NCERT Solutions for Class 12 Chemistry Chapter 9: Amines

Concept. An amine is classified by counting the number of carbon-containing groups (alkyl or aryl) attached directly to the nitrogen atom: one group → primary (1°), two groups → secondary (2°), three groups → tertiary (3°). This is different from alcohols and halides, which are classified by the carbon that bears the functional group — for amines it is always the nitrogen we look at. Branching in an alkyl group does not change the class.

Watch the trap in (iii). (C₂H₅)₂CHNH₂ looks as though it has “two ethyls”, but both ethyls are attached to the same carbon, which in turn carries the –NH₂; the nitrogen is bonded to only one carbon group (that branched carbon) and two hydrogens, so the amine is primary (it is pentan-3-amine). Contrast this with (iv) (C₂H₅)₂NH, where the two ethyl groups are attached directly to nitrogen, making it secondary. In (i) the single naphthyl group on nitrogen makes it a primary aromatic amine, and adding two methyls to that nitrogen in (ii) makes it tertiary.

Conclusion: (i) primary; (ii) tertiary; (iii) primary; (iv) secondary.

Concept. C₄H₁₁N is a saturated amine (no rings or double bonds). The isomers are generated by distributing the four carbons among the groups on nitrogen as primary (R–NH₂), secondary (R₂NH or R–NH–R′) and tertiary (R₃N) amines. Counting systematically gives eight isomeric amines: four primary, three secondary and one tertiary.

(iii) Types of isomerism. (a) The chain isomers — e.g. butan-1-amine and 2-methylpropan-1-amine (same primary class, different carbon skeleton) — and position isomers — butan-1-amine and butan-2-amine (–NH₂ on a different carbon). (b) Amines of different classes but the same formula (e.g. a primary butan-1-amine vs the secondary N-methylpropan-1-amine vs the tertiary N,N-dimethylethanamine) are related by metamerism (a kind of structural isomerism arising from different alkyl groups attached to the same polyvalent atom, here nitrogen) — they differ in the distribution of carbon atoms around nitrogen. All eight are structural (constitutional) isomers of one another.

Conclusion: Eight isomers — four 1° (butan-1-amine, butan-2-amine, 2-methylpropan-1-amine, 2-methylpropan-2-amine), three 2° (N-methylpropan-1-amine, N-methylpropan-2-amine, N-ethylethanamine) and one 3° (N,N-dimethylethanamine); they show chain, position and metamerism (all structural isomerism).

Concept. To put an –NH₂ on a benzene ring, we first introduce a nitro group (nitration) and then reduce it; to build a chain diamine, we convert each halogen to a –CN group (adding one carbon per end) and then reduce the nitriles to –CH₂NH₂. Alkylation of the amine then gives the tertiary amine in (ii).

(i) Benzene → aniline.

(ii) Benzene → N,N-dimethylaniline. Make aniline as in (i), then exhaustively methylate the nitrogen with two moles of methyl iodide (a base such as Na₂CO₃ mops up the HI).

(iii) Cl–(CH₂)₄–Cl → hexane-1,6-diamine. The target has six carbons; the starting dichloride has four, so two carbons must be added — one at each end through the nitrile. Treat the dichloride with KCN (both –Cl → –CN, adding a carbon each), then reduce both nitriles with H₂/Ni (or LiAlH₄).

Why these routes. In (i) and (ii) the amino group cannot be put on the ring directly, so nitration + reduction is the standard two-step entry; the Sn/HCl (or Fe/HCl) reduction gives the amine cleanly. In (iii) the key is the carbon count: reducing a nitrile R–CN gives R–CH₂NH₂, i.e. one carbon more, so adding –CN to each end of the 4-carbon dichloride and reducing gives a 6-carbon diamine (hexane-1,6-diamine, a monomer for nylon-6,6).

