ChemistryISC Class 12 · Organic Compounds Containing Nitrogen

Chemical Reactions and Tests of Amines (ISC Class 12)

Updated 15 October 2026 · Fact-checked

Amines react through the lone pair on nitrogen: they are bases, and they are nucleophiles that undergo alkylation and acylation. Nitrous acid, the carbylamine reaction and Hinsberg's reagent tell primary, secondary and tertiary amines apart. In aniline the –NH₂ group strongly activates the ring, so bromine water gives 2,4,6-tribromoaniline at once.

Notice that every reaction starts from the same lone pair on nitrogen, which either accepts H⁺ or attacks an electron-poor carbon.

Illustration drawn for QuestionBuddy.

Understand Chemical Reactions and Tests of Amines

The lone pair on nitrogen explains almost every reaction of amines: it accepts a proton (basicity) or attacks an electron-poor carbon (nucleophilicity).

Basic character. Amines form salts with acids, $C_{3} H_{7} N H_{2} + H C l \to C_{3} H_{7} N H_{3}^{+} C l^{-}$ , and are freed again by NaOH. Aliphatic amines are stronger bases than ammonia (alkyl groups push electron density onto N). In water the order is decided by +I effect, solvation of the cation and steric hindrance together: for methyl groups, (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃; for ethyl groups, (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃. Aniline is a much weaker base than ammonia because its lone pair is delocalised into the ring. Electron-releasing groups on the ring (–CH₃, –OCH₃) increase the basicity of aniline; electron-withdrawing groups (–NO₂) decrease it. Aniline forms salts such as anilinium chloride and anilinium hydrogensulphate.

Alkylation. Amines react with alkyl halides by SN2 (the amine is the nucleophile), giving higher amines and finally quaternary ammonium salts: $C_{6} H_{5} N H_{2} + C H_{3} I \to C_{6} H_{5} N H C H_{3} + H I$ ; $C_{6} H_{5} N H C H_{3} + C H_{3} I \to C_{6} H_{5} N (C H_{3})_{2} + H I$ .

Acylation (with mechanism). Primary and secondary amines react with acid chlorides, anhydrides or esters by nucleophilic acyl substitution, giving amides: $C_{2} H_{5} N H_{2} + C H_{3} C O C l \to C H_{3} C O N H C_{2} H_{5} + H C l$ (N-ethylethanamide). Mechanism: (1) the nitrogen lone pair attacks the carbonyl carbon of the acid chloride, giving a tetrahedral intermediate; (2) the C=O re-forms and Cl⁻ is expelled; (3) loss of H⁺ from nitrogen gives the amide. A stronger base such as pyridine removes the HCl formed and shifts the equilibrium. Tertiary amines have no H on N and are not acylated. Benzoylation with benzoyl chloride is similar: aniline gives benzanilide (N-phenylbenzamide), $C_{6} H_{5} N H_{2} + C_{6} H_{5} C O C l \to C_{6} H_{5} N H C O C_{6} H_{5} + H C l$ . Acylation reduces the reactivity of the lone pair because it is now shared with the C=O group.

Reaction with nitrous acid (NaNO₂ + HCl, cold). Primary aliphatic amines form unstable diazonium salts that decompose at once, giving brisk effervescence of N₂ and an alcohol: $C H_{3} C H_{2} C H_{2} N H_{2} + H N O_{2} \to C H_{3} C H_{2} C H_{2} O H + N_{2} + H_{2} O$ . Primary aromatic amines (aniline) give a diazonium salt stable at 273–278 K. Secondary amines give a yellow, oily N-nitrosoamine, $(C_{2} H_{5})_{2} N H + H N O_{2} \to (C_{2} H_{5})_{2} N N O + H_{2} O$ . Tertiary aliphatic amines form soluble nitrite salts (no visible change).

Carbylamine reaction. Only primary amines (aliphatic or aromatic), warmed with chloroform and alcoholic KOH, give isocyanides with an extremely offensive smell: $C_{6} H_{5} N H_{2} + C H C l_{3} + 3 K O H \to C_{6} H_{5} N C + 3 K C l + 3 H_{2} O$ .

Hinsberg's test with benzenesulphonyl chloride, C₆H₅SO₂Cl. Primary amines give an N-alkylbenzenesulphonamide whose N–H is acidic (strong electron withdrawal by SO₂), so it dissolves in alkali. Secondary amines give an N,N-dialkyl sulphonamide with no H on N, insoluble in alkali. Tertiary amines do not react.

