NCERT Solutions for Class 12 Chemistry Chapter 9: Amines – Free PDF Download

Get accurate, step-by-step NCERT Solutions for Class 12 Chemistry Chapter 9 – Amines, covering nomenclature, basicity trends, diazonium salt chemistry, and all the important name reactions from the CBSE 2026-27 syllabus. All 14 exercise questions are solved in detail below, and the complete chapter is also available as a free PDF download.

Last Updated: September 23, 2026

About NCERT Class 12 Chemistry Chapter 9: Amines

Amines are organic derivatives of ammonia (NH₃) formed by replacing one, two or three hydrogen atoms with alkyl or aryl groups, giving primary (1°), secondary (2°) and tertiary (3°) amines respectively. This chapter builds on the carbonyl and carboxylic-acid chemistry of Chapter 8 and introduces nitrogen-based functional group interconversions that are heavily tested in board exams: basicity comparisons (and why the naive “more alkyl groups = stronger base” rule fails in water), the Hinsberg test for identifying amine class, and the rich chemistry of aromatic diazonium salts — including the Sandmeyer, Gattermann, and Balz–Schiemann reactions, azo-dye coupling, Hofmann’s bromamide degradation, and Gabriel phthalimide synthesis. These reactions form the backbone of many multi-step organic conversion questions.

NCERT Solutions for Class 12 Chemistry Chapter 9: Amines – All Exercises

Q9.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines:
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) (CH₃CH₂)₂NCH₃
(vii) m-BrC₆H₄NH₂
Ans: (i) Propan-2-amine — primary. (ii) Propan-1-amine — primary. (iii) N-Methylpropan-2-amine — secondary. (iv) 2-Methylpropan-2-amine — primary. (v) N-Methylaniline (N-methylbenzenamine) — secondary. (vi) N-Ethyl-N-methylethanamine — tertiary. (vii) 3-Bromoaniline (3-bromobenzenamine) — primary.

Q9.2 Give one chemical test to distinguish between the following pairs of compounds:
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Ans: (i) Carbylamine test (CHCl₃+alc. KOH, heat): methylamine (1°) gives a foul-smelling isocyanide; dimethylamine (2°) gives no reaction. (ii) Hinsberg’s test (benzenesulfonyl chloride, C₆H₅SO₂Cl): the secondary amine forms an N,N-dialkylsulfonamide that is insoluble in KOH (no acidic N–H left); the tertiary amine does not react at all and separates as an unreacted oily layer. (iii) Diazotisation (NaNO₂/HCl, 273–278K): aniline forms a stable diazonium salt (clear solution, gives a coloured dye on coupling with alkaline β-naphthol); ethylamine’s diazonium salt decomposes instantly with brisk evolution of N₂ gas. (iv) Bromine water test: aniline gives an immediate white precipitate of 2,4,6-tribromoaniline (its –NH₂ is directly conjugated with the ring); benzylamine (C₆H₅CH₂NH₂) gives no such precipitate, since its –NH₂ is not attached to the ring. (v) Carbylamine test: aniline (1°) gives a positive test (foul-smelling phenyl isocyanide); N-methylaniline (2°) gives no reaction.

Q9.3 Account for the following:
(i) pK_b of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water, whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although the amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans: (i) In aniline, the nitrogen lone pair is delocalised into the benzene ring by resonance, making it far less available for protonation than methylamine’s fully localised lone pair — a higher pK_b means a weaker base. (ii) Ethylamine forms strong intermolecular hydrogen bonds with water; aniline’s bulky hydrophobic ring dominates its behaviour, and its weakly basic, resonance-delocalised nitrogen forms far fewer effective hydrogen bonds, so it is only sparingly soluble. (iii) Methylamine, though a weak base, generates enough hydroxide ion in water (CH₃NH₂+H₂O ⇄ CH₃NH₃⁺+OH⁻) to precipitate Fe³⁺ as hydrated Fe₂O₃ (ferric hydroxide). (iv) In the strongly acidic nitrating mixture, most aniline is protonated to the anilinium ion (–NH₃⁺), which is meta-directing and deactivating; the m-nitroaniline arises from this protonated form, while only the small residual free-aniline fraction gives the o-/p-products. (v) The Lewis acid catalyst AlCl₃ complexes with aniline’s basic lone pair, converting –NH₂ into a bulky, strongly electron-withdrawing –NH₂→AlCl₃ group that deactivates the ring rather than activating it, so the Friedel-Crafts reaction fails. (vi) The positive charge on an aromatic diazonium ion is delocalised into the ring by resonance, stabilising it enough to isolate at low temperature; aliphatic diazonium ions have no such stabilisation and decompose instantly, even at 273K, releasing N₂. (vii) Gabriel synthesis proceeds through an S_N2 attack of the phthalimide anion on an alkyl halide, which cleanly gives only a primary amine on hydrolysis with no risk of over-alkylation to secondary/tertiary amines (unlike direct ammonolysis of alkyl halides).

