Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is _________.

Solution

Chemistry ma’am · 4 steps

Only a free -NH₂ leaves an acidic N–H after sulfonylation, so count the primary amines.

1. What dissolves. Hinsberg's reagent, C₆H₅SO₂Cl, turns a primary amine into C₆H₅SO₂NHR. The SO₂ group makes the remaining N–H acidic, so this solid dissolves in NaOH as its sodium salt.
2. What does not. A secondary amine gives C₆H₅SO₂NR₂, a solid with no N–H, so it stays insoluble in NaOH. A tertiary amine has no N–H at all and does not form a sulfonamide.

   Compounds	Type of nitrogen	Solid soluble in NaOH?
   Aniline, 2-methoxyaniline, 4-aminodiphenylamine	primary -NH₂ on a benzene ring	yes, 3
   Propan-1-amine, cyclohexylamine	primary -NH₂ on an alkyl carbon	yes, 2
   N-methylaniline	secondary amine	no, solid is insoluble
   Triethylamine, the N-ethyl-N-methyl cyclohexylmethyl amine	tertiary amine	no reaction
   Benzamide, urea, semicarbazide	N attached to C=O, not amines	not counted

3. Mixed amine. 4-Aminodiphenylamine has a secondary N–H and a free -NH₂. The -NH₂ end still gives a sulfonamide with an acidic N–H, so its product dissolves in NaOH.
4. So. Five amines give NaOH-soluble solids, so the answer is 5.

The common slip: 4. Treating 4-aminodiphenylamine as a secondary amine because of its ring N–H gives 4. Its free -NH₂ still forms an alkali-soluble sulfonamide.

The idea to keep. In the Hinsberg test, a solid that dissolves in alkali means a primary amine: the sulfonamide still has one N–H, made acidic by the SO₂ group. Any free -NH₂ in a molecule is enough.

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