Reactions of aniline's ring. –NH₂ is strongly activating and ortho-para directing. Bromine water gives a white precipitate of 2,4,6-tribromoaniline at once: $C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$ . To get a monosubstituted product, the –NH₂ group is first acetylated (acetanilide), substituted, then hydrolysed. Nitration directly gives tarry oxidation products and much meta product, because in strong acid aniline is protonated to the meta-directing anilinium ion; acetylation first gives mainly 4-nitroacetanilide. Sulphonation: heating anilinium hydrogensulphate at 453–473 K gives sulphanilic acid (4-aminobenzenesulphonic acid), which exists as a zwitterion. Aniline does not undergo Friedel–Crafts reactions: as a Lewis base it forms a salt with the AlCl₃ catalyst, and the positively charged nitrogen then deactivates the ring.

Tests for aniline. (i) Carbylamine test (offensive smell). (ii) Azo dye test: diazotise in the cold and couple with alkaline phenol (or 2-naphthol) to give an orange-red dye. (iii) Bromine water: white precipitate of 2,4,6-tribromoaniline.

Chemical Reactions and Tests of Amines: formulas and key facts

Acylation: $R N H_{2} + C H_{3} C O C l pyridine C H_{3} C O N H R + H C l$; 1° and 2° amines only; nucleophilic acyl substitution.
Nitrous acid, 1° aliphatic: $R N H_{2} + H N O_{2} \to R O H + N_{2} + H_{2} O$; Brisk effervescence of N₂.
Nitrous acid, 2° amine: $R_{2} N H + H N O_{2} \to R_{2} N - N O + H_{2} O$; Yellow oily N-nitrosoamine.
Carbylamine reaction: $R N H_{2} + C H C l_{3} + 3 K O H \to R N C + 3 K C l + 3 H_{2} O$; Offensive smell; 1° amines only.
Hinsberg's test: $R N H_{2} \to C_{6} H_{5} S O_{2} N H R (alkali-soluble); R_{2} N H \to C_{6} H_{5} S O_{2} N R_{2} (insoluble); R_{3} N : no reaction$; Reagent: benzenesulphonyl chloride.
Bromination of aniline: $C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$; White precipitate of 2,4,6-tribromoaniline.

How to solve Chemical Reactions and Tests of Amines questions

Amine questions in ISC are distinguishing tests, reactions with equations, and identification of isomers. First classify each amine as 1°, 2° or 3°, and aliphatic or aromatic.

1Count the H atoms on nitrogen: two (1°), one (2°), none (3°).
2Choose the test: carbylamine (only 1°), nitrous acid (gas from 1° aliphatic, diazonium from 1° aromatic, yellow oil from 2°, no change for 3°), Hinsberg (solubility of the sulphonamide in alkali).
3For acylation and benzoylation, write the amide and HCl; tertiary amines do not react.
4For ring reactions of aniline, remember: bromine water gives the tribromo compound; protect –NH₂ by acetylation for mono-substitution; no Friedel–Crafts.
5For identification problems, combine the formula (CnH2n+3N for saturated amines) with test results and write structures and IUPAC names.
6Balance equations and give by-products (HCl, HBr, N₂, H₂O).

Chemical Reactions and Tests of Amines in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Titration and nitrous acid identify an amine

Suppose a quality-control chemist receives a liquid labelled 'C₃H₉N' (M = 59). Imagine that 0.59 g of it is exactly neutralised by 100 mL of 0.10 M HCl, and that a second sample gives brisk effervescence with NaNO₂ and dilute HCl in the cold.

Show that the amine has one basic nitrogen per molecule, identify its class, and suggest the two possible structures.

Try it first, then show the solutionSolution

Moles of HCl $= 0.100 \times 0.10 = 0.010$ mol; moles of amine $= \frac{0.59}{59} = 0.010$ mol; ratio 1 : 1, so one basic –N per molecule.
Brisk effervescence (N₂) with nitrous acid in the cold is given by primary aliphatic amines: $R N H_{2} + H N O_{2} \to R O H + N_{2} + H_{2} O$ .
Primary amines of formula C₃H₉N: propan-1-amine, CH₃CH₂CH₂NH₂, and propan-2-amine, (CH₃)₂CHNH₂.
The alcohol formed (propan-1-ol or propan-2-ol) could be told apart by the iodoform test, which only propan-2-ol gives.

Answer: A monoacidic primary aliphatic amine: propan-1-amine or propan-2-amine (distinguish via the alcohol formed).

