NCERT Solutions Class 12 Chemistry Chapter 9 Amines PDF

Concept used. An amine is a derivative of ammonia (NH3) in which one, two, or three of its hydrogens are replaced by alkyl or aryl groups, denoted R-NH2, R2NH, R3N respectively. The number of carbon-containing groups directly bonded to the nitrogen decides the class:

• Primary (1^∘): one R on N, so structure is R-NH2.
• Secondary (2^∘): two R groups on N, so R2NH.
• Tertiary (3^∘): three R groups on N, so R3N.

For IUPAC naming we use the substitutive system: replace the -e of the parent alkane name with -amine and place the lowest possible locant before the suffix. For secondary and tertiary amines, the largest alkyl chain is the parent, and the smaller groups are named as N-substituents (the letter N written in italics). Arylamines based on benzene use aniline as the retained IUPAC name (or benzenamine).

1. (i) (CH3)2CHNH2. The carbon attached to NH2 is a propan-2-yl group (isopropyl). Parent chain has 3 carbons, so parent alkane is propane → propane  ends with the amino group on C-2. IUPAC name: propan-2-amine. Only one R (the isopropyl carbon skeleton) is on N ⇒ primary amine.
2. (ii) CH3(CH2)2NH2. Expand: CH3-CH2-CH2-NH2. Three carbons in the longest chain, -NH2 on C-1. IUPAC name: propan-1-amine. One R on N ⇒ primary amine.
3. (iii) CH3NHCH(CH3)2. The N has two carbons bonded directly: a CH3 group and a (CH3)2CH- group. Choose the larger one (the 3-carbon isopropyl) as the parent → propan-2-amine. The methyl on N is an N-methyl substituent. IUPAC name: N-methylpropan-2-amine. Two R on N ⇒ secondary amine.
4. (iv) (CH3)3CNH2. The C attached to NH2 is a tert-butyl carbon (C(CH3)3). Parent is butane, -NH2 on C-2 of the 4-carbon skeleton CH3-C(CH3)2-NH2. Numbering: the parent chain that contains the amino group with lowest locant is 2-methylpropane. The NH2 is on C-2 of 2-methylpropane. IUPAC name: 2-methylpropan-2-amine. One R on N ⇒ primary amine.
5. (v) C6H5NHCH3. The N carries a phenyl group (C6H5) and a methyl group. Treat as aniline with an N-methyl substituent. IUPAC name: N-methylaniline (also N-methylbenzenamine). Two R on N ⇒ secondary amine.
6. (vi) (CH3CH2)2NCH3. The N has two ethyl groups and one methyl group. Take the larger group as parent: ethane → ethanamine. The second ethyl and the methyl become N-substituents. IUPAC name: N-ethyl-N-methylethanamine. Three R on N ⇒ tertiary amine.
7. (vii) m-BrC6H4NH2. A bromine on the meta position of aniline. Number the ring with C-1 carrying the principal group (NH2); meta is C-3. IUPAC name: 3-bromoaniline (also 3-bromobenzenamine). One R (the benzene ring) on N ⇒ primary amine.

(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 used. Three signature tests distinguish amine classes:

• Carbylamine test: 1^∘ amines (both alkyl and aryl) on warming with chloroform and alcoholic KOH give an isocyanide (R-NC) with a foul, penetrating smell. 2^∘ and 3^∘ amines give no reaction. R-NH2 + CHCl3 + 3KOH Δ R-NC + 3KCl + 3H2O.
• Hinsberg's test: amines react with benzenesulphonyl chloride (C6H5SO2Cl). 1^∘ amines give a sulphonamide soluble in alkali; 2^∘ amines give a sulphonamide insoluble in alkali; 3^∘ amines do not react (no N–H).
• Diazotisation: 1^∘ aromatic amines with NaNO2/HCl at 0–5 ^∘C form a stable diazonium salt ArN2+Cl-. 1^∘ aliphatic amines under the same conditions liberate N2 (vigorous effervescence). The diazonium salt then couples with phenol or 2-naphthol in alkaline medium to give an orange/red azo dye.

