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RNH2 reacts with C6H5SO2Cl in aqueous KO...

`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

A

`underset(H)underset(|)underset(N)overset(H)overset(|)(R-N^(+))-SO_2C_6H_5OH^-`

B

`C_6H_5SO_2NH_2`

C

`R-N^(-) SO_2C_6H_5K^+`

D

`R-NHSO_2C_6H_5`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze the reaction between RNH₂ (a primary amine) and C₆H₅SO₂Cl (benzenesulfonyl chloride) in the presence of aqueous KOH. ### Step-by-Step Solution: 1. **Identify the Reactants**: - The reactants are RNH₂ (a primary amine) and C₆H₅SO₂Cl (benzenesulfonyl chloride). 2. **Reaction with Aqueous KOH**: - When RNH₂ reacts with C₆H₅SO₂Cl, the amine attacks the sulfur atom in the sulfonyl chloride, leading to the formation of a sulfonamide. - The reaction can be represented as: \[ RNH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR + HCl \] - This reaction results in the formation of a sulfonamide (C₆H₅SO₂NHR) and hydrochloric acid (HCl). 3. **Formation of a Clear Solution**: - The product, C₆H₅SO₂NHR, is soluble in aqueous KOH, thus resulting in a clear solution. 4. **Acidification**: - Upon acidification (adding an acid, such as HCl or H₂SO₄), the sulfonamide can precipitate out of the solution. - The precipitate formed is due to the conversion of the sulfonamide back to its original amine form, which is less soluble in acidic conditions. 5. **Identify the Precipitate**: - The precipitate formed upon acidification is the sulfonamide itself (C₆H₅SO₂NHR), which is a white solid. ### Final Answer: The precipitate obtained upon acidification is due to the formation of **C₆H₅SO₂NHR** (the sulfonamide). ---
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