Amines Class 12: Preparation, Basicity, Tests and Diazonium Salts

What Amines is about

Weightage: This chapter is Unit IX, Amines, which carries 6 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit IX Amines

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6 marks per paper. See the questions

Key concepts

Structure, classification and nomenclature

NCERT section 9.1 Structure of Amines; 9.2 Classification; 9.3 Nomenclature

Structure: the nitrogen in amines is sp³ hybridised. Three of its orbitals bond to H or C, and the fourth holds a lone pair, so the shape is pyramidal. Because the lone pair repels the bonding pairs, the C–N–C angle is less than 109.5°; it is 108° in trimethylamine.

Classification: one H of NH₃ replaced gives a primary amine (RNH₂); two gives a secondary amine (R₂NH); three gives a tertiary amine (R₃N). Secondary and tertiary amines are simple if all groups are the same and mixed if they differ.

Naming:

• Common names: alkyl group + amine, written as one word: methylamine, dimethylamine.
• IUPAC: replace the -e of the alkane with -amine: CH₃CH₂NH₂ is ethanamine. For secondary and tertiary amines, the largest group is the parent and the others are given with the locant N: (CH₃)₂NCH₂CH₃ is N,N-dimethylethanamine.
• For two –NH₂ groups, -diamine is added to the full alkane name: H₂NCH₂CH₂NH₂ is ethane-1,2-diamine.
• Aryl amines: C₆H₅NH₂ is aniline, accepted in IUPAC as benzenamine; C₆H₅NHCH₃ is N-methylbenzenamine.

Worked example

Classify as primary, secondary or tertiary and give the IUPAC name: (a) (CH₃)₂CHNH₂ (b) CH₃NHCH₂CH₃ (c) (C₂H₅)₃N (d) C₆H₅N(CH₃)₂. (2 marks)

Try it first, then show the solutionSolution

1. (a) Primary: propan-2-amine. (½)
2. (b) Secondary: N-methylethanamine. (½)
3. (c) Tertiary: N,N-diethylethanamine. (½)
4. (d) Tertiary: N,N-dimethylbenzenamine (N,N-dimethylaniline). (½)

Answer: (a) 1°, propan-2-amine (b) 2°, N-methylethanamine (c) 3°, N,N-diethylethanamine (d) 3°, N,N-dimethylbenzenamine

Preparation of amines

NCERT section 9.4 Preparation of Amines

1. Reduction of nitro compounds: with H₂ over Ni, Pt or Pd, or with a metal and acid. Iron scrap and HCl is preferred: the FeCl₂ formed is hydrolysed and releases HCl, so only a small amount of acid is needed to start the reaction.
2. Ammonolysis of alkyl halides: an alcoholic solution of ammonia attacks the alkyl halide (SN2) to give a primary amine salt. The free amine is released by a strong base. Disadvantage: the primary amine is itself a nucleophile, so it reacts further, giving a mixture of 1°, 2° and 3° amines and the quaternary ammonium salt. A large excess of ammonia favours the primary amine. Order of halide reactivity: RI > RBr > RCl.
3. Reduction of nitriles with LiAlH₄ or H₂/Ni gives primary amines with one more carbon than the starting halide: useful for ascending the series.
4. Reduction of amides with LiAlH₄ gives amines.
5. Gabriel phthalimide synthesis: phthalimide with ethanolic KOH gives potassium phthalimide, which is heated with an alkyl halide to give N-alkylphthalimide; alkaline hydrolysis gives a pure primary amine. Aromatic primary amines cannot be made this way, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
6. Hoffmann bromamide degradation: an amide with Br₂ and aqueous or ethanolic NaOH gives a primary amine with one carbon fewer (the alkyl or aryl group migrates from the carbonyl carbon to nitrogen):
$RCON{H}_2+B{r}_2+4NaOH\to RN{H}_2+N{a}_{2}C{O}_3+2NaBr+2H_{2}O$

Worked example

(a) Write the equation for the Hoffmann bromamide degradation of ethanamide. (b) Why can aniline not be prepared by the Gabriel phthalimide synthesis? (c) Why is Fe/HCl preferred for reducing nitrobenzene? (3 marks)

Try it first, then show the solutionSolution

1. (a) $C{H}_{3}CON{H}_2+B{r}_2+4NaOH\to C{H}_{3}N{H}_2+N{a}_{2}C{O}_3+2NaBr+2H_{2}O$: methanamine, one carbon fewer. (1)
2. (b) Gabriel synthesis needs an SN2 reaction of the phthalimide anion with a halide; aryl halides do not undergo nucleophilic substitution easily, so no N-arylphthalimide forms. (1)
3. (c) FeCl₂ formed during the reaction is hydrolysed and releases HCl, so only a small amount of HCl is needed. (1)

Answer: (a) CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O (b) aryl halides resist nucleophilic substitution (c) HCl is regenerated.