1. (i) Methylamine vs dimethylamine: carbylamine test. CH3NH2 (1^∘) on heating with CHCl3 + alcoholic KOH gives methyl isocyanide CH3NC (offensive smell). CH3NH2 + CHCl3 + 3KOH Δ CH3NC + 3KCl + 3H2O. Dimethylamine (CH3)2NH (2^∘) gives no reaction. Smell test confirms 1^∘.
2. (ii) Secondary vs tertiary amines: Hinsberg's test. Shake with C6H5SO2Cl in KOH. The 2^∘ amine forms a sulphonamide R2N-SO2C6H5 which has no N–H and is therefore insoluble in NaOH. The 3^∘ amine has no N–H to start with and does not react; on acidifying, the tertiary amine dissolves in dilute HCl while the sulphonamide from the secondary amine does not.
3. (iii) Ethylamine vs aniline: azo-dye test. Aniline (1^∘ aromatic) with cold NaNO2/HCl at 0–5 ^∘C gives benzenediazonium chloride, which couples with alkaline 2-naphthol to give a bright orange/red azo dye. Ethylamine (1^∘ aliphatic) under the same conditions evolves N2 gas (effervescence) and gives ethanol; no dye forms. C6H5NH2 + HNO2 + HCl 273--278 K C6H5N2+Cl- + 2H2O, C2H5NH2 + HNO2 -> C2H5OH + N2 + H2O.
4. (iv) Aniline vs benzylamine: same azo-dye test. Aniline is aryl-1^∘ and forms a stable diazonium salt → orange dye on coupling with 2-naphthol/NaOH. Benzylamine C6H5CH2NH2 is alkyl-1^∘ (the amino group is on the CH2, not on the ring) and behaves like ethylamine: liberates N2, no dye.
5. (v) Aniline vs N-methylaniline: carbylamine test. Aniline is 1^∘ and gives the foul smell of phenyl isocyanide C6H5NC: C6H5NH2 + CHCl3 + 3KOH Δ C6H5NC + 3KCl + 3H2O. N-methylaniline C6H5NHCH3 is 2^∘ and gives no reaction.

Use the carbylamine test for 1^∘ vs 2^∘ (i, v); Hinsberg's test for 2^∘ vs 3^∘ (ii); and cold NaNO2/HCl + 2-naphthol coupling to distinguish aryl-1^∘ from alkyl-1^∘ amines (iii, iv).

Concept used. The basicity of an amine in water reflects the position of the equilibrium R-NH2 + H2O <=> R-NH3+ + OH-, K_b = [RNH3+][OH-][RNH2], with pK_b = -₁₀ K_b. A larger pK_b means a weaker base. Three effects decide K_b:

• Inductive effect (+I): alkyl groups push electron density onto N, raising the availability of its lone pair for protonation ⇒ more basic.
• Resonance/mesomeric effect: in arylamines the lone pair of N is delocalised into the ring through resonance, making it less available ⇒ less basic.
• Solvation by water: the conjugate acid RNH3+ is stabilised by hydrogen bonding with water. More N–H bonds on the cation ⇒ better H-bond solvation ⇒ more basic.

