that react with Hinsberg's reagent are A, C, and E, which aligns with the provided correct answer.

Correct Answer: A, C, and E only

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Approach Solution -2

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 $A, C, E$.

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