Q.Differentiate between the following pairs of compounds by writing chemical equations of one chemical test:
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The Hinsberg Test: Distinguishing Amines by Reactivity
Imagine you have three types of amines — primary (RNH2), secondary (R2NH), and tertiary (R3N) — and you need to tell them apart. They all smell fishy, they all are basic, but they differ in one crucial way: how many hydrogens are attached to the nitrogen.
A primary amine has two hydrogens on the nitrogen. A secondary amine has one. A tertiary amine has none. The Hinsberg reagent, benzenesulphonyl chloride (C6H5SO2Cl), exploits exactly this difference.
The reagent is a sulfonyl chloride — think of it as a very reactive molecule that loves to swap its chlorine atom for a bond to nitrogen, but only if that nitrogen still has a hydrogen to give up during the reaction.
What Actually Happens
When you shake the amine with benzenesulphonyl chloride in the presence of aqueous alkali (usually KOH or NaOH), three different things happen:
Primary amine (RNH2)
The nitrogen attacks the sulfur, kicking out chlorine. The product is a sulfonamide that still has one hydrogen on the nitrogen — C6H5SO2NHR. That hydrogen is acidic enough to be pulled off by the strong base, forming a soluble salt: C6H5SO2NR−K+. So the primary amine dissolves in the alkaline solution.
Secondary amine (R2NH)
Same attack, but now the product is C6H5SO2NR2 — a sulfonamide with no hydrogen on the nitrogen. It cannot form a salt with base. It simply precipitates out as a solid.
Tertiary amine (R3N)
No hydrogen on the nitrogen at all. The reaction cannot proceed to form a sulfonamide. The tertiary amine does not react — it either remains as an oily layer or, if it is a small molecule, may dissolve in the acid work-up later.
A common mistake: students think tertiary amines do nothing at all. They do not form a sulfonamide, but they are still basic. In the alkaline medium they remain as free amines, often floating as an immiscible oil.
The Practical Workflow
Here is how you actually perform the test in the lab:
- Take a small sample of the amine mixture in a test tube.
- Add benzenesulphonyl chloride and a few mL of aqueous KOH.
- Shake well. If the mixture warms up, cool it under the tap.
- Observe:
| Observation | What it means |
|---|---|
| Clear solution forms | Primary amine (soluble salt) |
| Solid precipitate forms | Secondary amine (insoluble sulfonamide) |
| Oily layer remains, no precipitate | Tertiary amine (no reaction) |
To confirm the primary amine, you can acidify the clear solution with dilute HCl. The sulfonamide salt breaks down, and the free sulfonamide (C6H5SO2NHR) precipitates out — a white solid.
If you have a mixture, you can separate them: filter the secondary amine precipitate, acidify the filtrate to get the primary amine, and the tertiary amine stays as the oil layer that never reacted.
Why This Works — The Key Insight …
- Methylamine (1°) vs dimethylamine (2°): carbylamine test — only the 1° amine gives a foul isocyanide: CH3NH2+CHCl3+3KOH→CH3NC+3KCl+3H2O; dimethylamine gives no reaction.
- Ethylamine vs aniline: bromine water — aniline gives a white ppt of 2,4,6-tribromoaniline, ethylamine does not.
- Aniline vs benzylamine: diazotisation/dye test — aniline (aromatic 1°) gives a stable diazonium salt at 0–5 °C that couples to an azo dye; benzylamine (aliphatic 1°) evolves N2.
- Aniline vs N-methylaniline (2°): carbylamine test positive only for aniline. …
Use a chemical test that responds to only one member of each pair: carbylamine (1° vs others), bromine water/diazotisation (aromatic vs aliphatic), and nitrous acid (2° vs 3°).
- Methylamine (1°) vs Dimethylamine (2°) — carbylamine test. Only primary amines give the foul-smelling isocyanide: CH3NH2+CHCl3+3KOHΔCH3NC (offensive smell)+3KCl+3H2O Dimethylamine (2°) → no carbylamine.
- Ethylamine vs Aniline — bromine water. The aromatic ring of aniline is highly activated, so it gives an instant white precipitate; ethylamine has no ring: C6H5NH2+3Br2⟶white ppt2,4,6-tribromoaniline+3HBr Ethylamine → no reaction with bromine water.
- Aniline vs Benzylamine — diazotisation / azo-dye test. With NaNO2/HCl at 0–5 °C, the aromatic amine aniline forms a stable benzenediazonium salt that couples with 2-naphthol to a bright orange azo dye: C6H5NH2NaNO2/HCl, 273−278 KC6H5N2+Cl−2-naphtholorange azo dye Benzylamine (C6H5CH2NH2, aliphatic 1°) instead gives an unstable diazonium that decomposes with effervescence of N2: C6H5CH2NH2+HNO2→C6H5CH2OH+N2↑+H2O.
