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Question: $\text{Ph-NH-SO}_3\text{Ph} \xrightarrow[]{\text{OH}^{\ominus}\text{/H}_2\text{O}}$ product of given...

Ph-NH-SO3Ph→OH⊖/H2O\text{Ph-NH-SO}_3\text{Ph} \xrightarrow[]{\text{OH}^{\ominus}\text{/H}_2\text{O}} product of given reaction is

A

Ph-NH-Ph

B

Ph-N¨\ddot{\text{N}}:

C

None of these

Answer

Ph–NH–Ph

Explanation

Solution

The sulfonamide (Ph–NH–SO₃Ph) undergoes base‐mediated hydrolysis. In the presence of OH⁻ (in H₂O) the S–N bond is cleaved. As a result, the sulfonyl group is removed (as a sulfonate ion) and the free amine (diphenylamine, Ph–NH–Ph) is formed.

Minimal Explanation:

  1. OH⁻ attacks the sulfonamide and cleaves the S–N bond.

  2. The sulfonyl group leaves as a sulfonate ion.

  3. The resulting product is diphenylamine (Ph–NH–Ph).