Conclusion: (i) nitrate benzene then reduce (Sn/HCl); (ii) make aniline, then dimethylate with 2 CH₃I/Na₂CO₃; (iii) Cl(CH₂)₄Cl + 2 KCN → the dinitrile, then H₂/Ni → hexane-1,6-diamine.

Concept. The basic strength of an amine depends on the availability of the lone pair on nitrogen and the stability of the ammonium ion formed. Three factors act together in aqueous solution: (1) the +I (electron-releasing) effect of alkyl groups increases the electron density on N, so aliphatic amines are stronger bases than ammonia; (2) solvation of the protonated amine by water (more N–H bonds → better solvation → more stable cation → stronger base), which favours 1° and 2° over 3°; (3) in aromatic amines, resonance delocalises the nitrogen lone pair into the ring, making it much less available, so aniline is a far weaker base than ammonia. Benzylamine (C₆H₅CH₂NH₂) has the –NH₂ on an sp³ carbon away from the ring, so the lone pair is not delocalised and it behaves like an aliphatic amine (stronger than aniline).

(i) Aniline is weakest (lone pair in the ring); ammonia next; benzylamine (aliphatic-type) stronger; ethylamine stronger still (+I of ethyl); diethylamine strongest (two +I ethyls, good solvation).

(ii) Aniline weakest; among the aliphatic ethylamines the aqueous order (balancing +I, steric and solvation) is 1° < 3° < 2°, i.e. C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH.

(iii) Aniline weakest; benzylamine next (aliphatic-type but a single –CH₂– ); then the methylamines in the aqueous order 3° < 1° < 2°, i.e. (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH.

Why the aliphatic order is irregular in water. On the +I effect alone the order would be 3° > 2° > 1°, but the protonated 3° amine has fewer N–H bonds and is poorly solvated (and its three groups crowd the nitrogen), so in water the secondary amine usually comes out strongest. For methyl amines the result is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N; for ethyl amines (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ (the bigger ethyl groups shift the 3° amine up relative to the 1°).

Conclusion: (i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH; (ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH; (iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH.

Concept. Amines are bases: the lone pair on nitrogen accepts a proton from an acid (HCl) to form a substituted ammonium salt (R–NH₃⁺Cl⁻, etc.). A primary amine gives a primary ammonium chloride; a tertiary amine gives a tertiary (trialkyl) ammonium chloride. The salt is named as the “…ammonium chloride”.

Working. In (i) propan-1-amine accepts a proton on its nitrogen to form the propylammonium ion, isolated as its chloride (propylammonium chloride). In (ii) triethylamine, although tertiary, still has a lone pair on nitrogen and so is a base; it accepts a proton to give the triethylammonium ion (triethylammonium chloride). These salts are water-soluble; treating them with a strong base (NaOH) regenerates the free amine — the basis of separating amines from neutral and acidic compounds.

Conclusion: (i) CH₃CH₂CH₂NH₃⁺Cl⁻ (propylammonium chloride); (ii) (C₂H₅)₃NH⁺Cl⁻ (triethylammonium chloride).

Concept. Aniline’s nitrogen is nucleophilic and attacks methyl iodide (SN2), replacing an N–H by an N–CH₃; with excess methyl iodide this exhaustive methylation continues until the nitrogen has no more hydrogens and then adds one further methyl to become a positively charged quaternary ammonium salt. Sodium carbonate is a mild base that neutralises the HI formed at each step, keeping the amine free to react further.

The reactions. Step 1: C₆H₅NH₂ + CH₃I → C₆H₅NHCH₃ + HI (N-methylaniline). Step 2: C₆H₅NHCH₃ + CH₃I → C₆H₅N(CH₃)₂ + HI (N,N-dimethylaniline). Step 3 (excess CH₃I): C₆H₅N(CH₃)₂ + CH₃I → C₆H₅N⁺(CH₃)₃ I⁻ (N,N,N-trimethylanilinium iodide, a quaternary ammonium salt). The Na₂CO₃ removes HI as NaI, CO₂ and water at each stage. The final product is therefore the quaternary salt, N,N,N-trimethylanilinium iodide (phenyltrimethylammonium iodide).