- Aniline vs N-methylaniline — carbylamine test. Aniline is a primary amine; N-methylaniline (C6H5NHCH3) is secondary. Only aniline gives the isocyanide: C6H5NH2+CHCl3+3KOHΔC6H5NC (foul smell)+3KCl+3H2O …
Showing the 12 most recent of 15 on this concept.
- CBSE 2026Set DZ1 markMCQQ.Which test is used to differentiate between primary, secondary and tertiary amines?(a) Brady test(b) Tollen's test(c) Hinsberg test(d) Lucas test
›Reveal solutionSolution
The Hinsberg test (with benzenesulphonyl chloride, C6H5SO2Cl) tells 1°, 2° and 3° amines apart, so option (c) is correct.
In the Hinsberg test the amine is treated with benzenesulphonyl chloride (C6H5SO2Cl, Hinsberg's reagent) in the presence of KOH:
- Primary amine → gives an N-substituted sulphonamide that has an acidic N–H; it dissolves in KOH (soluble product).
- Secondary amine → gives an N,N-disubstituted sulphonamide with no N–H; it is insoluble in KOH. …
- CBSE 2026Set ANNUAL1 markMCQQ.Identify the Hinsberg reagent:(a) benzene ring - CONH2 (benzamide)(b) benzene ring - O - COCl (phenyl chloroformate)(c) benzene ring - SO2Cl (benzenesulfonyl chloride)(d) CH3 - benzene ring - COCl (4-methylbenzoyl chloride)
›Reveal solutionSolution
Hinsberg's reagent is benzenesulfonyl chloride; its differing reaction behaviour with 1 degree, 2 degree, and 3 degree amines is the basis of the classic Hinsberg test.
Benzenesulfonyl chloride, C6H5-SO2Cl, reacts differently depending on the class of amine:
- Primary amines form an N-substituted sulfonamide that has an acidic N-H, which dissolves in excess alkali (KOH) - giving a clear solution.
- Secondary amines form an N,N-disubstituted sulfonamide with no acidic N-H, which is insoluble in alkali - it stays as a precipitate. …
- CBSE 2026Set ANNUAL1 markMCQQ.Hinsberg reagent is:(a) C6H5SO2Cl(b) C6H5N2Cl(c) C6H5NO(d) C6H5SO3H
›Reveal solutionSolution
Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl, used to distinguish 1°, 2°, and 3° amines.
Hinsberg's reagent is benzenesulphonyl chloride (C₆H₅SO₂Cl). In the Hinsberg test, it reacts with a primary amine to give an N-alkylbenzenesulphonamide that is soluble in alkali (the remaining N–H is acidic due to the electron-withdrawing SO₂ group); with a secondary amine it gives a sulphonamide that is insoluble in alkali (no acidic N–H le …
- CBSE 2025Set D1 markMCQQ.Which of the following is Hinsberg reagent?(a) Benzene sulphonyl chloride(b) Benzene sulphonic acid(c) Ethyl oxalate(d) Acetyl chloride
›Reveal solutionSolution
Hinsberg's reagent = benzenesulphonyl chloride, C6H5SO2Cl.
Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl). It is used to distinguish the three classes of amines:
- Primary amines form a sulphonamide that is soluble in alkali (has an acidic N-H). …
- CBSE 2025Set A1 markQ.Write True or False: Hinsberg's reagent reacts with primary and secondary amines to form sulphonamides.
›Reveal solutionSolution
Hinsberg's test uses benzenesulphonyl chloride to distinguish primary, secondary, and tertiary amines by whether/how a sulphonamide forms.
Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl). When treated with:
- A primary amine: it forms an N-alkylbenzenesulphonamide (C6H5SO2NHR), which has an acidic N–H (due to the electron-withdrawing SO2 group) and so dissolves in alkali (KOH) to give a clear solution.
- A secondary amine: it forms an N,N-dialkylbenzenesulphonamide (C6H5SO2NR2), which has no acidic hydrogen and stays insoluble in alkali. …
- CBSE 2025Set ANNUAL1 markMCQQ.Hinsberg's reagent is -(a) C6H5SO2Cl(b) C6H5SO2Cl2(c) C6H5SOCl2(d) C6H5SO3Cl
›Reveal solutionSolution
Hinsberg's reagent is benzenesulfonyl chloride, C6H5SO2Cl.
In Hinsberg's test, an amine is treated with benzenesulfonyl chloride (Hinsberg's reagent) in the presence of aqueous KOH:
- A primary amine forms an N-alkylbenzenesulfonamide with an acidic N-H, which dissolves in excess KOH (forms a soluble salt).