Physical properties

NCERT section 9.5 Physical Properties

• Lower aliphatic amines are gases with a fishy smell; primary amines with three or more carbons are liquids and higher ones are solids. Aniline and other aryl amines are usually colourless but become coloured on storage, because they are oxidised by air.
• Lower aliphatic amines are soluble in water, because they form hydrogen bonds with water; solubility falls as the hydrophobic alkyl part grows. Higher amines are almost insoluble. Amines are soluble in organic solvents.
• Boiling points: primary and secondary amines form intermolecular hydrogen bonds (N–H···N), but these are weaker than O–H···O bonds, because N is less electronegative than O. Primary amines have two H atoms on N and form more hydrogen bonds than secondary amines; tertiary amines have no H on N and cannot. So for isomers: primary > secondary > tertiary. Amines boil lower than alcohols of similar mass.

Worked example

Arrange in increasing order of boiling point and explain: C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂. (2 marks)

Try it first, then show the solutionSolution

1. Ethanol forms the strongest hydrogen bonds (O–H···O); ethanamine has two N–H bonds; dimethylamine has only one N–H and so forms fewer hydrogen bonds. (1)
2. Order: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH. (1)

Answer: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

Basic character of amines

NCERT section 9.6 Chemical Reactions (1) Basic character of amines

The lone pair on nitrogen accepts a proton, so amines are bases: $RN{H}_2+H_{2}O\rightleftharpoons RN{H}_3^{+}+O{H}^{-}$. They react with acids to form salts, which are soluble in water and give back the amine with NaOH.

Basic strength is given by $K_b$ or $p{K}_b=-\log K_b$: the smaller the pKb, the stronger the base. NCERT values in water: ammonia 4.75 and aniline 9.38; for the methyl series, primary 3.38, secondary 3.27, tertiary 4.22; for the ethyl series, primary 3.29, secondary 3.00, tertiary 3.25.

Aliphatic amines are stronger bases than ammonia: alkyl groups push electrons (+I) onto nitrogen, and the substituted ammonium ion is stabilised.

Order of basicity:

• Gas phase (inductive effect only): 3° > 2° > 1° > NH₃.
• Water: three factors act together: +I effect, solvation of the ammonium ion by hydrogen bonding (more N–H, better solvation), and steric hindrance to hydrogen bonding. So the order is not regular:
• Methyl: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
• Ethyl: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃

Aryl amines are much weaker bases (aniline pKb 9.38): the lone pair is delocalised into the ring by resonance, so it is less available; aniline has five resonance structures, but the anilinium ion has only two, so protonation loses resonance stability. Electron-releasing groups (–OCH₃, –CH₃) on the ring increase basicity; electron-withdrawing groups (–NO₂, –SO₃H, –COOH, –X) decrease it.

Worked example

Using pKb values (methanamine 3.38, ammonia 4.75, aniline 9.38), arrange them in decreasing order of basic strength and explain why aniline is the weakest. (3 marks)

Try it first, then show the solutionSolution

1. Smaller pKb means a stronger base: methanamine > ammonia > aniline. (1)
2. Methanamine is stronger than ammonia because the methyl group (+I) increases electron density on nitrogen. (1)
3. In aniline the lone pair on N is delocalised into the benzene ring by resonance, so it is less available for a proton; also the anilinium ion has fewer resonance structures than aniline. (1)

Answer: Methanamine > ammonia > aniline; resonance makes aniline's lone pair less available.

Reactions of amines and distinguishing tests

NCERT section 9.6 Chemical Reactions (2) Alkylation (3) Acylation (4) Carbylamine reaction (5) Reaction with nitrous acid (6) Reaction with arylsulphonyl chloride

Alkylation: amines react with alkyl halides to give higher amines and finally quaternary ammonium salts.