Memory hook: Gas with cold nitrous acid means a primary aliphatic amine.

Tricks to remember Chemical Reactions and Tests of Amines

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hinsberg: Primary Pours into alkali
Primary sulphonamide still has an acidic N–H, so it Pours into (dissolves in) alkali; Secondary Stays solid; Tertiary Takes no part.
Use it when: Hinsberg results.
Two H's for carbylamine
Carbylamine needs both H atoms on N (they leave with the chloroform's Cl as HCl), so only RNH₂ reacts.
Use it when: Carbylamine test answers.
Nitrous acid ladder
1° aliphatic: Gas. 1° aromatic: Diazonium. 2°: Yellow oil. 3°: Nothing visible.
Use it when: Nitrous acid observations.

Chemical Reactions and Tests of Amines: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Effervescence with nitrous acid · 1 mark · MCQBoard level

Which amine gives brisk effervescence of nitrogen with NaNO₂ and dilute HCl at about 278 K?
(a) N,N-dimethylmethanamine
(b) N-methylethanamine
(c) propan-1-amine
(d) aniline

Try it first, then show the solutionSolution

Primary aliphatic amines form unstable diazonium salts that lose N₂ at once.
Aniline forms a stable diazonium salt in the cold; 2° and 3° amines give no gas.

Answer: (c) propan-1-amine.

Example 2: Acetylation of ethylamine · 2 marks · Short answerBoard level

Write the equation for the acetylation of ethanamine with ethanoic anhydride and name the product. Why is the reaction often carried out in the presence of pyridine when an acid chloride is used?

Try it first, then show the solutionSolution

$C_{2} H_{5} N H_{2} + (C H_{3} C O)_{2} O \to C H_{3} C O N H C_{2} H_{5} + C H_{3} C O O H$ : N-ethylethanamide (1 mark).
Pyridine, a stronger base, removes the HCl formed with an acid chloride and shifts the equilibrium to the right (1 mark).

Answer: N-Ethylethanamide; pyridine removes HCl.

Example 3: Mechanism of acylation · 3 marks · Short answerHarder board level

Write the mechanism of the reaction of methanamine with ethanoyl chloride.

Try it first, then show the solutionSolution

The lone pair on N of CH₃NH₂ attacks the carbonyl carbon of CH₃COCl; the C=O π electrons move onto oxygen, forming a tetrahedral intermediate CH₃C(O⁻)(Cl)(N⁺H₂CH₃) (1 mark).
The C=O bond re-forms and Cl⁻ is expelled (1 mark).
Loss of H⁺ from nitrogen (taken by Cl⁻ or pyridine) gives N-methylethanamide: $C H_{3} N H_{2} + C H_{3} C O C l \to C H_{3} C O N H C H_{3} + H C l$ (1 mark).

Answer: Nucleophilic addition to C=O, elimination of Cl⁻, deprotonation: N-methylethanamide.

Example 4: Benzoylation and sulphonation of aniline · 3 marks · Short answerHarder board level

Write equations for (a) the benzoylation of aniline and (b) the sulphonation of aniline, naming the products. Why does the sulphonation product exist as a zwitterion?

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} N H_{2} + C_{6} H_{5} C O C l \to C_{6} H_{5} N H C O C_{6} H_{5} + H C l$ : benzanilide (N-phenylbenzamide) (1 mark).
(b) $C_{6} H_{5} N H_{2} + H_{2} S O_{4} \to C_{6} H_{5} N H_{3} H S O_{4}$ , then heating at 453–473 K: $C_{6} H_{5} N H_{3} H S O_{4} \to H_{2} N C_{6} H_{4} S O_{3} H + H_{2} O$ : sulphanilic acid (4-aminobenzenesulphonic acid) (1 mark).
The strongly acidic –SO₃H group protonates the basic –NH₂ group in the same molecule, giving ⁺H₃N–C₆H₄–SO₃⁻ (1 mark).

Answer: (a) Benzanilide. (b) Sulphanilic acid, a zwitterion because –SO₃H protonates –NH₂.

