Question
Three bottles contain ethylamine (1° amine), diethylamine (2° amine), and triethylamine (3° amine) respectively. Using the Hinsberg test with benzenesulphonyl chloride (C₆H₅SO₂Cl), how would you identify each? Write the reactions and observations.
(NCERT Class 12, Amines)
Solution — Step by Step
Benzenesulphonyl chloride (Hinsberg’s reagent) reacts differently with each class of amine. Shake the amine with C₆H₅SO₂Cl in the presence of aqueous NaOH.
The sulphonamide product has an N-H hydrogen. In NaOH, this hydrogen is acidic enough to dissolve as a sodium salt → clear solution. On adding dilute HCl, the sulphonamide precipitates out as a white solid.
The product has no N-H hydrogen → it cannot form a sodium salt → insoluble precipitate appears immediately. This precipitate does not dissolve in NaOH.
Tertiary amines have no N-H bond, so they cannot react with sulphonyl chloride. The amine remains unchanged — it separates as an oily layer or stays dissolved.
Why This Works
The test exploits the number of N-H bonds in each amine class:
- 1° amine (2 N-H bonds): Reacts, product has 1 N-H → acidic enough to dissolve in NaOH
- 2° amine (1 N-H bond): Reacts, product has 0 N-H → cannot dissolve in NaOH
- 3° amine (0 N-H bonds): Cannot react at all
The N-H in the primary sulphonamide is acidic (pKa ~10) because the sulphonyl group () is strongly electron-withdrawing. It stabilises the conjugate base by delocalising the negative charge. This acidity is what makes the primary amine product NaOH-soluble.
Alternative Method
You can also distinguish these amines using the carbylamine (isocyanide) test: only primary amines give the foul-smelling isocyanide when heated with CHCl₃ and NaOH. But the Hinsberg test is more systematic as it classifies all three classes in one go.
This is one of the most frequently asked reactions in CBSE 12th board exams — typically a 3-mark question asking you to write reactions with all three amine types. NEET also asks this as an MCQ. The key observations to remember: 1° → clear solution (precipitates with acid), 2° → immediate precipitate, 3° → no reaction.
Common Mistake
Students write that the tertiary amine “forms a precipitate” with Hinsberg’s reagent. This is wrong — tertiary amines do not react with benzenesulphonyl chloride because they lack an N-H bond for the substitution reaction. If you see any product forming with a 3° amine in the Hinsberg test, your answer is incorrect. The whole point is that 3° amines are inert to this reagent.