1. (i) pK_b of aniline (≈ 9.38) > pK_b of methylamine (≈ 3.38). In methylamine the methyl group has a +I effect, pushing electron density toward N, so the lone pair on N is more available to bind a proton. In aniline the nitrogen lone pair is conjugated into the benzene ring through resonance (it forms part of the π-system), so it is much less available for protonation. The result: aniline is a much weaker base than methylamine, and a weaker base has a larger pK_b.
2. (ii) Solubility. Ethylamine has only two carbon atoms and one -NH2. The amino group hydrogen-bonds vigorously with water (both as donor: N–H⋯O, and as acceptor: N⋯H–O). The hydrophobic alkyl part is small, so ethylamine is freely soluble. In aniline the hydrophobic part is the entire benzene ring; the ring is large, non-polar, and cannot form H-bonds. The size of the hydrophobic part dominates, so aniline is sparingly soluble in water (about 3.5 g/100 mL).
3. (iii) Methylamine + FeCl3 → Fe(OH)3. Methylamine in water gives OH-: CH3NH2 + H2O <=> CH3NH3+ + OH-. These hydroxide ions react with the Fe³⁺ from FeCl3: Fe³⁺ + 3 OH- -> Fe(OH)3 v (brown ppt., hydrated ferric oxide). The strong basicity of methylamine raises [OH-] enough to exceed the solubility product of Fe(OH)3, so it precipitates.
4. (iv) Aniline + HNO3/H2SO4 → substantial meta-isomer. The -NH2 is strongly activating and o/p-directing. But the nitration mixture contains conc. H2SO4, which protonates the very basic -NH2 to form the anilinium ion C6H5NH3+. The -N⁺H3 group is now deactivating and m-directing (it has no lone pair to donate and carries a + charge that withdraws electrons through induction). So a portion of anilinium ion in the mixture gives m-nitroaniline while the unprotonated aniline gives the o/p-products. Overall the product mixture contains a substantial fraction of m-nitroaniline (∼ 47%).
5. (v) No Friedel–Crafts on aniline. Friedel–Crafts uses a Lewis acid catalyst, typically AlCl3. The basic lone pair on the nitrogen of aniline forms a complex with the Lewis acid: C6H5NH2 + AlCl3 -> C6H5NH2⁺-AlCl3^-. The nitrogen now carries a positive charge, so the -NH2 becomes strongly deactivating (like -NO2) and the ring is no longer nucleophilic enough to react with the carbocation electrophile generated from RX/RCOCl. Hence no Friedel–Crafts product forms.
6. (vi) Stability of ArN2+ vs R-N2+. Aromatic diazonium ions are stabilised by extensive resonance delocalisation: the positive charge on -N2+ spreads onto the ortho and para carbons of the benzene ring (the same ring can be drawn with charges at C-2, C-4 etc.). Aliphatic diazonium ions cannot delocalise this way, so they lose N2 almost immediately at room temperature to form an alcohol / alkene / alkyl halide mixture. Aromatic salts can be isolated and stored at 0–5 ^∘C.
7. (vii) Why Gabriel synthesis is preferred for 1^∘ amines. Direct ammonolysis (RX + NH3) gives a mixture: the 1^∘ amine formed first reacts further with RX to give 2^∘, 3^∘, and quaternary ammonium salts, so the yield of pure 1^∘ amine is low. In the Gabriel phthalimide synthesis, potassium phthalimide reacts with RX to give an N-alkylphthalimide. This intermediate has no free N–H, so further alkylation is impossible. Hydrolysis (or hydrazinolysis) then liberates a pure 1^∘ amine and phthalic acid. The route avoids any over-alkylation.

(i) Resonance delocalises N lone pair into the ring. (ii) Hydrophobic phenyl dominates, no H-bonding. (iii) Methylamine raises [OH-], precipitates Fe(OH)3. (iv) In H2SO4, -NH2 is protonated to -N⁺H3 which is m-directing. (v) Lone pair complexes with AlCl3, deactivating the ring. (vi) Aryl diazonium ion is resonance-stabilised. (vii) Phthalimide blocks over-alkylation, giving only 1^∘ amine.

Concept used. Comparing amine basicity in water requires balancing three effects:

• Inductive (+I) effect of alkyl groups. More or larger alkyl groups on N push more electron density onto N, raising basicity.
• Steric crowding around N. A nitrogen surrounded by bulky groups is harder to approach by a proton or by water: decreases effective basicity.
• Solvation of the conjugate acid R_nNH_(4-n)+ by water. Each remaining N–H of the cation hydrogen-bonds with water. More N–H bonds ⇒ better solvation ⇒ more stable cation ⇒ equilibrium shifts forward, raising basicity. So the order of H-bonds is 1^∘ (3 N–H) > 2^∘ (2 N–H) > 3^∘ (1 N–H).