Q9.4 Arrange the following:
(i) In decreasing order of the pK_b values: C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength: C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) In increasing order of basic strength: aniline, p-nitroaniline and p-toluidine; also C₆H₅NH₂, C₆H₅NHCH₃ and C₆H₅CH₂NH₂
(iv) In decreasing order of basic strength in gas phase: C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point: C₂H₅OH, (CH₃)₂NH and C₂H₅NH₂
(vi) In increasing order of solubility in water: C₆H₅NH₂, (C₂H₅)₂NH and C₂H₅NH₂
Ans: (i) Decreasing pK_b (weakest base first): C₆H₅NH₂>C₆H₅NHCH₃>C₂H₅NH₂>(C₂H₅)₂NH. (ii) Increasing basicity: C₆H₅NH₂<C₆H₅N(CH₃)₂<CH₃NH₂<(C₂H₅)₂NH. (iii) p-Nitroaniline<aniline<p-toluidine (the –NO₂ group withdraws electron density and weakens basicity; –CH₃ donates and strengthens it); and C₆H₅NH₂<C₆H₅NHCH₃<C₆H₅CH₂NH₂ (benzylamine’s nitrogen is not conjugated with the ring, so it behaves like a simple aliphatic amine and is the most basic). (iv) In the gas phase, with no solvation effects, basicity rises steadily with alkyl substitution: (C₂H₅)₃N>(C₂H₅)₂NH>C₂H₅NH₂>NH₃. (v) Increasing boiling point: (CH₃)₂NH<C₂H₅NH₂<C₂H₅OH (alcohols hydrogen-bond more strongly than amines; a secondary amine has one fewer N–H than a primary amine of similar mass, so it boils lowest). (vi) Increasing solubility: C₆H₅NH₂<(C₂H₅)₂NH<C₂H₅NH₂.

Q9.5 How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol into ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid
Ans: (i) CH₃COOH→(NH₃, Δ)→CH₃CONH₂→(Br₂/KOH, Hofmann degradation)→CH₃NH₂. (ii) CH₃(CH₂)₄CN→(H₂O/H⁺, partial hydrolysis)→CH₃(CH₂)₄CONH₂→(Br₂/KOH)→CH₃(CH₂)₃CH₂NH₂ (one carbon is lost in the Hofmann step, giving the pentylamine). (iii) CH₃OH→(PBr₃)→CH₃Br→(KCN)→CH₃CN→(H₂O/H⁺, hydrolysis)→CH₃COOH. (iv) CH₃CH₂NH₂→(HNO₂)→CH₃CH₂OH→(oxidation, alk. KMnO₄)→CH₃COOH→(NH₃, Δ)→CH₃CONH₂→(Br₂/KOH)→CH₃NH₂. (v) CH₃COOH→(LiAlH₄)→CH₃CH₂OH→(PBr₃)→CH₃CH₂Br→(KCN)→CH₃CH₂CN→(H₂O/H⁺)→CH₃CH₂COOH. (vi) CH₃NH₂→(HNO₂)→CH₃OH→(PBr₃)→CH₃Br→(KCN)→CH₃CN→(LiAlH₄)→CH₃CH₂NH₂. (vii) CH₃NO₂→(Fe/HCl or Sn/HCl, reduction)→CH₃NH₂→(CHCl₃/alc. KOH, carbylamine)→CH₃NC→(H₂/Ni, reduction)→CH₃NHCH₃. (viii) CH₃CH₂COOH→(NH₃, Δ)→CH₃CH₂CONH₂→(Br₂/KOH)→CH₃CH₂NH₂→(HNO₂)→CH₃CH₂OH→(oxidation)→CH₃COOH.

