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 12–14).
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:
The carbylamine test is specific to which class of amines?
In the Hinsberg test, a tertiary amine:
An unknown amine reacts with benzenesulfonyl chloride. The product does not dissolve in aqueous NaOH. What is the class of the amine?
Treatment of aniline (a primary aromatic amine) with NaNO₂ and dilute HCl at 0–5 °C gives:
A secondary aliphatic amine is treated with NaNO₂ and dilute HCl. The expected product is:
Why does the sulfonamide product of a secondary amine in the Hinsberg test fail to dissolve in aqueous NaOH?
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:
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, p. 121° amine + CHCl₃ + alc. KOH → RNC (isocyanide, foul-smelling). Specific test for primary amines only.
-- NCERT, p. 14Aqueous basicity order: 2° ≈ 1° > 3° > NH3. Aromatic amines much weaker due to resonance delocalisation of lone pair.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKb | -log Kb | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student treats aniline as stronger base than methylamine. Aniline weaker due to lone-pair delocalization into ring.
Basicity ordering across aromatic and aliphatic amines.
Aliphatic amines (CH3NH2) > NH3 > aromatic amines (C6H5NH2). Aniline lone pair delocalised into ring → less available for protonation.
Category: Organic Reaction Conditions
Quaternary ammonium hydroxide (Hofmann) elimination favours LESS substituted alkene (anti-Saytzeff). Halide elimination (E2) follows Saytzeff (more substituted).
E2 elimination question with quaternary ammonium hydroxide vs alkyl halide.
Hofmann: bulky leaving group + base → least substituted alkene (Hofmann product). Saytzeff: alkyl halide + base → most substituted alkene (Zaitsev product).
Root cause: concept gap
Gas phase: 3° > 2° > 1° > NH₃. Aqueous: 2° > 1° > 3° > NH₃ (steric + solvation effects).
Root cause: concept gap
Aniline's lone pair delocalised into ring → less basic. Order: aliphatic (CH3NH2) > NH3 > aromatic (PhNH2).
Root cause: concept gap
Sandmeyer: ArN₂⁺ + CuCl/CuBr/CuCN → ArCl/ArBr/ArCN. Gattermann: Cu powder + HX. Hydrolysis: H₂O → ArOH.
Root cause: concept gap
Hofmann (quaternary R₄N⁺ + OH⁻): least substituted alkene. Saytzeff (alkyl halide + base): most substituted.
11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
The highest number of helium atoms is in
The correct order of energies of molecular orbitals of N molecule, is 2
Which of the following reactions will NOT give primary amine as the product?
Which of the following sequence of reactions is suitable to synthesize chlorobenzene?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
uses gas phase order in aqueous
Doesn't account for solvation reversal
confuses sandmeyer gattermann
Mixes up Cu reagents
Test yourself on this topic with real past-paper questions:
Practice this topic →Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.