The observed aqueous order for methyl-substituted amines is 2^∘ > 1^∘ > 3^∘ > NH3 (a non-monotonic order: the +I effect would predict 3^∘ > 2^∘ > 1^∘, but steric crowding and weaker solvation pull tertiary amines down). In ethyl amines the order shifts: the ethyl groups are larger, so crowding matters more and the order can become 2^∘ > 3^∘ > 1^∘. In the gas phase, where there is no solvent, only the +I and steric effects matter, so the order reduces to 3^∘ > 2^∘ > 1^∘ > NH3.

For arylamines the lone pair is delocalised into the ring, so they are always weaker bases than alkylamines. Electron-donating substituents on the ring (-CH3) increase basicity; electron-withdrawing substituents (-NO2) decrease it.

1. (i) Decreasing pK_b = increasing basicity in reverse. (C2H5)2NH (2^∘ aliphatic) has the smallest pK_b (≈ 3.0). C2H5NH2 (1^∘ aliphatic) has the next-smallest pK_b (≈ 3.25). Then come the arylamines: C6H5NHCH3 has a methyl that donates electrons, so it is more basic than C6H5NH2. Hence C6H5NHCH3 has a smaller pK_b than aniline. Decreasing pK_b (weakest to strongest base): C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH.
2. (ii) Increasing basic strength. Aniline is the weakest because its lone pair is delocalised. Adding two methyls on N (giving C6H5N(CH3)2) donates more electrons but the lone pair is still partly tied up in the ring, so it is more basic than aniline but still weaker than any alkylamine. Among alkyls, CH3NH2 (1^∘) is weaker than (C2H5)2NH (2^∘) because the second ethyl raises the +I effect. Increasing basic strength: C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH.
3. (iii)(a) Aniline vs p-toluidine vs p-nitroaniline. -CH3 on the para position donates electrons by +I and +H (hyperconjugation), increasing electron density at N ⇒ p-toluidine (i.e. 4-methylaniline) is the most basic. -NO2 at the para position withdraws electrons strongly by -I and -R, pulling the lone pair away from N ⇒ p-nitroaniline is the least basic. Aniline lies in the middle. p-nitroaniline < aniline < p-toluidine.
4. (iii)(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2. In C6H5CH2NH2 (benzylamine) the -NH2 is on the CH2, NOT directly on the ring, so the lone pair is not delocalised into the ring. It behaves like an aliphatic amine ⇒ most basic of the three. Between C6H5NH2 and C6H5NHCH3, the methyl on N donates electrons by +I, raising basicity ⇒ C6H5NHCH3 > C6H5NH2. C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2.
5. (iv) Gas-phase decreasing basicity. No solvent, no solvation; only +I and steric effects survive. The +I effect rises monotonically with the number of alkyl groups, so the order is determined purely by alkyl count: (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3. This is the ``natural'' order of basicity, recovered when the complications of aqueous solvation are removed.
6. (v) Increasing boiling point. (CH3)2NH is 2^∘ and has only one N–H, so weak H-bonding. C2H5NH2 is 1^∘ and has two N–H bonds, so stronger H-bonding ⇒ higher b.p. C2H5OH has an O–H whose H-bond is even stronger (O is more electronegative than N) ⇒ highest b.p. (CH3)2NH (7 ) < C2H5NH2 (17 ) < C2H5OH (78 ).
7. (vi) Increasing solubility in water. Aniline is sparingly soluble (large hydrophobic ring). Among the two alkylamines, the smaller, more polar one solvates better: C2H5NH2 (1^∘, two N–H) hydrogen-bonds with water more vigorously than (C2H5)2NH (2^∘, larger hydrophobic part, only one N–H). C6H5NH2 < (C2H5)2NH < C2H5NH2.

