Hinsberg Test: How to Distinguish Primary, Secondary and Tertiary Amines
Chemistry · Amines · NEET
The Hinsberg test uses benzenesulphonyl chloride (C6H5SO2Cl, called Hinsberg's reagent) to tell apart the three types of amines. A primary amine gives a solid that dissolves in alkali (NaOH), a secondary amine gives a solid that does NOT dissolve in alkali, and a tertiary amine does not react at all. Memory hook: "1 = soluble, 2 = not soluble, 3 = no reaction."
Count the N-H bonds: primary (2 N-H) gives an alkali-soluble solid, secondary (1 N-H) gives an alkali-insoluble solid, and tertiary (0 N-H) does not react.
Your doubts, answered
What exactly is Hinsberg's reagent?
Hinsberg's reagent is benzenesulphonyl chloride, C6H5SO2Cl. It is a benzene ring joined to an -SO2Cl group. The whole test is based on how this reagent reacts with the N-H bonds of an amine. Tip: p-toluenesulphonyl chloride can also be used for the same test.
Why does the primary amine product dissolve in alkali (NaOH)?
A primary amine (R-NH2) has two N-H bonds. One reacts with the reagent to form C6H5SO2-NH-R (an N-substituted sulphonamide). The remaining N-H is strongly acidic because the -SO2- group pulls electron density away from nitrogen. So this N-H loses its proton to NaOH and forms a soluble salt. That is why the solid dissolves in alkali.
Why is the secondary amine product insoluble in alkali?
A secondary amine (R2NH) has only one N-H. That single N-H reacts with the reagent, giving C6H5SO2-NR2 (an N,N-disubstituted sulphonamide). Now there is NO N-H left on nitrogen. With no acidic hydrogen, it cannot form a salt with NaOH, so the solid stays insoluble in alkali.
Why does a tertiary amine give no reaction?
A tertiary amine (R3N) has no N-H bond at all. The Hinsberg test works by replacing an N-H, so with no N-H there is nothing to react. The tertiary amine simply stays unreacted (it may just dissolve or float, no new solid forms). This is how you spot a 3° amine.
How do I remember the three results for NEET?
Count the N-H bonds. Primary has 2 N-H, so after one reacts there is still one acidic N-H left, product dissolves in alkali. Secondary has 1 N-H, after it reacts none is left, product does not dissolve. Tertiary has 0 N-H, so no reaction. NEET loves the words 'alkali-soluble solid' (=1°) and 'alkali-insoluble solid' (=2°).
⚠️ The NEET trap ✗ Thinking the secondary amine product dissolves in alkali because a sulphonamide 'should be acidic'. ✓ Only the PRIMARY amine product dissolves in alkali. The secondary amine product (C6H5SO2NR2) has NO N-H left, so it has no acidic hydrogen and stays insoluble. 🧠 Alkali-SOLUBLE solid = primary. Alkali-INSOLUBLE solid = secondary. No reaction = tertiary.
Real NEET questions
NEET 2021
Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.
A · (CH3)2CH-NH2 (isopropylamine, a primary amine) ✓
B · (CH3)2N-CH3 (a tertiary amine)
C · (CH3)2CH-NO2 (2-nitropropane)
D · (CH3)3N (trimethylamine, a tertiary amine)
Solution: A solid that dissolves in alkali is the signal for a PRIMARY amine. Isopropylamine (A) reacts with benzenesulphonyl chloride to form C6H5SO2-NH-CH(CH3)2. The leftover N-H is acidic (the -SO2- group pulls electrons away), so it dissolves in NaOH/KOH as a salt. Options B and D are tertiary amines with no N-H, so no reaction. Option C is a nitro compound, not an amine, so it does not respond. Answer: A.
NEET 2019 (Odisha)
The amine that reacts with Hinsberg's reagent to give an alkali-insoluble product is:
A · (CH3)2CH-NH-CH(CH3)2 (diisopropylamine, a secondary amine) ✓
B · (C2H5)3N (triethylamine, a tertiary amine)
C · a primary amine
D · a primary amine
Solution: An alkali-INSOLUBLE product is the signal for a SECONDARY amine. Diisopropylamine (A) has one N-H; it reacts to form C6H5SO2-NR2, which has no N-H left. With no acidic hydrogen it cannot form a salt with alkali, so it stays insoluble. Triethylamine (B) is tertiary (no N-H, no reaction), and primary amines (C, D) give alkali-soluble products. Answer: A.
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Can the Hinsberg test separate a mixture of amines?
Yes. Because the three products behave differently (primary product dissolves in alkali, secondary product stays as an insoluble solid, tertiary stays unreacted), you can physically separate a mixture of 1°, 2° and 3° amines. NCERT mentions this separation use.
Is Hinsberg's reagent the same as benzenesulphonyl chloride?
Yes, exactly. Hinsberg's reagent = benzenesulphonyl chloride, C6H5SO2Cl. p-Toluenesulphonyl chloride works too, so do not get confused if NEET uses that name.
Does the primary amine product form a precipitate first, then dissolve?
Yes. First a solid sulphonamide forms. When you add NaOH, that solid dissolves because its acidic N-H makes a soluble salt. So the sequence is: solid appears, then it dissolves in alkali. That two-step behaviour confirms a primary amine.
Why is the N-H in the sulphonamide acidic but the N-H in the amine is not?
The -SO2- group is a strong electron-withdrawing group. In the sulphonamide it pulls electron density off the nitrogen, weakening the N-H bond and making that hydrogen easy to remove as H+. A plain amine has no such group, so its N-H is not acidic.
