Hinsberg reagent is benzenesulfonyl chloride. In aqueous KOH this reacts with primary amine `(RNH_(2))` to give a clear solution which on acidification gives a white on acidification gives a white precipitate, which is due to the formation of

Options

A. `R-NHSO_(2)C_(6)H_(5)`
B. `C_(6)H_(5)SO_(2)NH_(2)`
C. `R-N-SO_(2)C_(6)H_(5)K^(+)`
D. `R-overset(N)overset(|)underset(H)underset(|)(N^(+))-SO_(2)C_(6)H_(5)OH^(-)`

Answer

To solve the problem regarding the reaction of Hinsberg reagent (benzenesulfonyl chloride) with a primary amine (RNH₂) in aqueous KOH, we can follow these steps:

### Step 1: Understand the Reaction
Hinsberg reagent is benzenesulfonyl chloride (C₆H₅SO₂Cl). When it reacts with a primary amine (RNH₂), it forms a sulfonamide.

### Step 2: Reaction Mechanism
1. The primary amine (RNH₂) attacks the electrophilic sulfur atom in the benzenesulfonyl chloride.
2. This leads to the formation of an intermediate sulfonamide (RNH-SO₂-C₆H₅) and the release of chloride ion (Cl⁻).

### Step 3: Formation of a Clear Solution
In the presence of aqueous KOH, the sulfonamide formed is soluble in the alkaline medium, resulting in a clear solution.

### Step 4: Acidification
When the solution is acidified (for example, with HCl), the sulfonamide becomes insoluble and precipitates out of the solution. This precipitate is the sulfonamide derivative.

### Step 5: Identify the White Precipitate
The white precipitate formed upon acidification is due to the formation of the sulfonamide (RNH-SO₂-C₆H₅). The general formula for the sulfonamide formed is RNH-SO₂-C₆H₅, where R is the alkyl group from the primary amine.

### Conclusion
The white precipitate formed upon acidification is due to the formation of a sulfonamide.

### Final Answer
The white precipitate is due to the formation of **sulfonamide (RNH-SO₂-C₆H₅)**.

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