(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 used. Three key chain-length operations recur in amine syntheses:

• Hofmann bromamide rearrangement: an amide is degraded by Br2/NaOH to give a primary amine with one carbon less: R-CONH2 + Br2 + 4 NaOH -> R-NH2 + Na2CO3 + 2 NaBr + 2 H2O.
• Nitrile reduction (Mendius / LiAlH₄ / catalytic H₂): R-C#N + 2 H2 Ni or LiAlH4 R-CH2-NH2. This adds one carbon compared to the starting alkyl halide (because RX → RCN adds the C of CN^-).
• Carbylamine elimination and chain shortening of acids: a carboxylic acid is converted to its amide (with NH3 / Δ), then Hofmann-degraded to the next lower amine; or an acid is converted to its potassium salt and treated with NaOH/CaO (decarboxylation, soda-lime) to lose one carbon as CO2.

1. (i) Ethanoic acid → methanamine. Ethanoic acid has 2 C; methanamine has 1 C ⇒ chain shortens by one. Use Hofmann bromamide. CH3COOH NH3, Δ CH3CONH2 Br2, NaOH CH3NH2.
2. (ii) Hexanenitrile → 1-aminopentane. Hexanenitrile is CH3(CH2)4-CN (5 C in the chain + 1 C of CN = 6 C in total). 1-aminopentane = CH3(CH2)4-NH2 has only 5 C, so the route must lose the cyanide carbon. Hydrolyse the nitrile to the amide, then Hofmann-degrade: CH3(CH2)4-CN H2O/H+ CH3(CH2)4-CONH2 Br2, NaOH CH3(CH2)4-NH2. CH3(CH2)4-NH2 is 1-aminopentane (pentan-1-amine).
3. (iii) Methanol → ethanoic acid. Methanol has 1 C; ethanoic acid has 2 C. Need to add one C. First convert methanol to methyl iodide, then SN2 with cyanide, then hydrolyse:
   1. [label=., leftmargin=*]
   2. CH3OH → CH3I (PI3 or HI)
   3. CH3I → CH3CN (KCN)
   4. CH3CN → CH3COOH (H2O, H+).
4. (iv) Ethanamine → methanamine. Both are 1^∘ amines; we need to lose one C. Convert ethanamine to ethanenitrile is not possible directly; instead, oxidise ethanamine carefully to ethanoic acid (or hydrolyse via diazonium on the alkyl version): nitrous acid converts ethanamine to ethanol, which is oxidised to ethanoic acid; then proceed as in (i):
   1. [label=., leftmargin=*]
   2. C2H5NH2 → C2H5OH (using HNO2)
   3. C2H5OH → CH3COOH (using K2Cr2O7/H2SO4)
   4. CH3COOH → CH3CONH2 (using NH3, Δ)
   5. CH3CONH2 → CH3NH2 (using Br2, NaOH: Hofmann).
5. (v) Ethanoic acid → propanoic acid. Need to add one C. Reduce to ethanol, convert to ethyl bromide, substitute with cyanide, hydrolyse:
   1. [label=., leftmargin=*]
   2. CH3COOH → CH3CH2OH (LiAlH4)
   3. CH3CH2OH → CH3CH2Br (PBr3)
   4. CH3CH2Br → CH3CH2CN (KCN)
   5. CH3CH2CN → CH3CH2COOH (H2O/H+).
6. (vi) Methanamine → ethanamine. Need to add one C to the amine chain. Convert to methanol via nitrous acid, then as in (iii) and (v):
   1. [label=., leftmargin=*]
   2. CH3NH2 → CH3OH (HNO2)
   3. CH3OH → CH3I (HI)
   4. CH3I → CH3CN (KCN)
   5. CH3CN → CH3CH2NH2 (LiAlH4).
   (Direct alkylation of methylamine with CH3I would also form ethyl-substituted amines, but C2H5 on N means the C is on N not in the chain, so that route does not give ethanamine. Hence the longer sequence above.)
7. (vii) Nitromethane → dimethylamine. Reduce nitromethane to methanamine first, then alkylate: CH3NO2 Sn/HCl CH3NH2 CH3I, Δ (CH3)2NH.
8. (viii) Propanoic acid → ethanoic acid. Need to lose one C. Use Hofmann route via the amide, then re-oxidise the resulting amine back to the acid:
   1. [label=., leftmargin=*]
   2. C2H5COOH → C2H5CONH2 (NH3, Δ)
   3. C2H5CONH2 → C2H5NH2 (Br2, NaOH: Hofmann removes the -CONH2 carbon)
   4. C2H5NH2 → C2H5OH (HNO2, 273 K)
   5. C2H5OH → CH3COOH (K2Cr2O7/H2SO4).
   Carbon balance: propanoic acid (3 C) → propanamide (3 C) → ethanamine (2 C, Hofmann drops the carbonyl C) → ethanol (2 C) → ethanoic acid (2 C). Net change: -1 C as required.