Extra depth. The quaternary ammonium ion has no lone pair left on nitrogen, so it is no longer basic or nucleophilic; it is a stable ionic salt. Exhaustive methylation to the quaternary salt is the first step of the Hofmann elimination used to study amine structures.

Conclusion: Aniline is methylated successively to N-methylaniline, then N,N-dimethylaniline, and finally — with excess CH₃I — to the quaternary salt N,N,N-trimethylanilinium iodide, C₆H₅N⁺(CH₃)₃ I⁻.

Concept. Aniline is a primary amine with an N–H bond, so it undergoes acylation (here benzoylation): the nucleophilic nitrogen attacks the electrophilic carbonyl carbon of benzoyl chloride (C₆H₅COCl), the chloride leaves, and an amide (an anilide) is formed with loss of HCl. This is the Schotten–Baumann reaction when carried out in the presence of aqueous NaOH, which neutralises the HCl and drives the reaction.

Reaction and product. C₆H₅NH₂ + C₆H₅COCl → C₆H₅–NH–CO–C₆H₅ + HCl. The product is N-phenylbenzamide, commonly called benzanilide — a white crystalline solid. The –NH₂ group of aniline is converted to the amide –NH–CO–C₆H₅, so the nitrogen is no longer strongly basic or easily oxidised. (This acylation is used to protect the amino group of aniline during reactions such as nitration, after which the amide is hydrolysed back to the amine.)

Conclusion: C₆H₅NH₂ + C₆H₅COCl → C₆H₅NHCOC₆H₅ + HCl; the product is N-phenylbenzamide (benzanilide).

Concept. C₃H₉N is a saturated amine; distributing the three carbons among the groups on nitrogen gives four isomers — two primary, one secondary and one tertiary. Only primary aliphatic amines liberate nitrogen gas (N₂) with nitrous acid, because they form very unstable aliphatic diazonium salts that decompose immediately, evolving N₂ (and giving alcohols). Secondary amines give N-nitrosamines (no N₂) and tertiary amines do not give N₂ either.

Working. The four isomers are the two primary amines (propan-1-amine and propan-2-amine), the secondary amine N-methylethanamine, and the tertiary amine trimethylamine. With nitrous acid (NaNO₂ + HCl), a primary aliphatic amine R–NH₂ forms R–N₂⁺, which is too unstable to survive and at once loses N₂ (R–N₂⁺ → R⁺ + N₂, giving alcohols/alkenes). So propan-1-amine and propan-2-amine liberate nitrogen gas (seen as brisk effervescence). The secondary amine gives a yellow N-nitrosamine and the tertiary amine merely forms a salt, so neither evolves N₂.

Conclusion: Four isomers — propan-1-amine, propan-2-amine (both 1°), N-methylethanamine (2°) and trimethylamine (3°); the two that liberate N₂ with nitrous acid are propan-1-amine and propan-2-amine.

Concept. IUPAC: name an amine as an alkanamine (replace “-e” of the parent alkane with “-amine”, –NH₂ taking the lowest locant); for 2°/3° amines, the largest group is the parent and the other groups are cited with the N- prefix. Classify by the number of carbon groups on nitrogen (1° one, 2° two, 3° three).

Notes. In (iii) the larger group is propan-2-yl (so the parent is propan-2-amine) and the methyl on nitrogen is cited as N-methyl. In (iv) (CH₃)₃CNH₂ has the –NH₂ on a carbon bonded to three methyls, but nitrogen still carries only one group → primary (2-methylpropan-2-amine). In (vi) nitrogen bears three groups (two ethyl, one methyl) → tertiary; the parent is ethanamine and the other groups are N-ethyl and N-methyl. In (v) and (vii) aniline is the aromatic parent.