- A secondary amine forms an N,N-dialkylbenzenesulfonamide with no acidic N-H, which is insoluble in KOH. …
- CBSE 2024Set A11 markMCQQ.The reagents used to separate the mixture of methylamine and dimethylamine are :(a) CHCl3 and HCl(b) C6H5SO2Cl and KOH(c) C6H5SO2Cl and HCl(d) CHCl3 and KOH
›Reveal solutionSolution
Hinsberg's reagent (C6H5SO2Cl) with KOH separates a 1° from a 2° amine, so the answer is (b).
Methylamine is a primary amine and dimethylamine is a secondary amine. Hinsberg's reagent (benzenesulphonyl chloride, C6H5SO2Cl) distinguishes them:
-
With methylamine (1°):
C6H5SO2Cl+H2NCH3→C6H5SO2NHCH3+HCl
The product still has an N–H whose proton is acidic (activated by the −SO2− group), so it dissolves in KOH to give a soluble salt.
-
With dimethylamine (2°):
C6H5SO2Cl+HN(CH3)2→C6H5SO2N(CH3)2+HCl …
-
- CBSE 2024Set B1 markQ.Match the pairs correctly. Column A item: "Hinsberg's reagent". Column B options:(a) C6H5SO2Cl(b) Keratin(c) C6H5NH2(d) Rickets(e) C6H5N2Cl(f) Cobalt. Which option from Column B matches "Hinsberg's reagent"?
›Reveal solutionSolution
Hinsberg's reagent is benzenesulphonyl chloride (C6H5SO2Cl), used in Hinsberg's test to distinguish primary, secondary and tertiary amines.
In the Hinsberg test, an amine is treated with benzenesulphonyl chloride (C6H5SO2Cl) in the presence of KOH:
- 1 degree amines give a sulphonamide that dissolves in excess KOH (the N-H is acidic enough to be deprotonated). …
- CBSE 2024Set ANNUAL1 markMCQQ.Hinsberg's reagent which is used to test amines is(a) Benzene sulphonamide(b) Benzene diazonium chloride(c) Benzene sulphonyl chloride(d) Acetanilide
›Reveal solutionSolution
Hinsberg's reagent is benzenesulphonyl chloride; it reacts differently with 1-degree, 2-degree, and 3-degree amines, which is the basis of the Hinsberg test used to distinguish them.
C6H5SO2Cl (benzenesulphonyl chloride) reacts with:
- Primary amines -> forms an N-alkyl benzenesulphonamide that has an acidic N-H (soluble in alkali, since the sulphonyl group makes the remaining N-H acidic).
- Secondary amines -> forms an N,N-dialkyl benzenesulphonamide with no N-H left (insoluble in alkali). …
- CBSE 2024Set ANNUAL1 markQ.Name a test to distinguish between aniline and N-methylaniline.
›Reveal solutionSolution
The carbylamine (isocyanide) test is specific to primary amines, so it cleanly distinguishes aniline (primary) from N-methylaniline (secondary).
When a primary amine is warmed with chloroform and alcoholic potassium hydroxide, it is converted into an extremely foul-smelling alkyl/aryl isocyanide (carbylamine):
C6H5NH2+CHCl3+3KOHΔC6H5NC(phenyl isocyanide, offensive odour)+3KCl+3H2O
This reaction requires an −NH2 group with two replaceable hydrogens on nitrogen to form the isocyanide's N≡C linkage — a mechanistic requirement that only primary amines satisfy.
- Aniline (C6H5NH2) is a primary amine ⇒ gives a positive carbylamine test (pungent, unpleasant smell of phenyl isocyanide). …
- CBSE 2024Set ANNUAL1 markMCQQ.The primary, secondary and tertiary amines can be distinguished by -(a) Fehling's test(b) Tollen's test(c) Libermann test(d) Hinsberg's test
›Reveal solutionSolution
Hinsberg's reagent (benzenesulphonyl chloride) reacts differently with 1°, 2°, and 3° amines, giving products with distinct solubility behaviour — this difference is the basis of the test.
Hinsberg's test: the amine is treated with benzenesulphonyl chloride (C6H5SO2Cl) in the presence of KOH:
- 1° amine: RNH2+C6H5SO2Cl→C6H5SO2NHR (N-alkylbenzenesulphonamide). This product still has an acidic N–H (due to the electron-withdrawing SO2 group), so it dissolves in excess KOH to give a soluble salt; on acidification, the sulphonamide is regenerated as a precipitate.
- 2° amine: R2NH+C6H5SO2Cl→C6H5SO2NR2. No N–H remains, so this product is insoluble in KOH and separates out directly as an oily/solid precipitate. …
- CBSE 2023Set BZ1 markMCQQ.Hinsberg reagent is:(a) Ethyl oxalate(b) Trimethyl amine(c) Benzene sulphonyl chloride(d) Benzyl chloride
›Reveal solutionSolution
Hinsberg's reagent is benzenesulphonyl chloride, C6H5SO2Cl, used to distinguish primary, secondary and tertiary amines.
…
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