Q9.6 Describe a method for the identification of primary, secondary and tertiary amines.
Ans: Hinsberg’s test uses benzenesulfonyl chloride (C₆H₅SO₂Cl) with KOH. A primary amine forms an N-alkylbenzenesulfonamide (C₆H₅SO₂NHR) that still has one acidic N–H (made acidic by the strongly electron-withdrawing –SO₂– group), so it dissolves in excess KOH as its potassium salt. A secondary amine forms a fully N,N-disubstituted sulfonamide (C₆H₅SO₂NR₂) with no acidic hydrogen left, so it remains insoluble in KOH. A tertiary amine has no N–H to begin with and does not react with benzenesulfonyl chloride at all, separating out unreacted as an oily layer.

Q9.7 Write short notes on the following:
(i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis
Ans: (i) Carbylamine reaction: a primary amine (aliphatic or aromatic) heated with CHCl₃ and alcoholic KOH forms a foul-smelling isocyanide (carbylamine), RNC — a test specific to primary amines only. (ii) Diazotisation: an aromatic primary amine treated with NaNO₂+HCl at 273–278K forms a diazonium salt, ArN₂⁺Cl⁻. (iii) Hofmann’s bromamide reaction: an amide treated with Br₂ and NaOH/KOH degrades to a primary amine with one carbon fewer than the starting amide, via an isocyanate intermediate. (iv) Coupling reaction: a diazonium salt reacts with a phenol (mildly alkaline medium) or an aromatic amine (mildly acidic medium) to give a brightly coloured azo compound (Ar–N=N–Ar’), the basis of azo dyes. (v) Ammonolysis: an alkyl halide reacts with excess ammonia (in a sealed tube) to give an amine, via nucleophilic substitution; the product can further react with more alkyl halide, so ammonolysis often gives a mixture of 1°, 2°, 3° amines and a quaternary salt. (vi) Acetylation: an amine reacts with acetic anhydride or acetyl chloride to form an N-substituted acetamide, converting a reactive amine into a less nucleophilic, protected amide (used, e.g., to moderate aniline’s ring-activating power before nitration/halogenation). (vii) Gabriel phthalimide synthesis: potassium phthalimide reacts with a primary alkyl halide (S_N2), and the resulting N-alkylphthalimide is hydrolysed to give a pure primary amine with no over-alkylation.

Q9.8 Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol
Ans: (i) C₆H₅NO₂→(Fe/HCl)→C₆H₅NH₂→(NaNO₂/HCl, 273K)→C₆H₅N₂Cl→(CuCN, then H₂O/H⁺)→C₆H₅COOH. (ii) C₆H₆→(Br₂/FeBr₃)→C₆H₅Br→(fuming H₂SO₄)→m-BrC₆H₄SO₃H→(fuse with NaOH, then H⁺)→m-bromophenol. (iii) C₆H₅COOH→(NH₃, Δ)→C₆H₅CONH₂→(Br₂/NaOH, Hofmann)→C₆H₅NH₂. (iv) C₆H₅NH₂→(Br₂(aq), excess)→2,4,6-tribromoaniline→(NaNO₂/HCl, 273K)→2,4,6-tribromobenzenediazonium chloride→(HBF₄, then Δ, Balz–Schiemann)→2,4,6-tribromofluorobenzene. (v) C₆H₅CH₂Cl→(KCN)→C₆H₅CH₂CN→(LiAlH₄ or H₂/Ni)→C₆H₅CH₂CH₂NH₂. (vi) C₆H₅Cl→(conc. HNO₃/H₂SO₄)→p-O₂NC₆H₄Cl→(Fe/HCl or Sn/HCl, reduction)→p-ClC₆H₄NH₂. (vii) C₆H₅NH₂→((CH₃CO)₂O, acetylation)→C₆H₅NHCOCH₃→(Br₂/CH₃COOH, then H₃O⁺/OH⁻ hydrolysis)→p-BrC₆H₄NH₂ (acetylation moderates –NH₂‘s activating power so bromination stays mono-substituted and para-selective). (viii) C₆H₅CONH₂→(Br₂/NaOH, Hofmann)→C₆H₅NH₂→(NaNO₂/HCl, 273K)→C₆H₅N₂Cl→(H₃PO₂/H₂O)→C₆H₆→(CH₃Cl/anhyd. AlCl₃, Friedel-Crafts)→C₆H₅CH₃. (ix) C₆H₅NH₂→(NaNO₂/HCl, 273K)→C₆H₅N₂Cl→(CuCN, then H₂O/H⁺)→C₆H₅COOH→(LiAlH₄)→C₆H₅CH₂OH.

Q9.9 Give the structures of A, B and C in the following reactions:
(i) CH₃CH₂I→(NaCN)→A→(partial hydrolysis, OH⁻)→B→(NaOH+Br₂)→C
(ii) C₆H₅N₂Cl→(CuCN)→A→(H₂O/H⁺)→B→(NH₃, Δ)→C
(iii) CH₃CH₂Br→(KCN)→A→(LiAlH₄)→B→(HNO₂, 273K)→C
(iv) C₆H₅NO₂→(Fe/HCl)→A→(NaNO₂+HCl, 273K)→B→(H₂O/H⁺, Δ)→C
(v) CH₃COOH→(NH₃, Δ)→A→(NaOBr)→B→(NaNO₂/HCl)→C
(vi) C₆H₅NO₂→(Fe/HCl)→A→(HNO₂, 273K)→B→(C₆H₅OH)→C
Ans: (i) A=CH₃CH₂CN (propanenitrile), B=CH₃CH₂CONH₂ (propanamide), C=CH₃CH₂NH₂ (ethanamine, via Hofmann degradation). (ii) A=C₆H₅CN (benzonitrile), B=C₆H₅COOH (benzoic acid), C=C₆H₅CONH₂ (benzamide). (iii) A=CH₃CH₂CN (propanenitrile), B=CH₃CH₂CH₂NH₂ (propan-1-amine), C=CH₃CH₂CH₂OH (propan-1-ol, via deamination). (iv) A=C₆H₅NH₂ (aniline), B=C₆H₅N₂Cl (benzenediazonium chloride), C=C₆H₅OH (phenol). (v) A=CH₃CONH₂ (acetamide), B=CH₃NH₂ (methanamine, via Hofmann degradation with NaOBr), C=CH₃OH (methanol, via deamination). (vi) A=C₆H₅NH₂ (aniline), B=C₆H₅N₂Cl (benzenediazonium chloride), C=p-HOC₆H₄N=NC₆H₅ (p-hydroxyazobenzene, via coupling with phenol).

Q9.10 An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’, which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Ans: A = C₆H₅COOH (benzoic acid) → B = C₆H₅CONH₂ (benzamide) → C = C₆H₅NH₂ (aniline, IUPAC name benzenamine). The Hofmann bromamide degradation of benzamide (C₇H₇NO) removes the carbonyl carbon, giving aniline (C₆H₇N) — matching the molecular formula given.

Q9.11 Complete the following reactions:
(i) C₆H₅NH₂+CHCl₃+alc. KOH→
(ii) C₆H₅N₂Cl+H₃PO₂+H₂O→
(iii) C₆H₅NH₂+H₂SO₄(conc.)→
(iv) C₆H₅N₂Cl+C₂H₅OH→
(v) C₆H₅NH₂+Br₂(aq)→
(vi) C₆H₅NH₂+(CH₃CO)₂O→
(vii) C₆H₅N₂Cl+HBF₄, then Δ; and C₆H₅N₂Cl+NaNO₂/Cu, Δ→
Ans: (i) C₆H₅NC (phenyl isocyanide)+3KCl+3H₂O (carbylamine reaction). (ii) C₆H₆ (benzene)+N₂+H₃PO₃+HCl (reduction of the diazonium group). (iii) C₆H₅NH₃⁺HSO₄⁻ (anilinium hydrogen sulphate — simple acid–base salt formation). (iv) C₆H₆ (benzene)+N₂+CH₃CHO+HCl (reduction of the diazonium group by ethanol). (v) 2,4,6-tribromoaniline (white precipitate)+3HBr. (vi) C₆H₅NHCOCH₃ (acetanilide)+CH₃COOH. (vii) With HBF₄ then heat: C₆H₅F (fluorobenzene)+BF₃+N₂ (Balz–Schiemann reaction). With NaNO₂/Cu and heat: C₆H₅NO₂ (nitrobenzene)+N₂ (a Cu-catalysed replacement of the diazonium group by –NO₂, alongside the more commonly tested halogen/cyanide Sandmeyer substitutions).