Acylation: primary and secondary amines react with acid chlorides, anhydrides or esters by nucleophilic substitution to give amides; a stronger base such as pyridine removes the HCl formed. Tertiary amines (no H on N) do not react. Benzoylation uses benzoyl chloride: $C{H}_{3}N{H}_2+C_6{H}_{5}COCl\to C_6{H}_{5}CONHC{H}_3+HCl$.

Carbylamine reaction (isocyanide test): aliphatic and aromatic primary amines, warmed with chloroform and ethanolic KOH, give isocyanides (carbylamines), which have a very unpleasant smell. Secondary and tertiary amines do not react, so this is a test for primary amines:
$RN{H}_2+CHC{l}_3+3KOH\to RNC+3KCl+3H_{2}O$

Nitrous acid (made in situ from NaNO₂ and HCl):

• Primary aliphatic amines give unstable aliphatic diazonium salts, which decompose to alcohols with quantitative evolution of nitrogen (used to estimate amino acids and proteins).
• Aromatic primary amines at 273–278 K give stable diazonium salts.

Hinsberg's reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl):

• Primary amines give an N-alkylbenzenesulphonamide, whose N–H is acidic (strongly electron-withdrawing sulphonyl group), so it dissolves in alkali.
• Secondary amines give an N,N-dialkylbenzenesulphonamide with no H on N, which is insoluble in alkali.
• Tertiary amines do not react.

This distinguishes and separates the three classes; today p-toluenesulphonyl chloride is often used instead.

Worked example

Describe a chemical test to distinguish (a) ethanamine and N-methylmethanamine (dimethylamine), (b) the three isomers propan-1-amine, N-methylethanamine and N,N-dimethylmethanamine using Hinsberg's reagent. (3 marks)

Try it first, then show the solutionSolution

1. (a) Carbylamine test: warm with CHCl₃ and alcoholic KOH; ethanamine (1°) gives a foul-smelling isocyanide; dimethylamine (2°) does not. (1)
2. (b) Propan-1-amine (1°) gives a sulphonamide that dissolves in alkali. (1)
3. N-methylethanamine (2°) gives a sulphonamide insoluble in alkali; N,N-dimethylmethanamine (3°) does not react. (1)

Answer: (a) Carbylamine test (b) Hinsberg: 1° product soluble in alkali, 2° product insoluble, 3° no reaction.

Electrophilic substitution of aniline

NCERT section 9.6 Chemical Reactions (7) Electrophilic substitution

The –NH₂ group is a strongly activating, ortho and para directing group (its lone pair increases electron density at o and p by resonance).

Bromination: shaken with bromine water, even at room temperature, aniline is substituted at all three activated positions and 2,4,6-tribromoaniline separates as a white solid. To get a monosubstituted product, the activating effect is reduced by acetylation: aniline → acetanilide (with acetic anhydride). In acetanilide, the nitrogen lone pair is also delocalised into the C=O group by resonance, so it activates the ring less. Bromination then gives mainly p-bromoacetanilide, which is hydrolysed to p-bromoaniline.

Nitration: direct nitration gives tarry oxidation products, and in strongly acidic medium aniline is protonated to the anilinium ion, which is meta directing. So besides ortho and para products, a significant amount (about 47%) of m-nitroaniline forms. Protecting –NH₂ by acetylation first gives mainly p-nitroaniline after hydrolysis.

Sulphonation: aniline + concentrated H₂SO₄ gives anilinium hydrogensulphate, which on heating at 453–473 K gives sulphanilic acid (p-aminobenzenesulphonic acid) as the major product.

No Friedel–Crafts reaction: aniline, a Lewis base, forms a salt with the Lewis acid AlCl₃; the nitrogen gets a positive charge and strongly deactivates the ring.

Worked example

(a) Why does aniline give 2,4,6-tribromoaniline with bromine water? (b) How can p-bromoaniline be prepared from aniline? (c) Why does aniline not undergo Friedel–Crafts reaction? (3 marks)

Try it first, then show the solutionSolution

1. (a) –NH₂ strongly activates the ring at all ortho and para positions, so all three are substituted: $C_6{H}_{5}N{H}_2+3B{r}_2\to C_6{H}_{2}B{r}_{3}N{H}_2+3HBr$. (1)
2. (b) Acetylate with acetic anhydride to acetanilide (less activating), brominate to p-bromoacetanilide, then hydrolyse to p-bromoaniline. (1)
3. (c) Aniline's lone pair forms a salt with the Lewis acid AlCl₃; the positively charged nitrogen deactivates the ring. (1)

Answer: (a) Strong activation at o and p (b) protect by acetylation, brominate, hydrolyse (c) salt formation with AlCl₃ deactivates the ring.

