How do you tell whether an amine is primary, secondary, or tertiary? NEET occasionally asks you to identify amine class from a chemical test or to predict the outcome of a distinguishing reaction. The classic tool is the Hinsberg test, and two supporting tests — the carbylamine (isocyanide) test and the nitrous acid test — complete the identification toolkit.
Hinsberg test. Benzenesulfonyl chloride (C₆H₅SO₂Cl, Hinsberg reagent) reacts differently with each class:
The critical reasoning: primary sulfonamide retains one N–H bond (acidic, pKₐ ~ 10), making it base-soluble. Secondary sulfonamide has no N–H, so it precipitates and stays insoluble. This NaOH-solubility distinction is the decisive test.
Carbylamine test. Only 1° amines respond. Heating with CHCl₃ and alcoholic KOH produces an isocyanide (R–NC) with a characteristic foul smell. 2° and 3° amines give no isocyanide. This test is specific but only confirms primary class — it cannot distinguish 2° from 3°.
Nitrous acid test (NaNO₂ + dil. HCl). Behaviour depends on class: 1° aliphatic amines yield unstable diazonium salts that immediately decompose to alcohols with N₂ gas evolution. 1° aromatic amines form stable diazonium salts (below 5 °C). 2° amines (both aliphatic and aromatic) form yellow oily N-nitrosamines. 3° aliphatic amines form nitrite salts; 3° aromatic amines undergo ring nitrosation (C-nitrosation at para position).
These three tests, applied together, allow unambiguous classification of an unknown amine as 1°, 2°, or 3° — a fact NCERT states explicitly (NCERT Class 12 Chemistry Chapter 9, Part 2, pages 270–272).
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In the Hinsberg test, benzenesulfonyl chloride reacts with a primary amine to form a product that is:
Answer: C. C is correct. The N-alkylbenzenesulfonamide formed from a 1° amine retains one N–H bond, making it acidic enough to dissolve in aqueous NaOH (NCERT Class 12 Chemistry Chapter 9, Part 2, page 270).
Why A is wrong: A is wrong because insolubility in NaOH is the characteristic of the secondary amine sulfonamide, which lacks an acidic N–H proton.
Why B is wrong: B is wrong because the sulfonamide from a primary amine is not an amine salt — it dissolves in NaOH (base), not dilute HCl.
Why D is wrong: D is wrong because the product does dissolve in NaOH due to the acidic N–H proton on the sulfonamide nitrogen.
The carbylamine test is specific to which class of amines?
Answer: D. D is correct. Only primary amines react with CHCl₃ and alcoholic KOH to form isocyanides (carbylamines) with a characteristic foul odour (NCERT Class 12 Chemistry Chapter 9, Part 2, page 270).
Why A is wrong: A is wrong because secondary amines do not have the required N–H₂ group to form the isocyanide intermediate.
Why B is wrong: B is wrong because tertiary amines have no N–H bonds at all and cannot undergo this reaction.
Why C is wrong: C is wrong because only primary amines respond; secondary amines lack the two N–H bonds needed for the reaction mechanism.
In the Hinsberg test, a tertiary amine:
Answer: B. B is correct. Tertiary amines have no N–H bond to react with benzenesulfonyl chloride, so they remain unreacted under Hinsberg test conditions (NCERT Class 12 Chemistry Chapter 9, Part 2, page 270).
Why A is wrong: A is wrong because forming a NaOH-soluble sulfonamide is the behaviour of primary amines in the Hinsberg test.
Why C is wrong: C is wrong because forming an NaOH-insoluble sulfonamide is the behaviour of secondary amines in the Hinsberg test.
Why D is wrong: D is wrong because yellow oily products (N-nitrosamines) form when secondary amines react with nitrous acid, not in the Hinsberg test.
An unknown amine reacts with benzenesulfonyl chloride. The product does not dissolve in aqueous NaOH. What is the class of the amine?