Conclusion: (i) propan-2-amine (1°); (ii) propan-1-amine (1°); (iii) N-methylpropan-2-amine (2°); (iv) 2-methylpropan-2-amine (1°); (v) N-methylaniline (2°); (vi) N-ethyl-N-methylethanamine (3°); (vii) 3-bromoaniline (1°).

Concept. The standard tests that separate amine classes are the carbylamine test (only 1° amines give the foul isocyanide), the Hinsberg test (1°, 2°, 3° behave differently with benzenesulphonyl chloride), the azo-dye test (an aromatic 1° amine forms a diazonium salt at 273 K that couples with 2-naphthol to a coloured dye), and the reaction with nitrous acid.

Why these tests work. The carbylamine reaction needs an N–H₂ on a primary amine; secondary and tertiary amines cannot form the isocyanide, so it separates 1° from the rest. The Hinsberg test separates all three classes by whether the sulphonamide has an acidic N–H (1° → alkali-soluble, 2° → alkali-insoluble, 3° → no reaction). The azo-dye (diazotisation–coupling) test is specific for an aromatic primary amine, so it tells an aromatic 1° amine (aniline) from an aliphatic 1° amine (ethylamine, benzylamine), whose diazonium salts are unstable and just release N₂.

Conclusion: Use the carbylamine test for (i) and (v), the Hinsberg test (or HNO₂) for (ii), and the azo-dye (diazotisation–coupling) test for (iii) and (iv).

Concept. Most of these follow from two facts about aniline: its nitrogen lone pair is delocalised into the benzene ring (resonance), and the whole molecule is dominated by a large hydrophobic ring. The last two concern diazonium-salt stability and the selectivity of the Gabriel synthesis.

Extra depth. The protonation argument in (iv) is confirmed by the fact that if the –NH₂ is first acetylated (to acetanilide, a non-basic group that is still o/p-directing), nitration then gives predominantly the para product — showing that the meta product in free aniline comes from the anilinium ion. The resonance stabilisation of the arenediazonium ion in (vi) is the same effect that makes aromatic diazonium salts useful synthetic intermediates.

Conclusion: All seven follow from aniline’s lone-pair delocalisation and hydrophobic ring (weaker base/higher pK_b, insoluble, meta-nitration via the anilinium ion, no Friedel–Crafts), the base-generated OH⁻ precipitating Fe(OH)₃, the ring-resonance stabilisation of arenediazonium ions, and the Gabriel route giving pure primary amines.

Concept. Basic strength rises with lone-pair availability (+I of alkyl groups up, ring delocalisation and electron-withdrawing groups down); a stronger base has a smaller pK_b. Boiling point depends on hydrogen bonding (O–H > N–H; 1°/2° amines H-bond, 3° do not). Water solubility depends on H-bonding with water (small aliphatic amines high, aromatic low).

Reasoning for the key ones. (i) Aniline and N-methylaniline (ring-delocalised lone pair) are the weakest bases → highest pK_b; the aliphatic amines are stronger (lower pK_b), diethylamine strongest. (iii)(a) The electron-withdrawing p-NO₂ makes p-nitroaniline the weakest base; the electron-releasing p-CH₃ (p-toluidine) the strongest. (iii)(b) Benzylamine is aliphatic-type (lone pair not in the ring), so it is the strongest; aniline weakest. (iv) In the gas phase, with no solvation, the +I effect alone operates, so 3° > 2° > 1° > NH₃. (v) Alcohols H-bond more strongly (O–H) than amines (N–H), and among the two amines the 1° (two N–H) boils higher than the 2° (one N–H). (vi) Aniline’s big ring makes it least soluble; among the aliphatic amines the smaller, less hydrophobic ethylamine is more soluble than diethylamine.