Diazonium salts: preparation and replacement reactions

NCERT section 9.7 Method of Preparation of Diazonium Salts; 9.8 Physical Properties; 9.9 Chemical Reactions (A) Reactions involving displacement of nitrogen

Diazonium salts have the general formula $R{N}_2^{+}{X}^{-}$ (the –N₂⁺ group is the diazonium group). Arenediazonium salts are stable for a short time in solution at low temperature, because the C–N bond is stabilised by resonance with the ring; aliphatic diazonium salts are very unstable.

Diazotisation: aniline with nitrous acid (NaNO₂ + HCl) at 273–278 K gives benzenediazonium chloride:
$C_6{H}_{5}N{H}_2+NaN{O}_2+2HCl\to C_6{H}_5{N}_2^{+}C{l}^{-}+NaCl+2H_{2}O$
It is used immediately; it is a colourless crystalline solid, soluble in water, and the fluoroborate salt is stable at room temperature.

Replacement of nitrogen (N₂ is an excellent leaving group):

• By Cl⁻, Br⁻, CN⁻: with Cu(I) salts, the Sandmeyer reaction (CuCl/HCl, CuBr/HBr, CuCN/KCN); with copper powder and HCl or HBr, the Gatterman reaction.
• By iodide: simply warm with KI.
• By fluoride: with fluoroboric acid (HBF₄) the diazonium fluoroborate precipitates; heating it gives fluorobenzene.
• By H: mild reducing agents such as hypophosphorous acid (phosphinic acid) or ethanol give benzene.
• By OH: warming the solution (283 K or above) gives phenol.
• By NO₂: the fluoroborate heated with NaNO₂ and copper gives nitrobenzene.

These reactions let us put groups on a benzene ring that cannot be introduced directly (F, I, CN, OH), or remove an –NH₂ group after it has done its job as a director.

Worked example

Write the reagents and equations to convert benzenediazonium chloride into (a) benzonitrile, (b) iodobenzene, (c) benzene. (3 marks)

Try it first, then show the solutionSolution

1. (a) Sandmeyer: $C_6{H}_5{N}_{2}Cl+CuCN\to C_6{H}_{5}CN+N_2+CuCl$ (CuCN/KCN). (1)
2. (b) Warm with KI: $C_6{H}_5{N}_{2}Cl+KI\to C_6{H}_{5}I+KCl+N_2$. (1)
3. (c) Hypophosphorous acid: $C_6{H}_5{N}_{2}Cl+H_{3}P{O}_2+H_{2}O\to C_6{H}_6+N_2+H_{3}P{O}_3+HCl$. (1)

Answer: (a) CuCN/KCN (b) KI (c) H₃PO₂ (or ethanol)

Coupling reactions and importance of diazonium salts

NCERT section 9.9 Chemical Reactions (B) Reactions involving retention of diazo group: coupling reactions; 9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds

In coupling reactions the diazonium ion acts as an electrophile and the nitrogen atoms are kept: it attacks the para position of an activated ring (phenol or aniline), giving an azo compound, –N=N–, joining two aromatic rings. The extended conjugation makes these compounds coloured, and they are used as dyes.

• With phenol in alkaline medium (the phenoxide ion is more reactive): p-hydroxyazobenzene, an orange dye:

$C_6{H}_5{N}_{2}Cl+C_6{H}_{5}OH\stackrel{O{H}^{-}}{\to }HO{C}_6{H}_{4}N=N{C}_6{H}_5+HCl$

• With aniline in mildly acidic medium: p-aminoazobenzene, a yellow dye:

$C_6{H}_5{N}_{2}Cl+C_6{H}_{5}N{H}_2\stackrel{H^{+}}{\to }{H}_{2}N{C}_6{H}_{4}N=N{C}_6{H}_5+HCl$

Importance in synthesis: diazonium salts are excellent intermediates for introducing –F, –Cl, –Br, –I, –CN, –OH and –NO₂ into an aromatic ring. Aryl fluorides and iodides, which cannot be made by direct halogenation, and cyanobenzene, which cannot be made from chlorobenzene by nucleophilic substitution, are obtained easily this way.