Answer: D. D is correct. In the Hinsberg test, a secondary amine forms N,N-dialkylbenzenesulfonamide that has no acidic N–H, so it is insoluble in NaOH. A primary amine's product would dissolve in NaOH, and a tertiary amine would not react at all (NCERT Class 12 Chemistry Chapter 9, Part 2, page 270).
Why A is wrong: A is wrong because a primary amine's sulfonamide retains an N–H bond and dissolves in NaOH — the opposite of what is observed here.
Why B is wrong: B is wrong because a tertiary amine does not react with benzenesulfonyl chloride at all, so no sulfonamide product would form.
Why C is wrong: C is wrong because the Hinsberg test is designed to distinguish all three classes: NaOH-soluble product = 1°, NaOH-insoluble product = 2°, no reaction = 3°.
Treatment of aniline (a primary aromatic amine) with NaNO₂ and dilute HCl at 0–5 °C gives:
Answer: A. A is correct. Primary aromatic amines form stable diazonium salts (ArN₂⁺Cl⁻) at 0–5 °C because the aromatic ring stabilises the diazonium ion through resonance. Primary aliphatic amines, by contrast, would decompose immediately to give an alcohol and N₂ gas (NCERT Class 12 Chemistry Chapter 9, Part 2, page 272).
Why B is wrong: B is wrong because yellow oily N-nitrosamines form from secondary amines reacting with nitrous acid, not from primary amines.
Why C is wrong: C is wrong because immediate decomposition to an alcohol with N₂ evolution is the behaviour of primary aliphatic amines, not aromatic amines. Aniline's diazonium salt is stabilised by the aromatic ring.
Why D is wrong: D is wrong because aniline does react with nitrous acid — it forms a diazonium salt that is stable at low temperature.
A secondary aliphatic amine is treated with NaNO₂ and dilute HCl. The expected product is:
Answer: C. C is correct. Secondary amines (both aliphatic and aromatic) react with nitrous acid to form N-nitrosamines, which are characteristically yellow and oily. This distinguishes them from primary amines, which form diazonium salts (NCERT Class 12 Chemistry Chapter 9, Part 2, page 272).
Why A is wrong: A is wrong because unstable diazonium salts that decompose to alcohols are formed by primary aliphatic amines, not secondary amines.
Why B is wrong: B is wrong because isocyanide formation is the carbylamine test (CHCl₃ + alc. KOH), specific to primary amines — it is a different reaction entirely.
Why D is wrong: D is wrong because stable diazonium salts are formed by primary aromatic amines at low temperature, not by secondary amines.
Why does the sulfonamide product of a secondary amine in the Hinsberg test fail to dissolve in aqueous NaOH?
Answer: B. B is correct. In the secondary amine's sulfonamide, both hydrogens on nitrogen have been replaced — one by the sulfonyl group and one by the second alkyl group. With no N–H bond, there is no acidic proton to be removed by NaOH, so the product remains insoluble (NCERT Class 12 Chemistry Chapter 9, Part 2, page 270).
Why A is wrong: A is wrong because molecular weight is not the decisive factor. The primary amine's sulfonamide has a comparable molecular weight yet dissolves in NaOH because of its acidic N–H.
Why C is wrong: C is wrong because the sulfonamide does not undergo significant hydrolysis under the mild NaOH conditions of the Hinsberg test.
Why D is wrong: D is wrong because both primary and secondary sulfonamides contain the same aromatic sulfonyl group — the difference lies in the N–H bond, not the ring.
An unknown compound X gives a positive carbylamine test and also forms a product soluble in NaOH when treated with benzenesulfonyl chloride. When X is treated with NaNO₂/dil. HCl, nitrogen gas evolves immediately. X is most likely:
Answer: A. A is correct. Positive carbylamine test → primary amine. NaOH-soluble sulfonamide (Hinsberg test) → confirms primary. Immediate N₂ evolution with NaNO₂/HCl → aliphatic primary amine (its diazonium salt is unstable and decomposes at once). A primary aromatic amine would instead give a stable diazonium salt at 0–5 °C without immediate gas evolution (NCERT Class 12 Chemistry Chapter 9, Part 2, pages 270–272).
