An organic compound "A" on treatment with benzene sulphonyl chloride gives compound B. B is soluble in dil. NaOH solution. Compound A is :

Solution

Chemistry ma’am · 3 steps

Soluble in alkali means the sulphonamide still has an N–H, so A must be a primary amine.

1. Hinsberg reagent. Benzene sulphonyl chloride, C₆H₅SO₂Cl, reacts with an N–H of the amine and forms a sulphonamide. Each N–H that reacts loses its hydrogen to the sulphonyl group.
2. Why NaOH matters. A primary amine gives C₆H₅SO₂NHR. Its one remaining N–H is acidic because the SO₂ group pulls electrons strongly, so NaOH removes that H and forms a soluble sodium salt.

   Compound	Type of amine	Product with C₆H₅SO₂Cl	In dil. NaOH
   (a) C₆H₅CH(CH₃)NH₂	primary	C₆H₅SO₂NH-CH(CH₃)C₆H₅, one N–H left	soluble
   (b) C₆H₅NHCH₂CH₃	secondary	C₆H₅SO₂N(C₂H₅)C₆H₅, no N–H	insoluble
   (c) C₆H₅N(CH₃)₂	tertiary	no reaction, no N–H	insoluble
   (d) C₆H₅CH₂NHCH₃	secondary	C₆H₅SO₂N(CH₃)CH₂C₆H₅, no N–H	insoluble

3. So. Only the primary amine (a), 1-phenylethanamine, gives a sulphonamide B that dissolves in dil. NaOH. Compound A is (a).

The idea to keep. In the Hinsberg test, count the N–H bonds left in the product. One left: dissolves in alkali (primary). None left: insoluble (secondary). None to begin with: no reaction (tertiary).

Answer from the NTA final answer key. Solution by PrepWiser.