Conclusion: (i) C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH; (ii) C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH; (iii)(a) p-nitroaniline < aniline < p-toluidine; (b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂; (iv) (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃; (v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH; (vi) C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Concept. The carbon count is the key. To shorten by one carbon: Hofmann bromamide degradation of an amide (–CONH₂ → –NH₂, one carbon lost) for amines, or descend the acid series. To lengthen by one carbon: go through the nitrile (R–X → R–CN → reduce to R–CH₂NH₂, or hydrolyse to R–COOH). Reactions used: amide formation (NH₃), Hofmann degradation (Br₂/NaOH), nitrile formation (KCN) and reduction (LiAlH₄/H₂).

Working logic. “Descending” an amine (iv, and the amine step of viii) or making an amine with fewer carbons (i, ii) uses the Hofmann bromamide degradation, which removes the carbonyl carbon. “Ascending” (iii, v, vi) uses the nitrile to add a carbon, then either reduction (→ amine) or hydrolysis (→ acid). The nitrous-acid step (R–NH₂ → R–OH) converts an aliphatic amine to the alcohol so that the chain can be re-functionalised (needed in iv, vi, viii).

Conclusion: Each conversion is achieved as tabulated — Hofmann degradation for the carbon-shortening amine steps (i, ii, iv, viii) and the nitrile route for the carbon-lengthening steps (iii, v, vi); (vii) is reduction of nitromethane then methylation.

Concept — Hinsberg’s test. The best single method to tell apart all three classes of amine is Hinsberg’s test, which uses benzenesulphonyl chloride (C₆H₅SO₂Cl, “Hinsberg’s reagent”). The three classes behave differently because of the number of N–H bonds available: a 1° amine gives a sulphonamide that still has one N–H (acidic, alkali-soluble), a 2° amine gives a sulphonamide with no N–H (alkali-insoluble), and a 3° amine (no N–H at all) does not react.

How to carry it out. Shake the amine with benzenesulphonyl chloride, then add aqueous NaOH: (1) if a clear solution forms that gives a precipitate on acidification → primary amine; (2) if a precipitate forms that is insoluble in NaOH → secondary amine; (3) if the amine does not react (no sulphonamide forms) → tertiary amine. An alternative first screen is the carbylamine test (only primary amines give the foul isocyanide with CHCl₃ + alc. KOH: RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O), which quickly flags the primary amine.

Conclusion: Hinsberg’s test with benzenesulphonyl chloride identifies the three classes — the 1° amine gives an alkali-soluble sulphonamide, the 2° amine an alkali-insoluble sulphonamide, and the 3° amine does not react (the carbylamine test confirms the primary amine).

(i) Carbylamine reaction. A test for primary amines: heating a primary amine (aliphatic or aromatic) with chloroform and alcoholic KOH gives a foul-smelling isocyanide (carbylamine). RNH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O. Secondary and tertiary amines do not respond.

(ii) Diazotisation. The conversion of a primary aromatic amine into a diazonium salt by treatment with nitrous acid (NaNO₂ + HCl) at 273–278 K: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O. The low temperature is essential because the diazonium salt is unstable when warm.

(iii) Hofmann’s bromamide reaction (degradation). An amide treated with bromine and a strong alkali gives a primary amine with one carbon fewer: R–CONH₂ + Br₂ + 4NaOH → R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The R group migrates from the carbonyl carbon to the nitrogen (via an isocyanate intermediate).

(iv) Coupling reaction. An arenediazonium salt reacts with an electron-rich aromatic compound — a phenol (in mild alkali) or an aromatic amine (in mild acid) — at the para position to give a brightly coloured azo compound (Ar–N=N–Ar′): C₆H₅N₂⁺Cl⁻ + C₆H₅OH → p-HO–C₆H₄–N=N–C₆H₅ (p-hydroxyazobenzene, an orange dye). These are the azo dyes.

(v) Ammonolysis. The cleavage of the C–X bond of an alkyl (or benzyl) halide by ammonia: R–X + NH₃ → R–NH₂ (as its salt, freed by alkali). Because the amine formed is itself nucleophilic, the reaction gives a mixture of 1°, 2°, 3° amines and the quaternary ammonium salt; a large excess of ammonia favours the primary amine.