Worked example

Benzenediazonium chloride is treated with phenol in alkaline medium. (a) Write the equation. (b) What is the colour of the product, and why is it coloured? (c) Is nitrogen lost in this reaction? (2 marks)

Try it first, then show the solutionSolution

1. (a) $C_6{H}_5{N}_{2}Cl+C_6{H}_{5}OH\stackrel{O{H}^{-}}{\to }HO{C}_6{H}_{4}N=N{C}_6{H}_5+HCl$ (p-hydroxyazobenzene). (1)
2. (b) Orange; the –N=N– group joins two rings into an extended conjugated system that absorbs visible light. (c) No: both nitrogen atoms are retained (coupling). (1)

Answer: p-Hydroxyazobenzene, an orange dye; nitrogen is retained.

Chemistry in real life: Amines

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.

Topics in this chapter

• Preparation of Amines
• Basicity of Amines
• Diazonium Salts and Their Reactions

Amines formula sheet

Reduction of nitrobenzene

$C_6{H}_{5}N{O}_2+3H_2\stackrel{Pd}{\to }{C}_6{H}_{5}N{H}_2+2H_{2}O$

Fe scrap + HCl is preferred industrially (HCl is regenerated).

Ammonolysis (freeing the amine)

$RN{H}_3^{+}{X}^{-}+NaOH\to RN{H}_2+H_{2}O+NaX$

Ammonolysis gives a mixture of 1°, 2°, 3° amines and quaternary salt.

Reduction of nitriles

$C{H}_{3}CN+2H_2\stackrel{Ni}{\to }C{H}_{3}C{H}_{2}N{H}_2$

Adds one carbon to the chain (from the halide used to make the nitrile).

Hoffmann bromamide degradation

$RCON{H}_2+B{r}_2+4NaOH\to RN{H}_2+N{a}_{2}C{O}_3+2NaBr+2H_{2}O$

Amine has one carbon fewer than the amide.

Basicity

$RN{H}_2+H_{2}O\rightleftharpoons RN{H}_3^{+}+O{H}^{-},p{K}_b=-\log K_b$

Smaller pKb, stronger base. NH₃ 4.75; CH₃NH₂ 3.38; (C₂H₅)₂NH 3.00; C₆H₅NH₂ 9.38.

Acylation

$C_2{H}_{5}N{H}_2+C{H}_{3}COCl\to C_2{H}_{5}NHCOC{H}_3+HCl$

Pyridine removes HCl; 3° amines do not react.

Carbylamine reaction

$RN{H}_2+CHC{l}_3+3KOH\to RNC+3KCl+3H_{2}O$

Foul-smelling isocyanide: test for primary amines.

Primary aliphatic amine + nitrous acid

$RN{H}_2+HN{O}_2\to ROH+N_2+H_{2}O$

Via an unstable diazonium salt; N₂ evolved quantitatively.

Diazotisation

$C_6{H}_{5}N{H}_2+NaN{O}_2+2HCl\to C_6{H}_5{N}_2^{+}C{l}^{-}+NaCl+2H_{2}O$

At 273–278 K.

Hinsberg reaction (primary amine)

$C_6{H}_{5}S{O}_{2}Cl+C_2{H}_{5}N{H}_2\to C_6{H}_{5}S{O}_{2}NH{C}_2{H}_5+HCl$

Product soluble in alkali (acidic N–H); 2° product insoluble; 3° no reaction.

Bromination of aniline

$C_6{H}_{5}N{H}_2+3B{r}_2\to C_6{H}_{2}B{r}_{3}N{H}_2+3HBr$

White 2,4,6-tribromoaniline.

Acetylation of aniline

$C_6{H}_{5}N{H}_2+(C{H}_{3}CO{)}_{2}O\to C_6{H}_{5}NHCOC{H}_3+C{H}_{3}COOH$

Protects –NH₂ and lowers its activating effect.