Example 5: Sorting C₄H₁₁N amines · 5 marks · Long answerHOTS / competency

Three isomeric amines X, Y and Z have the formula C₄H₁₁N. X gives an offensive smell with CHCl₃ and alcoholic KOH, and its Hinsberg product dissolves in NaOH; X is optically active. Y gives a yellow oily liquid with nitrous acid, and its Hinsberg product is insoluble in alkali; Y has two identical alkyl groups. Z does not react with benzenesulphonyl chloride and has two methyl groups.
(a) Identify X, Y and Z with IUPAC names. [3]
(b) Write the equation for Y with nitrous acid. [1]
(c) Write the equation for X with benzenesulphonyl chloride. [1]

Try it first, then show the solutionSolution

X is a primary amine that is chiral: butan-2-amine, CH₃CH₂CH(NH₂)CH₃ (1 mark).
Y is a secondary amine with two identical groups: N-ethylethanamine, (C₂H₅)₂NH (1 mark).
Z is tertiary with two methyl groups: N,N-dimethylethanamine, (CH₃)₂NC₂H₅ (1 mark).
(b) $(C_{2} H_{5})_{2} N H + H N O_{2} \to (C_{2} H_{5})_{2} N N O + H_{2} O$ (N-nitrosodiethylamine) (1 mark).
(c) $C H_{3} C H_{2} C H (N H_{2}) C H_{3} + C_{6} H_{5} S O_{2} C l \to C_{6} H_{5} S O_{2} N H C H (C H_{3}) C_{2} H_{5} + H C l$ ; the product dissolves in NaOH (1 mark).

Answer: X butan-2-amine, Y N-ethylethanamine, Z N,N-dimethylethanamine.

Where students lose marks in Chemical Reactions and Tests of Amines

The slips examiners see most often, and what to write instead.

Mistake: Writing alkylation when acetylation is asked
ISC examiners reported alkylation shown instead of acetylation of ethylamine, or the by-product omitted (2024).
Wrong working
$C_{2} H_{5} N H_{2} + C H_{3} C l \to C_{2} H_{5} N H C H_{3}$
Right working
$C_{2} H_{5} N H_{2} + C H_{3} C O C l \to C H_{3} C O N H C_{2} H_{5} + H C l$
Do this: Acetylation adds CH₃CO– (an acyl group) and gives an amide.
Usually costs: 1 mark.
Mistake: Giving resonance as the reason aniline fails Friedel–Crafts
Examiners noted that candidates missed the key words 'Lewis base' and 'forms a salt with the Lewis acid AlCl₃' (2024).
Do this: Aniline (a Lewis base) forms a salt with AlCl₃; the positive nitrogen then strongly deactivates the ring.
Usually costs: 1 mark.
Mistake: Missing by-products in bromination
Examiners reported HBr omitted and unbalanced equations for aniline with bromine water (2025).
Do this: $C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$ .
Usually costs: 1 mark.
Mistake: Choosing the wrong amine for the carbylamine test
In 2025 candidates chose aniline or dimethylamine instead of methylamine to form methyl isocyanide.
Do this: The isocyanide carries the amine's own R group: methyl isocyanide needs methylamine; secondary amines never react.
Usually costs: 1 mark.

Exam tips

In distinguishing questions, give the reagent and the observation for every compound named.
Write IUPAC names of amines (N-methylethanamine, N,N-dimethylethanamine), as examiners expect them.
For aniline ring reactions, mention the activating effect of –NH₂ and the need for protection by acetylation.
Learn the 'key words' examiners look for: Lewis base, lone pair, delocalised, acidic N–H, zwitterion.
Balance every equation and include HCl, HBr, N₂ or H₂O as by-products.