Why B is wrong: B is wrong because secondary amines do not give a positive carbylamine test (carbylamine is specific to primary amines) and their Hinsberg product is insoluble in NaOH.
Why C is wrong: C is wrong because a primary aromatic amine would form a stable diazonium salt (especially below 5 °C) rather than immediately releasing N₂ gas. The immediate gas evolution indicates an aliphatic primary amine.
Why D is wrong: D is wrong because tertiary amines give neither a positive carbylamine test nor a Hinsberg reaction product — they fail both tests.
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Given
An organic compound Y is known to be an amine. The following observations are recorded:• Hinsberg test: Y reacts with C₆H₅SO₂Cl; the product is insoluble in aqueous NaOH.• Nitrous acid test (NaNO₂ + dil. HCl): Y gives a yellow oily product.• Carbylamine test: Negative (no foul-smelling isocyanide produced).
Required
Classify Y as primary, secondary, or tertiary amine.
Concept
The Hinsberg test, nitrous acid test, and carbylamine test together provide an unambiguous classification of amines. Each class gives a unique combination of outcomes across these three tests.
Identification logic (in place of formula)
| Test | 1° amine | 2° amine | 3° amine |
|------|----------|----------|----------|
| Hinsberg | Reacts; product soluble in NaOH | Reacts; product insoluble in NaOH | No reaction |
| Nitrous acid | Diazonium salt (stable if aromatic; N₂ if aliphatic) | Yellow oily N-nitrosamine | Nitrite salt (aliphatic) or C-nitrosation (aromatic) |
| Carbylamine | Positive (foul smell) | Negative | Negative |
Matching observations to table
• Hinsberg: product formed but insoluble in NaOH → matches 2° amine column.• Nitrous acid: yellow oily product → matches 2° amine (N-nitrosamine).• Carbylamine: negative → consistent with 2° amine (also consistent with 3°, but Hinsberg already rules out 3°).
Reasoning
All three tests converge on the secondary amine classification. The Hinsberg result alone distinguishes 2° from 1° (whose product would dissolve in NaOH) and from 3° (which would not react at all). The nitrous acid and carbylamine results provide independent confirmation.
Final answer
Y is a secondary amine.
Common trap
A frequent error is confusing "no reaction in Hinsberg test" (3° amine) with "product insoluble in NaOH" (2° amine). Both might superficially seem like "negative results," but they are fundamentally different: 2° amines DO react — they form a precipitate that simply won't dissolve in base.
Similar NEET-style question
Compound Z gives a positive carbylamine test. Its Hinsberg product dissolves in NaOH. Treatment with NaNO₂/HCl at 0–5 °C gives a clear solution that couples with alkaline β-naphthol to give an orange dye. Identify the class and type (aliphatic/aromatic) of Z. *(Answer: primary aromatic amine — carbylamine confirms 1°, Hinsberg confirms 1°, stable diazonium + azo coupling confirms aromatic.)*
---
Amine + benzenesulphonyl chloride (C₆H₅SO₂Cl, Hinsberg's reagent). 1°: forms sulphonamide soluble in NaOH. 2°: forms sulphonamide insoluble in NaOH. 3°: no reaction. Distinguishes 1°/2°/3° amines.
-- NCERT Class 12 Chemistry, Ch. 9, p. 2711° amine + CHCl₃ + alc. KOH → RNC (isocyanide, foul-smelling). Specific test for primary amines only.
-- NCERT Class 12 Chemistry, Ch. 9, p. 271More in Amines: 10 exam traps and mistakes · 1 formula · 3 question patterns from its other lessons.
2 questions from NEET 2020, 2021. Answers verified against NTA official keys.
Which of the following amine will give the carbylamine test ?
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →