NCERT Solutions for Class 12 Chemistry Chapter 9: Amines
These Class 12 Chemistry Chapter 9 solutions cover Amines from the NCERT textbook (session 2026–27). You get every end-of-chapter Exercise (9.1–9.14) and every Intext Question (9.1–9.9) reproduced word-for-word and solved in full — IUPAC naming, basic-strength order, named reactions, distinguishing tests and complete reaction sequences for diazonium chemistry, all in exam-ready CBSE style.
Chapter 9, Amines, deals with the nitrogen-containing organic compounds formed by replacing one, two or three hydrogen atoms of ammonia with alkyl and/or aryl groups, giving primary (1°), secondary (2°) and tertiary (3°) amines. The nitrogen is sp3 hybridised with a lone pair, so amines are pyramidal and act as Lewis bases and nucleophiles. The chapter covers nomenclature (common and IUPAC), six methods of preparation (reduction of nitro compounds, nitriles and amides, ammonolysis, Gabriel phthalimide synthesis and Hoffmann bromamide degradation), physical properties, and the chemical reactions that explain basic strength, acylation, the carbylamine test, reaction with nitrous acid and Hinsberg’s reagent, and electrophilic substitution in aniline. The second half introduces diazonium salts — their preparation by diazotisation and their use (Sandmeyer, Gattermann, coupling reactions) to introduce –F, –Cl, –Br, –I, –CN, –OH and –NO2 groups into the aromatic ring and to make azo dyes.
Key Concepts & Definitions
Classification: 1° (RNH2/ArNH2), 2° (R2NH/RNHR′), 3° (R3N). “Simple” if all groups are the same, “mixed” if different.
Basicity: amines are bases because the lone pair on N accepts a proton. Larger Kb (smaller pKb) means a stronger base. Aliphatic amines are stronger than NH3 (+I effect); aromatic amines (aniline) are weaker than NH3 because the lone pair is delocalised into the ring.
Ascent of series: reduction of a nitrile (R–CN → R–CH2–NH2) adds one carbon; Hoffmann bromamide and decarboxylation routes remove carbon.
Tests: carbylamine test & reaction with HNO2 identify 1° amines; Hinsberg’s reagent (C6H5SO2Cl) distinguishes 1°, 2° and 3° amines.
Diazotisation: primary aromatic amine + NaNO2/HCl at 273–278 K gives an arenediazonium salt — a versatile aromatic intermediate.
Basicity in water (methyl): (CH3)2NH > CH3NH2 > (CH3)3N > NH3
Basicity in water (ethyl): (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
Boiling point (isomeric amines): Primary > Secondary > Tertiary (H-bonding falls as N–H atoms decrease)
pKb: ammonia 4.75; methanamine 3.38; aniline (benzenamine) 9.38 — aniline is a much weaker base.
NCERT Intext Questions (9.1–9.9) – Solutions
Questions reproduced verbatim from the NCERT textbook. Note: Intext Q 9.1, the first part of Q 9.2 and Q 9.3 in the book are based on structures shown in figures; these are reproduced in words below and solved fully.
9.1 Classify the following amines as primary, secondary or tertiary:
(The amines shown in the textbook figure are: (i) a primary 3°-carbon amine with the structure (CH3)2CH–CH2–NH2; (ii) a benzene ring bearing one –CH3 and one –NH2 (a toluidine); (iii) the cyclic tertiary amine N-methyl substituted on a ring; (iv) (C2H5)2NH.)
ANSWERAn amine is classified by the number of carbon atoms directly bonded to nitrogen (1° = one, 2° = two, 3° = three), not by the nature of the carbon chain.(i) (CH3)2CH–CH2–NH2 — the N is attached to only one carbon ⇒ primary (1°) amine.(ii) Methyl-substituted aniline (toluidine), Ar–NH2 — N attached to one (ring) carbon ⇒ primary (1°) amine.(iii) N-methyl amine on the ring, >N–CH3 with N bonded to three carbons ⇒ tertiary (3°) amine.(iv) (C2H5)2NH — N attached to two carbons ⇒ secondary (2°) amine.
9.2 (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N.
(ii) Write IUPAC names of all the isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
ANSWER(i) & (ii) Eight isomers of C4H11N:Primary (4): CH3CH2CH2CH2NH2 (butan-1-amine); CH3CH2CH(NH2)CH3 (butan-2-amine); (CH3)2CHCH2NH2 (2-methylpropan-1-amine); (CH3)3CNH2 (2-methylpropan-2-amine).Secondary (3): CH3CH2CH2NHCH3 (N-methylpropan-1-amine); (CH3)2CHNHCH3 (N-methylpropan-2-amine); CH3CH2NHCH2CH3 (N-ethylethanamine).Tertiary (1): (CH3)2N–CH2CH3 (N,N-dimethylethanamine).(iii) Pairs differing in carbon skeleton (e.g. butan-1-amine and 2-methylpropan-1-amine) show chain isomerism; pairs differing in the position of –NH2 (e.g. butan-1-amine and butan-2-amine) show position isomerism; and a primary, secondary or tertiary amine of the same formula (e.g. butan-1-amine, N-methylpropan-1-amine, N,N-dimethylethanamine) are related by metamerism (different alkyl groups around the N atom).
9.3 How will you convert (i) Benzene into aniline (ii) Benzene into N,N-dimethylaniline (iii) Cl–(CH2)4–Cl into hexan-1,6-diamine?
ANSWER(i) Benzene → aniline: nitrate benzene (conc. HNO3 + conc. H2SO4) to nitrobenzene, then reduce: C6H6 ⟶ (HNO3/H2SO4) C6H5NO2 ⟶ (Sn/HCl or Fe/HCl, then OH–) C6H5NH2.(ii) Benzene → N,N-dimethylaniline: first make aniline as above, then exhaustively methylate: C6H5NH2 + 2CH3I → C6H5N(CH3)2 + 2HI (in presence of a base to remove HI).(iii) Cl(CH2)4Cl → hexane-1,6-diamine: first convert both –Cl to –CN with KCN to lengthen the chain by two carbons, then reduce both nitrile groups with LiAlH4 (or H2/Ni): Cl(CH2)4Cl ⟶ (2KCN) NC–(CH2)4–CN ⟶ (LiAlH4) H2N–CH2(CH2)4CH2–NH2 (hexane-1,6-diamine).
9.4 Arrange the following in increasing order of their basic strength:
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH
ANSWERAromatic amines are the weakest (lone pair delocalised into the ring); aliphatic amines are stronger because of the +I effect, modified in water by solvation and steric effects.(i) C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH.(ii) C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH.(iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH.
9.5 Complete the following acid-base reactions and name the products:
9.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
ANSWERAniline is exhaustively methylated. Successive alkylation gives N-methylaniline, then N,N-dimethylaniline; with excess CH3I the final product is the quaternary salt N,N,N-trimethylanilinium iodide.C6H5NH2 ⟶ (CH3I) C6H5NHCH3 ⟶ (CH3I) C6H5N(CH3)2 ⟶ (excess CH3I) C6H5N+(CH3)3I–. The Na2CO3 neutralises the HI released at each step and keeps the medium mildly basic.
9.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
ANSWERAniline undergoes benzoylation (acylation) of its –NH2 group:C6H5NH2 + C6H5COCl ⟶ (pyridine) C6H5NH–CO–C6H5 + HCl.The product is N-phenylbenzamide (benzanilide).
9.8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
ANSWERFour isomers of C3H9N:(a) CH3CH2CH2NH2 — propan-1-amine (1°);(b) (CH3)2CHNH2 — propan-2-amine (1°);(c) CH3CH2NHCH3 — N-methylethanamine (2°);(d) (CH3)3N — N,N-dimethylmethanamine (3°).Only primary amines liberate N2 gas with nitrous acid (aliphatic 1° amines form unstable diazonium salts that decompose, releasing N2). Hence the answers are propan-1-amine and propan-2-amine.
9.9 Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene.
ANSWER(i) 3-Methylaniline → 3-nitrotoluene: diazotise the amine, then replace the diazonium group by –NO2 using sodium nitrite/copper. 3-CH3C6H4NH2 ⟶ (NaNO2/HCl, 273–278 K) 3-CH3C6H4N2+Cl– ⟶ (HBF4; then NaNO2/Cu, Δ) 3-CH3C6H4NO2 (3-nitrotoluene).(ii) Aniline → 1,3,5-tribromobenzene: first brominate (the strongly activating –NH2 gives 2,4,6-tribromoaniline), then remove the amino group by diazotisation and reduction (hypophosphorous acid). C6H5NH2 ⟶ (Br2/H2O) 2,4,6-Br3C6H2NH2 ⟶ (NaNO2/HCl, 273–278 K) 2,4,6-Br3C6H2N2+Cl– ⟶ (H3PO2/H2O) 1,3,5-tribromobenzene.
NCERT Exercises (9.1–9.14) – Solutions
All questions reproduced verbatim from the NCERT textbook (Exercises). Answers are original and exam-ready; IUPAC names cross-checked with the official NCERT answer key.
9.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
9.2 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.
ANSWER(i) Methylamine vs dimethylamine — carbylamine test: heated with CHCl3 + alc. KOH, the 1° amine methylamine gives a foul-smelling isocyanide (CH3NC); dimethylamine (2°) gives no such smell.(ii) Secondary vs tertiary — Hinsberg’s test: with C6H5SO2Cl, a 2° amine forms an N,N-disubstituted sulphonamide (insoluble in alkali); a 3° amine does not react at all.(iii) Ethylamine vs aniline — azo dye test: diazotise each and couple with alkaline 2-naphthol; only aniline (aromatic 1° amine) gives a bright orange-red azo dye. Aniline also gives a white precipitate with bromine water; ethylamine does not.(iv) Aniline vs benzylamine — azo dye / nitrous acid test: with HNO2 at 273–278 K aniline forms a stable diazonium salt that couples to give an orange dye; benzylamine (aliphatic 1°) gives an unstable salt and brisk N2 evolution with an alcohol, no coupling.(v) Aniline vs N-methylaniline — carbylamine test: aniline (1°) gives a foul-smelling isocyanide with CHCl3/alc. KOH; N-methylaniline (2°) does not.
9.3 Account for the following:
(i) pKb 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 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.
ANSWER(i) In aniline the lone pair on nitrogen is delocalised into the benzene ring (resonance), so it is less available for protonation; aniline is therefore a weaker base than methylamine and has a higher pKb. In methylamine the +I effect of –CH3 increases electron density on N, making it a stronger base (lower pKb).(ii) Ethylamine forms hydrogen bonds with water through its small –NH2 group and small alkyl part, so it is soluble. In aniline the large hydrophobic benzene ring dominates and hinders H-bonding with water, making it almost insoluble.(iii) Methylamine is a stronger base than NH3; in water it produces OH– ions (CH3NH2 + H2O → CH3N+H3 + OH–). These OH– ions react with Fe3+ from FeCl3 to precipitate hydrated ferric oxide, Fe2O3·xH2O (brown).(iv) Nitration is carried out in strongly acidic medium where aniline is largely protonated to the anilinium ion. The –N+H3 group is meta-directing (deactivating), so a substantial amount of m-nitroaniline is formed along with the o- and p-products from unprotonated aniline.(v) Friedel-Crafts uses AlCl3 (a Lewis acid). The basic nitrogen of aniline forms a salt with AlCl3, putting a positive charge on N. This makes the ring strongly deactivated, so aniline does not undergo Friedel-Crafts alkylation or acylation.(vi) Aryldiazonium ions are stabilised by resonance/dispersal of the positive charge into the benzene ring, so they are stable for a short while at 273–278 K. Alkyldiazonium ions have no such stabilisation and decompose immediately, releasing N2.(vii) Gabriel phthalimide synthesis gives pure primary amines only, without contamination by 2° or 3° amines (unlike ammonolysis, which gives a mixture). Hence it is preferred for making aliphatic primary amines.
9.4 Arrange the following:
(i) In decreasing order of the pKb 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: (a) Aniline, p-nitroaniline and p-toluidine (b) C6H5NH2, C6H5NHCH3, 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, C2H5NH2
(vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.
ANSWER(i) A stronger base has a lower pKb, so decreasing pKb = increasing basicity reversed: C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH.(ii) Increasing basic strength: C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH.(iii)(a) Electron-withdrawing –NO2 decreases, electron-releasing –CH3 increases basicity: p-nitroaniline < aniline < p-toluidine.(iii)(b)C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2 (in benzylamine the –NH2 is not attached to the ring, so the lone pair is fully available).(iv) In the gas phase only the +I effect matters, so basicity rises with the number of alkyl groups: (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3.(v) O–H bonds H-bond more strongly than N–H, and more N–H means more association: (CH3)2NH < C2H5NH2 < C2H5OH.(vi) Solubility falls with larger hydrophobic part and is lowest for the aromatic amine: C6H5NH2 < (C2H5)2NH < C2H5NH2.
9.5 How will you convert:
(i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to 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?
9.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
ANSWERHinsberg’s test identifies all three types. The amine is shaken with benzenesulphonyl chloride, C6H5SO2Cl (Hinsberg’s reagent), and the mixture is then treated with aqueous KOH.1° amine: forms an N-substituted sulphonamide whose N–H is acidic, so it dissolves in alkali. C6H5SO2Cl + H2N–R → C6H5SO2NH–R + HCl — soluble in KOH (forms C6H5SO2N–R).2° amine: forms an N,N-disubstituted sulphonamide with no N–H, so it is insoluble in alkali. C6H5SO2Cl + HN(R)(R′) → C6H5SO2N(R)(R′) + HCl — insoluble in KOH.3° amine: has no N–H and does not react with the reagent — it remains as an insoluble oily liquid that dissolves in acid. Thus: soluble product = 1°, insoluble solid product = 2°, no reaction = 3°.
9.7 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.
ANSWER(i) Carbylamine reaction: primary amines (aliphatic and aromatic) heated with CHCl3 and alcoholic KOH form foul-smelling isocyanides (carbylamines). R–NH2 + CHCl3 + 3KOH → R–NC + 3KCl + 3H2O. It is a test for 1° amines; 2° and 3° amines do not respond.(ii) Diazotisation: conversion of a primary aromatic amine into a diazonium salt with NaNO2/HCl at 273–278 K. C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl– + NaCl + 2H2O.(iii) Hofmann’s bromamide reaction: an amide treated with Br2 and aqueous/ethanolic NaOH gives a primary amine with one carbon less than the amide. R–CONH2 + Br2 + 4NaOH → R–NH2 + Na2CO3 + 2NaBr + 2H2O.(iv) Coupling reaction: a diazonium salt reacts with phenol or aniline at the para position to give a coloured azo compound (azo dye). C6H5N2+Cl– + C6H5OH → p-HO–C6H4–N=N–C6H5 (p-hydroxyazobenzene) + HCl.(v) Ammonolysis: cleavage of the C–X bond of an alkyl/benzyl halide by ammonia. R–X + NH3 → R–NH2 + HX; with excess NH3 the 1° amine is the major product (otherwise a mixture of 1°, 2°, 3° amines and a quaternary salt forms).(vi) Acetylation: replacement of an N–H of a 1° or 2° amine by an acetyl group using acetic anhydride or acetyl chloride (in pyridine), giving an amide. C6H5NH2 + (CH3CO)2O → C6H5NHCOCH3 (acetanilide) + CH3COOH.(vii) Gabriel phthalimide synthesis: phthalimide ⟶ (KOH) potassium phthalimide ⟶ (R–X) N-alkylphthalimide ⟶ (alkaline hydrolysis) pure primary amine R–NH2. Aromatic 1° amines cannot be made this way (aryl halides do not undergo the SN with phthalimide anion).
9.8 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.
9.9 Give the structures of A, B and C in the following reactions:
(i) CH3CH2I ⟶ (NaCN) A ⟶ (OH–, partial hydrolysis) B ⟶ (NaOH + Br2) C
(ii) C6H5N2Cl ⟶ (CuCN) A ⟶ (H2O/H+) B ⟶ (NH3, Δ) C
(iii) CH3CH2Br ⟶ (KCN) A ⟶ (LiAlH4) B ⟶ (HNO2, 0°C) C
(iv) C6H5NO2 ⟶ (Fe/HCl) A ⟶ (NaNO2 + HCl, 273 K) B ⟶ (H2O/H+) C
(v) CH3COOH ⟶ (NH3) A ⟶ (NaOBr) B ⟶ (NaNO2/HCl) C
(vi) C6H5NO2 ⟶ (Fe/HCl) A ⟶ (HNO2, 273 K) B ⟶ (C2H5OH) C
ANSWER(i) A = CH3CH2CN (propanenitrile); B = CH3CH2CONH2 (propanamide, from partial hydrolysis); C = CH3CH2NH2 (ethanamine, by Hoffmann bromamide degradation, one C less).(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, from the aliphatic 1° amine + HNO2, with N2 evolved).(iv) A = C6H5NH2 (aniline); B = C6H5N2+Cl– (benzenediazonium chloride); C = C6H5OH (phenol, by warming with H2O).(v) A = CH3CONH2 (acetamide); B = CH3NH2 (methanamine, by Hoffmann bromamide with NaOBr); C = CH3OH (methanol, from the aliphatic 1° amine + HNO2).(vi) A = C6H5NH2 (aniline); B = C6H5N2+Cl– (benzenediazonium chloride); C = C6H6 (benzene — ethanol reduces the diazonium salt, replacing –N2+ by –H).
9.10 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.
ANSWERC (C6H7N) is an aromatic primary amine = aniline, C6H5NH2 (benzenamine), formed by Hoffmann bromamide degradation (one C less), so B has 7 carbons.B = benzamide, C6H5CONH2 (formed from A + aq. NH3 on heating).A = benzoic acid, C6H5COOH (an acid + ammonia → ammonium salt → amide on heating).Sequence: C6H5COOH (A) ⟶ (NH3, Δ) C6H5CONH2 (B) ⟶ (Br2/KOH) C6H5NH2 (C).
9.12 Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
ANSWERGabriel synthesis needs the phthalimide anion (a nucleophile) to attack an alkyl halide by nucleophilic substitution. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion because the C–X bond in aryl halides has partial double-bond character (resonance with the ring) and the ring repels the incoming nucleophile. Hence aromatic primary amines such as aniline cannot be prepared by this method.
9.13 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
ANSWER(i) Aromatic 1° amine: reacts with HNO2 (NaNO2/HCl) at low temperature (273–278 K) to give a relatively stable diazonium salt. C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl– + NaCl + 2H2O.(ii) Aliphatic 1° amine: forms a very unstable diazonium salt that decomposes at once, liberating N2 gas quantitatively and giving an alcohol. R–NH2 + HNO2 → R–OH + N2↑ + H2O.
9.14 Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
ANSWER(i) Acidity depends on how easily the N–H or O–H bond releases H+ and how stable the resulting anion is. Nitrogen is less electronegative than oxygen, so the N–H bond is less polar and the amide ion (RNH–) is less stable than the alkoxide ion (RO–). Hence amines part with H+ less readily and are less acidic than alcohols of similar mass.(ii) Primary amines have two N–H bonds and so undergo extensive intermolecular hydrogen bonding. Tertiary amines have no N–H bond, so they cannot form intermolecular hydrogen bonds. More H-bonding means more energy is needed to separate the molecules, so primary amines boil at a higher temperature than tertiary amines.(iii) In aliphatic amines the alkyl group releases electrons (+I effect), increasing the electron density on nitrogen and making the lone pair readily available for protonation. In aromatic amines the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is much less available to accept a proton. Therefore aliphatic amines are stronger bases than aromatic amines.
Extra Practice Questions
Short Answer Type Questions
Q1. Why is aniline stored in dark/coloured bottles?
ANSWERAniline is colourless when pure but is readily oxidised by atmospheric oxygen on storage, turning brown. Dark bottles slow this oxidation and keep it colourless for longer.
Q2. Arrange in increasing order of pKb: aniline, p-toluidine, p-nitroaniline.
ANSWERA stronger base has a smaller pKb. Basicity order p-nitroaniline < aniline < p-toluidine, so pKb: p-toluidine < aniline < p-nitroaniline.
Q3. Why is the diazonium salt prepared below 5°C (273–278 K)?
ANSWERArenediazonium salts are unstable and decompose at higher temperatures (forming phenol and N2 when warmed). Low temperature keeps the salt stable long enough to be used in further reactions.
Q4. Why does methylamine react with water to form a basic solution?
ANSWERMethylamine accepts a proton from water (CH3NH2 + H2O ⇌ CH3N+H3 + OH–), releasing OH– ions that make the solution basic.
Q5. Name the reagent used to distinguish 1°, 2° and 3° amines and the modern substitute for it.
ANSWERBenzenesulphonyl chloride, C6H5SO2Cl (Hinsberg’s reagent). These days it is often replaced by p-toluenesulphonyl chloride.
Long Answer Type Questions
Q1. Explain why the order of basic strength of methyl-substituted amines in aqueous solution is (CH3)2NH > CH3NH2 > (CH3)3N > NH3.
ANSWERIn the gas phase basicity follows the +I effect (3° > 2° > 1° > NH3). In water, three factors act together: the +I effect (which raises electron density on N), solvation/stabilisation of the cation by H-bonding (a cation with more N–H bonds is better solvated), and steric hindrance of bulky groups. For methyl amines these effects balance so that the dimethyl (2°) amine is most basic; trimethylamine, with no N–H for solvation and crowded methyls, drops below the 1° amine, giving the observed order (CH3)2NH > CH3NH2 > (CH3)3N > NH3.
Q2. Describe the preparation of aniline from benzene and list two reasons aniline is a weaker base than ammonia.
ANSWERPreparation: benzene is nitrated with conc. HNO3 + conc. H2SO4 to nitrobenzene, which is then reduced with Sn/HCl (or Fe/HCl) and the salt freed with NaOH to give aniline. Reasons it is weaker: (1) the nitrogen lone pair is delocalised into the benzene ring (aniline is a resonance hybrid of five structures), so it is much less available for protonation; (2) the anilinium ion formed on protonation has only two resonance structures and is less stabilised relative to aniline, so proton acceptance is unfavourable. Hence aniline is a weaker base than ammonia (pKb 9.38 vs 4.75).
Q3. Discuss the synthetic importance of diazonium salts in preparing substituted aromatic compounds.
ANSWERArenediazonium salts are versatile intermediates because the –N2+ group is an excellent leaving group that can be replaced by many others: by Cl/Br/CN with Cu(I) (Sandmeyer) or Cu powder + HX (Gattermann); by I with KI; by F via the fluoroborate (Balz-Schiemann); by –OH on warming with water (phenol); by –H with H3PO2 or ethanol; and by –NO2 via the fluoroborate with NaNO2/Cu. They also couple with phenols and amines (retaining the diazo group) to give azo dyes. Many of these substituents — aryl –F, –I and –CN — cannot be introduced by direct substitution on benzene, so diazonium chemistry is the route of choice for such compounds.
MCQs & Assertion–Reason
1. The IUPAC name of (CH3)2CHNH2 is:
(a) propan-1-amine (b) propan-2-amine (c) N-methylethanamine (d) 2-aminopropane only
2. Which is the strongest base in aqueous solution?
(a) NH3 (b) C6H5NH2 (c) (CH3)2NH (d) (CH3)3N
3. The carbylamine reaction is given by:
(a) tertiary amines only (b) secondary amines only (c) primary amines only (d) all amines
4. Aniline on reaction with NaNO2/HCl at 273–278 K gives:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: Aniline is a weaker base than methylamine.
Reason: In aniline the nitrogen lone pair is delocalised into the benzene ring, lowering its availability for protonation.
A-R 2. Assertion: Primary amines have higher boiling points than tertiary amines of comparable mass.
Reason: Tertiary amines have no N–H bond and cannot form intermolecular hydrogen bonds.
A-R 3. Assertion: Aromatic primary amines can be prepared by Gabriel phthalimide synthesis.
Reason: Aryl halides readily undergo nucleophilic substitution with the phthalimide anion.
A-R 4. Assertion: Arenediazonium salts are prepared at low temperature (273–278 K).
Reason: These salts are unstable and decompose on warming, releasing nitrogen.
A-R 5. Assertion: Ammonolysis of an alkyl halide with excess ammonia mainly gives a primary amine.
Reason: Excess ammonia minimises further alkylation of the primary amine formed.
Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).
Common Mistakes & Exam Tips
Common mistakes to avoid
Classifying amines by the carbon (1°/2°/3° carbon) instead of by the number of carbons on nitrogen.
Writing the gas-phase basicity order [(C2H5)3N > (C2H5)2NH > …] when the question asks for the aqueous order, or vice versa.
Forgetting that aliphatic 1° amines + HNO2 give alcohol + N2, not a stable diazonium salt.
Using Gabriel synthesis to make aniline — it works only for aliphatic primary amines.
Omitting the low temperature (273–278 K) condition in diazotisation.
Not protecting –NH2 by acetylation before nitrating/brominating aniline for a mono-substituted product.
How to score full marks in this chapter
Always state reagents and conditions on the arrow (reagent, temperature, catalyst). For basicity questions, mention the deciding factor — +I effect, resonance/delocalisation, solvation or steric hindrance. Quote pKb values (ammonia 4.75, methanamine 3.38, aniline 9.38) to justify orders. For distinguishing tests, name the reagent, the observation, and which amine gives it. In conversions, count carbons carefully and pick the right ascent (nitrile + LiAlH4) or descent (Hoffmann bromamide) route.
Frequently Asked Questions
What is Class 12 Chemistry Chapter 9 Amines about?
Chapter 9, Amines, covers the classification, nomenclature, preparation, physical and chemical properties of amines, their basic strength, and the chemistry of diazonium salts — including diazotisation, Sandmeyer and coupling reactions used to make aryl halides, phenols and azo dyes.
Why is aniline a weaker base than aliphatic amines?
In aniline the lone pair on nitrogen is delocalised into the benzene ring by resonance, so it is less available to accept a proton. In aliphatic amines the +I effect of the alkyl group increases electron density on nitrogen, making them stronger bases.
How are primary, secondary and tertiary amines distinguished?
By Hinsberg’s test with benzenesulphonyl chloride: a 1° amine gives a sulphonamide soluble in alkali, a 2° amine gives one insoluble in alkali, and a 3° amine does not react. The carbylamine test (CHCl3 + alc. KOH) is specific for primary amines.
Are these Class 12 Chemistry Chapter 9 solutions free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026-27, with every Exercise and Intext question solved.