Sandmeyer reaction

$C_6{H}_5{N}_{2}Cl+CuCN\to C_6{H}_{5}CN+N_2+CuCl$

Also CuCl/HCl → C₆H₅Cl and CuBr/HBr → C₆H₅Br. Gatterman uses Cu powder + HX.

Replacement by iodide

$C_6{H}_5{N}_{2}Cl+KI\to C_6{H}_{5}I+KCl+N_2$

Iodobenzene cannot easily be made by direct iodination.

Replacement by fluoride

$C_6{H}_5{N}_2^{+}B{F}_4^{-}\stackrel{\Delta }{\to }{C}_6{H}_{5}F+B{F}_3+N_2$

Fluoroborate from the diazonium salt and HBF₄.

Replacement by hydrogen

$C_6{H}_5{N}_{2}Cl+C{H}_{3}C{H}_{2}OH\to C_6{H}_6+N_2+C{H}_{3}CHO+HCl$

Or H₃PO₂ (hypophosphorous acid).

Replacement by OH

$C_6{H}_5{N}_{2}Cl+H_{2}O\to C_6{H}_{5}OH+N_2+HCl$

On warming the solution.

Coupling with phenol

$C_6{H}_5{N}_{2}Cl+C_6{H}_{5}OH\stackrel{O{H}^{-}}{\to }HO{C}_6{H}_{4}N=N{C}_6{H}_5+HCl$

Orange dye. With aniline (H⁺): p-aminoazobenzene, yellow.

Amines: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Amines

The question patterns to expect, and the method that works for each.

Tricks to remember Amines

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

• Gabriel is Good for Primary aliphatic only

  Gabriel gives pure 1° aliphatic amines; never aniline (aryl halides do not do SN2).

  Use it when: Gabriel synthesis questions.

• Hoffmann Hacks off a carbon

  Amide (n carbons) + Br₂/NaOH → amine (n − 1 carbons). Nitrile reduction does the opposite: it keeps the extra carbon.

  Use it when: Counting carbons in conversions.

• Ethyl series: 2 > 3 > 1; Methyl series: 2 > 1 > 3

  Secondary wins in both. In the methyl series tertiary falls to third because of poor solvation; in the ethyl series tertiary stays second.

  Use it when: Aqueous basicity orders.

• Carbylamine smells only for primary

  CHCl₃ + KOH + 1° amine = stinking isocyanide. 2° and 3° stay odour-free.

  Use it when: Test for primary amines.

• Sandmeyer uses Salt, Gatterman uses Grains

  Sandmeyer: Cu(I) salts (CuCl, CuBr, CuCN). Gatterman: copper powder (grains) with HX.

  Use it when: Telling the two replacement reactions apart.

Amines: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Amines

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

• ✗Mistake: Writing gas-phase basicity order for aqueous solution

  Wrong working

  In water: (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃

  Right working

  In water: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

  ✓Do this: In water, add solvation and steric effects to the inductive effect.

  Usually costs: 1 mark in an ordering question.

• ✗Mistake: Wrong carbon count after Hoffmann degradation

  Wrong working

  CH₃CH₂CONH₂ + Br₂ + NaOH → CH₃CH₂CH₂NH₂

  Right working

  CH₃CH₂CONH₂ + Br₂ + 4NaOH → CH₃CH₂NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

  ✓Do this: The carbonyl carbon leaves as carbonate: one carbon fewer.

  Usually costs: The equation mark.

• ✗Mistake: Forgetting temperature in diazotisation

  ✓Do this: Always write 273–278 K (0–5 °C); at higher temperature the diazonium salt decomposes to phenol.

  Usually costs: ½ mark.

• ✗Mistake: Saying aniline nitrates only at ortho and para

  ✓Do this: In strong acid, aniline becomes anilinium ion (meta directing), so a large amount of m-nitroaniline forms; protect –NH₂ by acetylation to get p-nitroaniline.

  Usually costs: 1 mark.

• ✗Mistake: Losing nitrogen in coupling reactions

  Wrong working

  C₆H₅N₂Cl + C₆H₅OH → C₆H₅–C₆H₄OH + N₂ + HCl

  Right working

  C₆H₅N₂Cl + C₆H₅OH (OH⁻) → HOC₆H₄–N=N–C₆H₅ + HCl

  ✓Do this: In coupling the –N=N– is kept; only in replacement reactions is N₂ lost.

  Usually costs: The full equation mark.

