Number of compounds which give reaction with Hinsberg's reagent is _________.

Solution

Chemistry ma’am · 3 steps

Hinsberg's reagent needs an amine nitrogen with an N–H on it, so sort each compound by its N–H first.

1. What reacts. Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl. The lone pair of an amine nitrogen attacks sulphur and an N–H hydrogen leaves with Cl⁻. So only primary and secondary amines react. A tertiary amine has no N–H to lose.
2. Not every N–H counts. In urea and dibenzamide the nitrogen does carry H. But its lone pair is spread into the C=O groups, so the nitrogen is far too weak a nucleophile to attack the reagent. The diazonium ion has no N–H and no free lone pair, so it cannot react either.

   Compounds	Type of nitrogen	Reacts?
   Aniline, propylamine, ethane-1,2-diamine	primary amine, -NH₂	yes, 3
   Piperidine, N-methylbenzylamine	secondary amine, >NH	yes, 2
   N-phenylpiperidine, trimethylamine, pyridine	tertiary nitrogen, no N–H	no
   Urea, dibenzamide	amide or imide N–H, lone pair delocalised	no
   Benzenediazonium chloride	N₂⁺ group, no N–H	no

3. So. Three primary amines and two secondary amines react. Ethane-1,2-diamine has two -NH₂ groups but is still one compound. The count is 3 + 2 = 5.

The common slip: 7. Counting urea and dibenzamide because they carry N–H gives 7. Their nitrogen lone pair is tied up with the C=O, so they are not amines for this test.

The idea to keep. Hinsberg's reagent needs a real amine N–H: primary and secondary amines react, tertiary amines do not. Amide and imide N–H do not count, because the C=O pulls the lone pair away.

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