(vi) Acetylation. Introduction of an acetyl group (CH₃CO–) onto the nitrogen of a 1° or 2° amine, using acetyl chloride or acetic anhydride (with a base such as pyridine), to give an amide: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH. It is used to protect the –NH₂ group.

(vii) Gabriel phthalimide synthesis. A method for making pure primary amines: phthalimide is converted to its potassium salt (KOH), which is alkylated (R–X, SN2) to N-alkylphthalimide; alkaline hydrolysis (or reaction with hydrazine) then releases the primary amine R–NH₂ and phthalic acid (or phthalhydrazide). Aromatic primary amines cannot be made this way because aryl halides do not undergo the SN2 alkylation step.

Conclusion: The seven are, respectively: the carbylamine test for 1° amines; diazotisation (aromatic 1° amine → diazonium salt at 273 K); Hofmann degradation (amide → 1° amine, −1 C); coupling (diazonium + phenol/amine → azo dye); ammonolysis (R–X + NH₃ → amine mixture); acetylation (amine → amide, protecting –NH₂); and the Gabriel synthesis (pure 1° aliphatic amines only).

Concept. These multi-step syntheses use the amine/diazonium toolkit: nitration + reduction to make –NH₂; diazotisation + replacement (Sandmeyer, by –OH, by H, Balz–Schiemann for –F); acetylation to protect –NH₂ for controlled ring substitution; the nitrile route to lengthen a chain; and Hofmann degradation. The order of steps is chosen from the directing effects.

Note on (viii) Benzamide → toluene. A clean two/three-step route is: benzamide → (Br₂/NaOH, Hofmann degradation) aniline → (NaNO₂/HCl, 273 K) benzenediazonium chloride → (CuCN, Sandmeyer) benzonitrile → … this does not give toluene directly. The standard NCERT route to toluene is via the diazonium salt and a methyl source: benzamide → aniline → benzenediazonium chloride → (replace –N₂⁺ by –CH₃ is not a standard single step). In practice toluene is obtained from benzamide by first degrading to aniline, diazotising, and replacing –N₂⁺ by –H (H₃PO₂) to give benzene, then Friedel–Crafts methylation (CH₃Cl/AlCl₃) to toluene. The key handles are Hofmann degradation (amide → amine) and diazonium replacement.

Why the protecting step in (vii). Free aniline reacts with bromine water to give 2,4,6-tribromoaniline at once; to stop at the mono-para stage the –NH₂ is first acetylated to acetanilide (a weaker activator, and bulky), which directs bromination mainly to the para position; hydrolysis then regenerates the amine as p-bromoaniline.

Conclusion: Each target is reached through the amine/diazonium sequence as tabulated — reduction of –NO₂, diazotisation and replacement, protection by acetylation for selective substitution, Hofmann degradation, and the nitrile route for chain extension.

Concept. Each arrow is a named reaction; identifying the reagent fixes the product. NaCN/KCN makes a nitrile (+1 C); partial hydrolysis of a nitrile gives an amide; NaOH+Br₂ (or NaOBr) is Hofmann degradation (amide → amine, −1 C); CuCN on a diazonium salt is Sandmeyer (→ aryl nitrile); H₂O/H⁺ hydrolyses a nitrile to an acid; NH₃/Δ makes an amide; LiAlH₄ reduces a nitrile to a 1° amine; HNO₂ on an aliphatic 1° amine gives an alcohol, and on an aromatic 1° amine a diazonium salt; H₂O/H⁺ on a diazonium salt gives a phenol; coupling with phenol gives an azo dye.

