Question:easy

The correct formula of Hinsberg's reagent is :

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Primary amines produce sulphonamides that dissolve in base.
Secondary amines produce sulphonamides that do not dissolve in base.
Tertiary amines do not react with the reagent at all.
Updated On: Jul 22, 2026
  • \( \text{C}_6\text{H}_5\text{COCl} \)
  • \( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \)
  • \( \text{C}_6\text{H}_5\text{CONHCH}_3 \)
  • \( \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Break the name Hinsberg's reagent into its parts.
Hinsberg's reagent is another name for benzene sulphonyl chloride. The name itself tells us the structure: a benzene ring carrying a sulphonyl group that ends in a chlorine atom.
Step 2: Build the formula piece by piece.
Benzene contributes $\text{C}_6\text{H}_5-$. The sulphonyl group is $-\text{SO}_2-$. Attaching a chlorine at the end gives $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$.
Step 3: Check the other options are named differently.
$\text{C}_6\text{H}_5\text{COCl}$ is benzoyl chloride, an acid chloride, not a sulphonyl chloride. $\text{C}_6\text{H}_5\text{CONHCH}_3$ is an amide, and $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$ is simply benzylamine, an amine, not the test reagent.
Step 4: Confirm by its known use.
Hinsberg's reagent reacts with amines to give sulphonamide products, and the solubility of that product in alkali is what lets us tell primary, secondary and tertiary amines apart. This use matches $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$.
\[ \boxed{\text{C}_6\text{H}_5\text{SO}_2\text{Cl}} \]
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