Use Hofmann bromamide whenever chain shortens by one carbon, and a cyanide → nitrile hydrolysis whenever chain lengthens by one carbon.

Concept used. The standard laboratory method is Hinsberg's test, which uses benzenesulphonyl chloride (C6H5SO2Cl, also called Hinsberg's reagent). The reagent reacts only with N–H bonds, and the solubility of the resulting sulphonamide in alkali separates the three amine classes cleanly.

• Primary amine + Hinsberg's reagent. The two N–H bonds allow loss of one HCl and formation of a sulphonamide R-NH-SO2C6H5. The N–H still left on the sulphonamide is acidic (because of the electron-withdrawing sulphonyl) and is removed by KOH/NaOH to give a salt that is soluble in alkali.
• Secondary amine + Hinsberg's reagent. Only one N–H is available; after substitution, the sulphonamide R2N-SO2C6H5 has no N–H. Hence it does not dissolve in alkali ⇒ a solid insoluble in alkali.
• Tertiary amine + Hinsberg's reagent. No N–H to lose ⇒ no reaction. The amine forms a separate layer; on acidifying, the tertiary amine dissolves in dilute HCl as R3NH+Cl-.

1. Step 1: Mix. Add a few drops of Hinsberg's reagent (C6H5SO2Cl) to the unknown amine, then add aqueous KOH (excess) and shake.
2. Step 2: Reaction with 1^∘ amine. The amine substitutes on the sulphur, losing HCl: R-NH2 + C6H5SO2Cl -> R-NH-SO2C6H5 + HCl. The N–H of the sulphonamide is acidic. KOH deprotonates it: R-NH-SO2C6H5 + KOH -> R-N(K)-SO2C6H5 + H2O. The potassium salt is ionic and dissolves ⇒ clear solution.
3. Step 3: Reaction with 2^∘ amine. R2NH + C6H5SO2Cl -> R2N-SO2C6H5 + HCl. The sulphonamide has no N–H, so KOH cannot deprotonate it. It stays as an undissolved solid ⇒ insoluble in alkali.
4. Step 4: Reaction with 3^∘ amine. No N–H, so no reaction with the sulphonyl chloride. On acidifying with dilute HCl, R3N + HCl -> R3N+HCl- (dissolves), and the tertiary amine goes into the aqueous layer.
5. Step 5: Recover the pure amine. The 1^∘ amine salt in alkali, on acidifying with HCl, releases the sulphonamide solid; further hydrolysis with concentrated HCl regenerates the original 1^∘ amine. The 2^∘ sulphonamide is filtered off and similarly hydrolysed to give back the 2^∘ amine.

Hinsberg's test: 1^∘ → soluble in KOH; 2^∘ → insoluble in KOH; 3^∘ → no reaction with the reagent but dissolves in dilute HCl.

Concept used. Each named reaction has a single signature step plus a defining set of reagents.