NCERT exercise guide: Amines

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Amines: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6 marks per paper; 29 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Amines

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Amines question, by year

Practice questions on Amines

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

1. Q1 · 1 mark · MCQBoard level

   The best reagent for converting an amide into an amine with one carbon fewer is:
   (a) LiAlH₄ (b) Br₂ + NaOH (c) H₂/Ni (d) NaNO₂ + HCl

   Hint

   Hoffmann.

   Answer

   (b) Br₂ + NaOH (Hoffmann bromamide degradation).

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Aniline does not undergo Friedel–Crafts reaction.
   Reason (R): Aniline forms a salt with AlCl₃, which deactivates 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

   Lewis base meets Lewis acid.

   Answer

   (a)

3. Q3 · 2 marks · Very short answerBoard level

   Write the steps of the Gabriel phthalimide synthesis for preparing ethanamine. (2 marks)

   Hint

   KOH, then the alkyl halide, then hydrolysis.

   Answer

   Phthalimide + ethanolic KOH → potassium phthalimide (½); + C₂H₅Br on heating → N-ethylphthalimide (½); alkaline hydrolysis (NaOH) → C₂H₅NH₂ + sodium phthalate (1).

4. Q4 · 2 marks · Very short answerBoard level

   Why do primary amines have higher boiling points than tertiary amines of the same molar mass, but lower than alcohols? (2 marks)

   Hint

   Count N–H bonds; compare N and O.

   Answer

   Primary amines have two N–H bonds and form intermolecular hydrogen bonds; tertiary amines have none (1). N is less electronegative than O, so N–H···N hydrogen bonds are weaker than O–H···O bonds in alcohols (1).

5. Q5 · 3 marks · Short answerHarder board level

   Complete: (a) C₆H₅N₂Cl + KI → (b) C₆H₅N₂Cl + CH₃CH₂OH → (c) C₆H₅NH₂ + CHCl₃ + KOH (alc.) → (3 marks)

   Hint

   Iodide, reduction to H, isocyanide.

   Answer

   (a) C₆H₅I + KCl + N₂ (1). (b) C₆H₆ + N₂ + CH₃CHO + HCl (1). (c) C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O (1).

6. Q6 · 3 marks · Short answerHarder board level

   Arrange in increasing order of basic strength: aniline, p-nitroaniline, p-toluidine, p-methoxyaniline. Give reasons. (3 marks)

   Hint

   −NO₂ weakens; –CH₃ and –OCH₃ strengthen.

   Answer

   p-Nitroaniline < aniline < p-toluidine < p-methoxyaniline (2). –NO₂ withdraws electrons from N; –CH₃ (+I, hyperconjugation) and –OCH₃ (strong +R at para) push electron density towards N, making its lone pair more available (1).

7. Q7 · 2 marks · Very short answerHarder board level

   Convert (a) ethanoic acid to methanamine, (b) bromoethane to propan-1-amine. (2 marks)

   Hint

   One carbon fewer; one carbon more.

   Answer

   (a) CH₃COOH + SOCl₂ → CH₃COCl; + NH₃ → CH₃CONH₂; + Br₂/NaOH → CH₃NH₂ (1). (b) C₂H₅Br + KCN → C₂H₅CN; + LiAlH₄ (or H₂/Ni) → CH₃CH₂CH₂NH₂ (1).

8. Q8 · 5 marks · Long answerHarder board level

   (a) Explain the basicity of amines in terms of Kb and pKb. (1) (b) Why is the order of basicity of ethyl amines in water different from that in the gas phase? (2) (c) Describe the Hinsberg test for the three classes of amines. (2) (5 marks)

   Hint

   Three factors in water.

   Answer

   (a) Kb is the equilibrium constant for RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻; pKb = −log Kb; larger Kb (smaller pKb) means stronger base (1). (b) Gas phase: only the +I effect, so 3° > 2° > 1° > NH₃ (1); in water, solvation by hydrogen bonding and steric hindrance also act, giving (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃ (1). (c) With C₆H₅SO₂Cl: 1° amine gives a sulphonamide soluble in alkali; 2° gives an alkali-insoluble sulphonamide; 3° does not react (2).