Working through two of them. In (i) ethyl iodide + NaCN gives propanenitrile (A), partial hydrolysis gives propanamide (B), and NaOH + Br₂ (Hofmann bromamide) removes the carbonyl carbon to give ethanamine (C) — note the loss of one carbon. In (iii) the same propanenitrile (A) is instead fully reduced by LiAlH₄ to propan-1-amine (B, a 1° aliphatic amine), which with nitrous acid at 0 °C gives propan-1-ol (C) + N₂ (aliphatic diazonium salts are unstable). In (vi) aniline is diazotised (B) and then couples with phenol (not hydrolysed) to give the orange azo dye p-hydroxyazobenzene (C).

Conclusion: (i) propanenitrile, propanamide, ethanamine; (ii) benzonitrile, benzoic acid, benzamide; (iii) propanenitrile, propan-1-amine, propan-1-ol; (iv) aniline, benzenediazonium chloride, phenol; (v) acetamide, methanamine, methanol; (vi) aniline, benzenediazonium chloride, p-hydroxyazobenzene.

Concept. Work backwards from C. The formula C₆H₇N with four degrees of unsaturation (a benzene ring) and one nitrogen is aniline (C₆H₅NH₂). C is formed from B by “Br₂ and KOH”, which is the Hofmann bromamide degradation — so B must be the amide whose degradation gives aniline, namely benzamide (C₆H₅CONH₂). B is formed from A by “aqueous ammonia and heating”, which converts a carboxylic acid (via its ammonium salt) to an amide — so A is benzoic acid (C₆H₅COOH).

Confirmation. The Hofmann degradation removes the carbonyl carbon of the amide, so benzamide (C₆H₅CONH₂, 7 carbons) gives aniline (C₆H₅NH₂, 6 carbons) — matching the stated formula C₆H₇N exactly. The “aqueous ammonia + heat” step is the standard acid → amide conversion (RCOOH → RCOONH₄ → RCONH₂ + H₂O), confirming A as benzoic acid.

Conclusion: A = benzoic acid, B = benzamide, C = aniline.

Concept. This set covers the carbylamine test, three diazonium replacement reactions (by H, by F, and reductive replacement by ethanol), sulphonation of aniline, bromination of aniline, and acetylation.

Notes. In (iii) concentrated sulphuric acid first forms anilinium hydrogensulphate; on strong heating this rearranges to sulphanilic acid, which exists as an internal salt (zwitterion): the –SO₃⁻ and –N⁺H₃ on the same molecule. In (vii) route (i) HBF₄ makes the diazonium fluoroborate (the Balz–Schiemann route to fluorobenzene), while route (ii) NaNO₂/Cu replaces –N₂⁺ by –NO₂ to give nitrobenzene — the two bracketed reagents correspond to two different products, and the NCERT answer for the “NaNO₂/Cu” step is nitrobenzene. In (v) free aniline is so strongly activated that bromine water gives the 2,4,6-tribromo product directly.

Conclusion: (i) phenyl isocyanide; (ii) benzene + N₂; (iii) sulphanilic acid (via anilinium hydrogensulphate); (iv) benzene + N₂ + acetaldehyde; (v) 2,4,6-tribromoaniline; (vi) acetanilide; (vii) fluorobenzene (via HBF₄) and nitrobenzene (via NaNO₂/Cu).

Concept. The Gabriel synthesis makes a primary amine by alkylating potassium phthalimide with an alkyl halide in an SN2 reaction, and then releasing the amine by hydrolysis. The success of the method therefore depends on the alkyl halide being able to undergo nucleophilic substitution at its carbon.

Explanation. To make an aromatic primary amine (e.g. aniline), the phthalimide anion would have to react with an aryl halide (such as chlorobenzene) in the alkylation step. But aryl halides do not undergo SN2 (nucleophilic substitution) under these conditions: the C–X bond is on an sp² carbon, it is shortened and strengthened by resonance with the ring (partial double-bond character), and back-side attack by the nucleophile is blocked by the ring. The phthalimide nitrogen therefore cannot displace the halide from an aryl halide, so no N-aryl phthalimide forms, and the aromatic amine cannot be obtained. The method is thus limited to aliphatic (and benzylic) primary amines, where the halide carbon is sp³ and reacts readily by SN2.