1. (i) Carbylamine reaction. Test for 1^∘ amines. A 1^∘ amine (aliphatic or aromatic) on heating with chloroform and alcoholic KOH gives an isocyanide (carbylamine) with a foul, penetrating smell. 2^∘ and 3^∘ amines do not react. R-NH2 + CHCl3 + 3 KOH Δ R-NC + 3 KCl + 3 H2O. Mechanism summary: KOH dehydrohalogenates CHCl3 to give dichlorocarbene (CCl2 with a lone pair); the amine attacks the carbene, loses HCl twice to give the isocyanide.
   !%

   [See diagram in the PDF version]
2. (ii) Diazotisation. The conversion of a 1^∘ aromatic amine into an arenediazonium salt with cold NaNO2/HCl (the mixture supplies nitrous acid in situ, since HNO2 is too unstable to store): ArNH2 + HNO2 + HCl 273--278 K Ar-N#N⁺ Cl- + 2 H2O. Aliphatic primary amines undergo the same first step but the resulting diazonium ion is unstable and at once loses N2, giving an alcohol. Aryl diazonium salts can be isolated and stored cold (resonance-stabilised).
3. (iii) Hofmann's bromamide reaction. An amide is degraded with bromine and alkali to give a 1^∘ amine with one carbon less than the starting amide: R-CONH2 + Br2 + 4 NaOH Δ R-NH2 + Na2CO3 + 2 NaBr + 2 H2O. The mechanism proceeds through an N-bromoamide R-CONHBr, loss of HBr to a nitrene-like intermediate, migration of R from C to N (giving an isocyanate R-N=C=O), and finally hydrolysis of the isocyanate to the amine and CO2.
   !%

   [See diagram in the PDF version]
4. (iv) Coupling reaction. An aryl diazonium salt reacts with an electron-rich aromatic compound (phenol, naphthol, aniline) to give an azo compound containing the -N=N- bridge. The azo compound is intensely coloured (orange, red or yellow) because -N=N- extends the π-conjugation. C6H5N2+Cl- + C6H5OH NaOH/cold HO-C6H4-N=N-C6H5 + HCl. The reaction is electrophilic aromatic substitution: ArN2+ is a weak electrophile and attacks only highly activated rings (phenol, naphthol, N,N-dimethylaniline).
   0.92!%

   [See diagram in the PDF version]
5. (v) Ammonolysis (of alkyl halides). Direct heating of an alkyl halide with alcoholic ammonia under pressure gives a primary amine: R-X + NH3 -> R-NH2 + HX. Drawback: the 1^∘ amine formed is itself nucleophilic and competes with NH3 for the next R-X, giving 2^∘, 3^∘ and quaternary ammonium salts (a mixture). A large excess of NH3 biases the product toward 1^∘.
6. (vi) Acetylation. 1^∘ and 2^∘ amines react with acetic anhydride (or acetyl chloride) in pyridine to give an N-acetyl derivative (an amide): R-NH2 + (CH3CO)2O -> R-NHCOCH3 + CH3COOH. The amide is less reactive than the parent amine (the lone pair is delocalised onto the carbonyl), which is why acetylation is used as a protecting group for -NH2 during electrophilic aromatic substitution (e.g. bromination of aniline must protect -NH2 first or all three positions get brominated).
7. (vii) Gabriel phthalimide synthesis. A clean method for pure 1^∘ amines, free of 2^∘/3^∘ contamination. The steps:
   • Treat phthalimide with KOH to form potassium phthalimide (the N–H is acidic, pK_a ≈ 8, because of the two flanking -CO- groups).
   • React with R-X: an SN2 substitution gives an N-alkylphthalimide. The new C–N bond has no N–H left, so further alkylation cannot occur.
   • Hydrolyse with aqueous NaOH (or with hydrazine, the Ing–Manske variant) to release the pure 1^∘ amine and phthalic acid (or phthalhydrazide).
   Important caveat: aryl primary amines cannot be made this way because aryl halides do not undergo SN2 substitution with the soft phthalimide nucleophile.
   0.95!%

   [See diagram in the PDF version]

All seven name reactions concisely covered above with reagents, products, and a one-line mechanism.