9. Q9 · 3 marks · Short answerHOTS / competency

   Suppose a student wants to prepare m-bromotoluene. Direct bromination of toluene gives o- and p-isomers. Plan a route from p-toluidine (4-methylaniline) using the directing effect of –NH₂ and a diazonium step. (3 marks)

   Hint

   Let –NH₂ (protected) put Br ortho to itself, then remove the nitrogen group.

   Answer

   Acetylate p-toluidine to reduce activation, then brominate: Br goes ortho to the –NHCOCH₃ group, i.e. meta to CH₃ (1). Hydrolyse to the amine and diazotise with NaNO₂/HCl at 273–278 K (1). Replace –N₂⁺ by H with H₃PO₂ (or ethanol), leaving Br meta to CH₃: m-bromotoluene (1).

10. Q10 · 2 marks · Very short answerHOTS / competency

    Explain why the yield of a primary amine by ammonolysis of an alkyl halide is low, and how it can be improved. (2 marks)

    Hint

    The product is also a nucleophile.

    Answer

    The primary amine formed is itself nucleophilic and reacts further with the alkyl halide, giving 2° and 3° amines and a quaternary salt (1). Using a large excess of ammonia makes the reaction with NH₃ more likely, favouring the primary amine (1).

11. Q11 · 2 marks · Very short answerBoard level

    Write the reactions of benzenediazonium chloride with HBF₄ followed by heating, and with CuBr/HBr. Name the second reaction. (2 marks)

    Hint

    Fluoroborate; Cu(I) salt.

    Answer

    C₆H₅N₂Cl + HBF₄ → C₆H₅N₂⁺BF₄⁻ + HCl; heat → C₆H₅F + BF₃ + N₂ (1). C₆H₅N₂Cl + CuBr/HBr → C₆H₅Br + N₂: Sandmeyer reaction (1).

12. Q12 · 3 marks · Short answerHOTS / competency

    A compound X (C₃H₉N) reacts with benzenesulphonyl chloride to give a solid insoluble in alkali. Identify X and give its IUPAC name. Which isomer of C₃H₉N would not react at all? (3 marks)

    Hint

    Insoluble product = secondary amine.

    Answer

    An alkali-insoluble sulphonamide means a secondary amine (1). The only secondary amine C₃H₉N is CH₃NHCH₂CH₃: N-methylethanamine (1). The tertiary isomer, (CH₃)₃N (N,N-dimethylmethanamine), does not react (1).

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. 1.What is the shape of amines and the C–N–C angle in trimethylamine?Show answer

   Pyramidal; 108°.

2. 2.Give the IUPAC name of (CH₃)₂NCH₂CH₃.Show answer

   N,N-Dimethylethanamine.

3. 3.Which method gives pure primary amines from alkyl halides?Show answer

   Gabriel phthalimide synthesis.

4. 4.What is the pKb of aniline, and of ammonia?Show answer

   9.38 and 4.75.

5. 5.Give the order of basicity of ethylamines in water.Show answer

   (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.

6. 6.Which amines give the carbylamine test?Show answer

   Primary amines, aliphatic and aromatic.

7. 7.What is Hinsberg's reagent?Show answer

   Benzenesulphonyl chloride, C₆H₅SO₂Cl.

8. 8.At what temperature is aniline diazotised?Show answer

   273–278 K.

9. 9.Name the product of coupling of benzenediazonium chloride with aniline and its colour.Show answer

   p-Aminoazobenzene, a yellow dye.

10. 10.Why is sulphanilic acid the major product of sulphonation of aniline?Show answer

    Anilinium hydrogensulphate formed first rearranges on heating at 453–473 K to p-aminobenzenesulphonic acid (sulphanilic acid).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name primary, secondary and tertiary amines in IUPAC?
• ☐Can you write Gabriel synthesis and Hoffmann bromamide equations?
• ☐Can you explain why aliphatic amines are stronger and aniline weaker bases than ammonia?
• ☐Can you write the aqueous basicity orders for methyl and ethyl amines?
• ☐Can you describe the carbylamine and Hinsberg tests with observations?
• ☐Can you explain why acetylation is used before bromination or nitration of aniline?
• ☐Can you convert benzenediazonium chloride into chloro-, bromo-, iodo-, fluoro-, cyano-, hydroxy- and unsubstituted benzene?
• ☐Can you write coupling reactions with phenol and aniline and name the colours?

Amines: frequently asked questions