Conclusion: Because the Gabriel synthesis needs an SN2 alkylation of potassium phthalimide, and aryl halides (sp² C–X, resonance-strengthened, back-side attack blocked) do not undergo SN2, aromatic primary amines cannot be prepared by it.

Concept. Nitrous acid (HNO₂), generated in situ from sodium nitrite and a mineral acid (NaNO₂ + HCl), reacts with a primary amine to form a diazonium salt (R–N₂⁺ or Ar–N₂⁺). The crucial difference is stability: an aromatic diazonium ion is stabilised by resonance with the ring and survives at low temperature, whereas an aliphatic diazonium ion has no such stabilisation and decomposes instantly, losing N₂.

(i) Aromatic primary amines. At 273–278 K, aniline gives a stable benzenediazonium salt (diazotisation):

The diazonium salt is a stable, useful intermediate that undergoes replacement and coupling reactions (Sandmeyer, Balz–Schiemann, coupling to azo dyes). On warming above 283 K it is slowly hydrolysed to phenol.

(ii) Aliphatic primary amines. The diazonium salt formed is very unstable and decomposes at once, evolving nitrogen gas and giving a mixture of products (chiefly the alcohol, with some alkene and alkyl halide):

For example, ethylamine gives ethanol with brisk effervescence of N₂: C₂H₅NH₂ + HNO₂ → C₂H₅OH + N₂ + H₂O. The evolution of nitrogen gas is a characteristic test for an aliphatic primary amine.

Conclusion: Aromatic primary amines give stable arenediazonium salts at 273–278 K (diazotisation); aliphatic primary amines give alcohols with brisk evolution of N₂ because their diazonium ions decompose immediately.

(i) Amines are less acidic than alcohols. Acidity here means the tendency to lose the N–H (or O–H) proton. Nitrogen is less electronegative than oxygen, so the N–H bond is less polar and harder to break than the O–H bond; and the conjugate base of an amine, the amide ion (R–NH⁻), is less stable than the alkoxide ion (R–O⁻) because the negative charge is held on the less electronegative nitrogen. Therefore an amine gives up its proton less readily than an alcohol of comparable mass — amines are weaker acids (and, correspondingly, amide ions are stronger bases than alkoxide ions).

(ii) Primary amines boil higher than tertiary amines. Primary amines (R–NH₂) have two N–H bonds and secondary amines one, so their molecules associate through intermolecular N–H···N hydrogen bonds. Tertiary amines (R₃N) have no N–H bond, so they cannot form hydrogen bonds among themselves and are held only by weaker dipole–dipole and dispersion forces. Breaking the hydrogen-bonded network of a primary amine needs more energy, so a primary amine boils at a higher temperature than a tertiary amine of comparable molar mass (order 1° > 2° > 3°).

(iii) Aliphatic amines are stronger bases than aromatic amines. Basic strength depends on the availability of the nitrogen lone pair. In an aliphatic amine the lone pair is fully available and is further enriched by the electron-releasing (+I) alkyl group(s), so the amine readily accepts a proton. In an aromatic amine (aniline) the lone pair is delocalised into the benzene ring by resonance, so it is much less available to bond to a proton; moreover, protonation destroys this stabilising delocalisation, which further disfavours it. Hence aliphatic amines are much stronger bases than aromatic amines.

Conclusion: (i) the N–H bond is less polar and the amide ion R–NH⁻ less stable than the alkoxide ion, so amines are weaker acids than alcohols; (ii) primary amines hydrogen-bond (two N–H) while tertiary amines have no N–H, so 1° amines boil higher; (iii) the aliphatic amine lone pair is free and enriched by +I, whereas the aromatic amine lone pair is delocalised into the ring, so aliphatic amines are the stronger bases.