Question:
Identify the reagent used to distinguish primary and secondary amines by Hinsberg test.
Identify the reagent used to distinguish primary and secondary amines by Hinsberg test.
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Hinsberg reagent is benzenesulfonyl chloride.
Updated On: Oct 1, 2026
- \(\text{NaNO}_2\) and HCl
- \(\text{FeCl}_3\) (alc)
- \(\text{C}_6\text{H}_5\text{SO}_2\text{Cl}\)
- \(\text{CHCl}_3\) and KOH (alc)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The Hinsberg test separates primary, secondary and tertiary amines using benzenesulfonyl chloride, \(\text{C}_6\text{H}_5\text{SO}_2\text{Cl}\).
Step 2: What happens
A primary amine gives a sulfonamide that has an acidic N-H, so it dissolves in alkali. A secondary amine gives a sulfonamide with no acidic hydrogen, which is insoluble in alkali. A tertiary amine does not react in the same way.
Step 3: Check the options
(A) \(\text{NaNO}_2\)/HCl is used for the nitrous acid test. (B) Alcoholic \(\text{FeCl}_3\) is a test for phenols. (D) \(\text{CHCl}_3\) and alcoholic KOH is the carbylamine test for primary amines only. (C) Benzenesulfonyl chloride is the Hinsberg reagent.
Final Answer:
Hinsberg's reagent is benzenesulfonyl chloride, option (C). \[ \boxed{\text{C}_6\text{H}_5\text{SO}_2\text{Cl}} \]
The Hinsberg test separates primary, secondary and tertiary amines using benzenesulfonyl chloride, \(\text{C}_6\text{H}_5\text{SO}_2\text{Cl}\).
Step 2: What happens
A primary amine gives a sulfonamide that has an acidic N-H, so it dissolves in alkali. A secondary amine gives a sulfonamide with no acidic hydrogen, which is insoluble in alkali. A tertiary amine does not react in the same way.
Step 3: Check the options
(A) \(\text{NaNO}_2\)/HCl is used for the nitrous acid test. (B) Alcoholic \(\text{FeCl}_3\) is a test for phenols. (D) \(\text{CHCl}_3\) and alcoholic KOH is the carbylamine test for primary amines only. (C) Benzenesulfonyl chloride is the Hinsberg reagent.
Final Answer:
Hinsberg's reagent is benzenesulfonyl chloride, option (C). \[ \boxed{\text{C}_6\text{H}_5\text{SO}_2\text{Cl}} \]
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