`RNH_2` reacts with `C_6H_5SO_2Cl` in aqueous `KOH` to give a clear solution. On acidification a precepitate is obtained which is due to the formation of

Options

A.
B. `R-N-SO_2C_6K_5K^(+)`
C. `R-NHSO_2C_6H_5`
D. `C_6H_5SO_2NH_2`

Explanation

To solve the problem step by step, let's analyze the reaction between the primary amine (RNH₂) and benzene sulfonyl chloride (C₆H₅SO₂Cl) in the presence of aqueous KOH, followed by acidification.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The primary amine is represented as RNH₂.
- The reagent is benzene sulfonyl chloride, C₆H₅SO₂Cl.

2. **Reaction with Aqueous KOH**:
- When RNH₂ reacts with C₆H₅SO₂Cl in the presence of aqueous KOH, a reaction occurs where the amine attacks the sulfonyl chloride.
- This results in the formation of a sulfonamide (C₆H₅SO₂NHR) and hydrochloric acid (HCl) as a byproduct.
- The reaction can be summarized as:
\[ RNH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR + HCl \]

3. **Formation of a Clear Solution**:
- The product formed (C₆H₅SO₂NHR) is initially soluble in the aqueous KOH solution, leading to a clear solution.

4. **Acidification**:
- Upon acidification of the solution (adding an acid), the sulfonamide (C₆H₅SO₂NHR) precipitates out of the solution.
- This is because the sulfonamide becomes less soluble in acidic conditions.

5. **Precipitate Formation**:
- The precipitate obtained upon acidification is due to the formation of the sulfonamide, which is a white solid.
- The final product can be represented as:
\[ C_6H_5SO_2NHR \quad (\text{white precipitate}) \]

6. **Conclusion**:
- The precipitate formed upon acidification is due to the formation of the sulfonamide (C₆H₅SO₂NHR).

### Final Answer:
The precipitate obtained upon acidification is due to the formation of **C₆H₅SO₂NHR** (a sulfonamide).

---