Q9.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans: Gabriel synthesis requires an S_N2 attack of the phthalimide anion on an alkyl halide. An aryl halide cannot undergo this step: its C–X bond has partial double-bond character from resonance with the ring, and the sp² carbon is geometrically shielded against backside nucleophilic attack. Since aryl halides are unreactive towards nucleophilic substitution under these conditions, the phthalimide anion cannot be alkylated with them, so aromatic primary amines cannot be made this way.

Q9.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Ans: (i) Aromatic primary amines (e.g. aniline) react with HNO₂ (from NaNO₂+HCl) at 273–278K to form a reasonably stable diazonium salt: C₆H₅NH₂+HNO₂+HCl→C₆H₅N₂Cl+2H₂O, which can be isolated and used in a wide range of substitution and coupling reactions. (ii) Aliphatic primary amines react with HNO₂ to form an unstable diazonium salt that decomposes immediately, even in the cold, evolving nitrogen gas and forming a carbocation that goes on to give a mixture of an alcohol (major), alkene and alkyl halide: RNH₂+HNO₂→ROH+N₂+H₂O. This brisk, quantitative-with-effervescence reaction is itself used as a test to detect primary aliphatic amines.

Q9.14 Give plausible explanations for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular mass?
(ii) Why do primary amines have a higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Ans: (i) Nitrogen (electronegativity ≈3.0) is less electronegative than oxygen (≈3.5), so it holds the N–H bonding electrons less tightly; the resulting amide-ion-like conjugate base (R–NH⁻) is far less stable than an alkoxide ion (R–O⁻), making amines much weaker acids than alcohols. (ii) A primary amine (R–NH₂) has two N–H bonds available for intermolecular hydrogen bonding, so its molecules associate strongly; a tertiary amine (R₃N) has no N–H bonds at all and cannot hydrogen-bond with other amine molecules, so it has a noticeably lower boiling point than an isomeric primary amine. (iii) In aromatic amines, the nitrogen lone pair is delocalised into the ring by resonance, making it far less available to accept a proton; in aliphatic amines the lone pair remains fully localised on nitrogen (and is further pushed towards N by the +I effect of alkyl groups), so aliphatic amines are consistently stronger bases than aromatic ones.

Frequently Asked Questions (FAQs)

Q1. Why are secondary amines more basic than primary and tertiary amines in aqueous solution?
Ans: In water, base strength depends on both the electron-donating (+I) effect of alkyl groups and how well the resulting ammonium ion is solvated by water. A secondary amine has enough alkyl substitution to raise the nitrogen’s electron density while still leaving enough N–H bonds for effective hydrogen-bond solvation of R₂NH₂⁺; a tertiary amine, despite having the most electron-donating groups, is so sterically hindered that its ammonium ion is poorly solvated, which drags its basicity below both primary and secondary amines in water.

Q2. What is the difference between the Sandmeyer reaction and the Gattermann reaction?
Ans: Both replace a diazonium group with –Cl, –Br or –CN, but the Sandmeyer reaction uses a cuprous salt (Cu₂Cl₂, Cu₂Br₂ or CuCN) as the source of the nucleophile, while the Gattermann reaction uses copper powder together with the corresponding halogen acid (HCl or HBr) directly — achieving the same overall transformation but generally in somewhat lower yield.

Written by Satish

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