Practice questions on Chemical Reactions and Tests of Amines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following does NOT give the carbylamine test?
(a) aniline
(b) ethanamine
(c) N-methylaniline
(d) benzylamine
Hint
Only primary amines respond.
Answer
(c) N-methylaniline, a secondary amine.
Q2 · 2 marks · Short answerBoard level
Write the balanced equation for the reaction of aniline with bromine water, and give the observation.
Hint
Include the by-product.
Answer
$C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$ ; a white precipitate of 2,4,6-tribromoaniline forms at once.
Q3 · 2 marks · Short answerHarder board level
Write equations for the reaction of aniline with (a) one mole and (b) a second mole of iodomethane. What is formed if excess iodomethane is used?
Hint
Each step replaces an H on N by CH₃.
Answer
(a) $C_{6} H_{5} N H_{2} + C H_{3} I \to C_{6} H_{5} N H C H_{3} + H I$ (N-methylaniline). (b) $C_{6} H_{5} N H C H_{3} + C H_{3} I \to C_{6} H_{5} N (C H_{3})_{2} + H I$ (N,N-dimethylaniline). With excess, the quaternary salt C₆H₅N(CH₃)₃⁺I⁻ forms.
Q4 · 3 marks · Short answerHarder board level
Aniline is treated with NaNO₂ and dilute HCl (a) at 273–278 K and (b) at room temperature. Give the products in each case and explain the difference.
Hint
The diazonium salt is stable only in the cold.
Answer
(a) Benzenediazonium chloride: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ . (b) The diazonium salt formed decomposes as it forms: $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ , so phenol and nitrogen are obtained. Arenediazonium salts are stable only at low temperature.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Acetanilide is less reactive than aniline towards electrophilic substitution.
Reason (R): In acetanilide the nitrogen lone pair is also delocalised onto the carbonyl oxygen, so less of it is donated to the ring.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Competing resonance.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Describe three chemical tests for aniline, giving the reagent and the observation for each.
Hint
One uses CHCl₃, one uses a coupling partner, one uses bromine.
Answer
(i) Carbylamine: warm with CHCl₃ and alcoholic KOH; offensive smell of phenyl isocyanide. (ii) Azo dye: diazotise with NaNO₂/HCl at 273–278 K and add alkaline phenol (or 2-naphthol); orange-red dye. (iii) Bromine water: immediate white precipitate of 2,4,6-tribromoaniline.
Q7 · 3 marks · Short answerHOTS / competency
0.93 g of aniline (M = 93) reacts completely with bromine water. Calculate the mass of 2,4,6-tribromoaniline (M = 330) formed and the mass of bromine (Br₂, M = 160) used.
Hint
1 mol aniline needs 3 mol Br₂.
Answer
Aniline $= 0.010$ mol. Tribromoaniline $= 0.010 \times 330 = 3.30$ g. Br₂ $= 0.030$ mol $= 0.030 \times 160 = 4.80$ g.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a pharmaceutical plant makes paracetamol, N-(4-hydroxyphenyl)ethanamide, by warming 4-aminophenol (HO–C₆H₄–NH₂) with ethanoic anhydride.
(i) Write the equation. [1]
(ii) The molecule has both –OH and –NH₂ groups. Why is the nitrogen, not the oxygen, acetylated? [1]
(iii) Is paracetamol more or less basic than 4-aminophenol? Explain. [1]
Hint
Compare the nucleophilicity of N and O lone pairs.
Answer
(i) $H O C_{6} H_{4} N H_{2} + (C H_{3} C O)_{2} O \to H O C_{6} H_{4} N H C O C H_{3} + C H_{3} C O O H$ . (ii) Nitrogen is less electronegative than oxygen, so its lone pair is more available: the –NH₂ group is the stronger nucleophile and reacts first. (iii) Less basic: in the amide the nitrogen lone pair is delocalised onto the carbonyl oxygen and is hardly available to accept a proton.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of a 2° amine with nitrous acid?Show answer
A yellow, oily N-nitrosoamine.
2.Reagent in Hinsberg's test?Show answer
Benzenesulphonyl chloride.
3.Why is a primary sulphonamide soluble in alkali?Show answer
Its N–H is acidic because of the strongly electron-withdrawing SO₂ group.
4.Product of benzoylation of aniline?Show answer
Benzanilide (N-phenylbenzamide).
5.What is sulphanilic acid?Show answer
4-Aminobenzenesulphonic acid, existing as a zwitterion.
6.Why don't tertiary amines undergo acylation?Show answer
They have no H on nitrogen to be replaced.

Chemical Reactions and Tests of Amines: frequently asked questions

How do you distinguish primary, secondary and tertiary amines?

With Hinsberg's reagent: the primary amine's product dissolves in alkali, the secondary amine's product does not, and the tertiary amine does not react. Nitrous acid and the carbylamine test also help.

What is the mechanism of acylation of amines?

The nitrogen lone pair attacks the carbonyl carbon of the acid chloride, a tetrahedral intermediate forms, chloride is expelled and a proton is lost from nitrogen, giving an amide.

Why does aniline give 2,4,6-tribromoaniline with bromine water?

The –NH₂ group strongly activates all the ortho and para positions, so substitution happens at all three at once.

Why can't aniline undergo Friedel–Crafts reactions?

Aniline, a Lewis base, forms a salt with the AlCl₃ catalyst; the positively charged nitrogen deactivates the ring.

What happens when a primary aliphatic amine reacts with nitrous acid?

An unstable diazonium salt forms and decomposes, giving an alcohol and brisk effervescence of nitrogen.