Concept used. Three big handles run through all nine routes:

• Diazonium chemistry: ArN2+ is converted to ArCl (CuCl/HCl, Sandmeyer), ArBr (CuBr/HBr), ArF (HBF4, then heat - Balz–Schiemann), ArOH (warm water), ArCN (CuCN, Sandmeyer), or Ar-H (H3PO2, reduction).
• -NH2 protection (acetylation) before electrophilic aromatic substitution. The bare -NH2 is too activating and gives 2,4,6-trisubstitution. Acetylation masks it as -NHCOCH3, a moderate o/p-director.
• Hofmann (amide → amine, -1 C) and Clemmensen / Wolff reductions (C=O → CH2) for chain editing.

1. (i) Nitrobenzene → benzoic acid. Reduce to aniline, diazotise, replace by -CN (Sandmeyer), then hydrolyse the nitrile:
   1. [label=., leftmargin=*]
   2. C6H5NO2 → C6H5NH2 (Sn/HCl)
   3. C6H5NH2 → C6H5N2+Cl- (NaNO2, HCl, 273 K)
   4. C6H5N2+Cl- → C6H5CN (CuCN, Sandmeyer)
   5. C6H5CN → C6H5COOH (H2O, H+).
2. (ii) Benzene → m-bromophenol. The -Br and -OH are meta to each other. Use diazonium: start with benzene, nitrate to nitrobenzene, brominate (the -NO2 is m-directing) to get m-bromonitrobenzene, reduce to m-bromoaniline, diazotise, hydrolyse to phenol:
   1. [label=., leftmargin=*]
   2. C6H6 → C6H5NO2 (HNO3/H2SO4)
   3. C6H5NO2 → m-BrC6H4NO2 (Br2/FeBr3)
   4. m-BrC6H4NO2 → m-BrC6H4NH2 (Sn/HCl)
   5. m-BrC6H4NH2 → m-BrC6H4N2+Cl- (NaNO2, HCl, 273 K)
   6. m-BrC6H4N2+Cl- → m-BrC6H4OH (H2O, Δ).
3. (iii) Benzoic acid → aniline. Convert to amide, Hofmann-degrade: C6H5COOH NH3, Δ C6H5CONH2 Br2, NaOH C6H5NH2.
4. (iv) Aniline → 2,4,6-tribromofluorobenzene. Brominate aniline in aqueous Br2 (very activated ring, all three o/p-positions react) to get 2,4,6-tribromoaniline. Diazotise, treat with HBF4 (Balz–Schiemann):
   1. [label=., leftmargin=*]
   2. C6H5NH2 → 2,4,6-Br3C6H2NH2 (Br2/H2O)
   3. 2,4,6-Br3C6H2NH2 → 2,4,6-Br3C6H2N2+Cl- (NaNO2, HCl, 273 K)
   4. 2,4,6-Br3C6H2N2+Cl- → 2,4,6-Br3C6H2N2+BF4- (HBF4)
   5. 2,4,6-Br3C6H2N2+BF4- → 2,4,6-Br3C6H2F + N2 + BF3 (Δ, Balz–Schiemann).
5. (v) Benzyl chloride → 2-phenylethanamine. Substitute with cyanide, reduce the nitrile: C6H5CH2Cl KCN C6H5CH2CN H2/Ni or LiAlH4 C6H5CH2CH2NH2.
6. (vi) Chlorobenzene → p-chloroaniline. Direct amination of chlorobenzene needs very harsh conditions, so we go via nitration: the -Cl is mildly o/p-directing, so nitration gives a mixture of o- and p-isomers; separate and reduce: C6H5Cl HNO3/H2SO4 p-ClC6H4NO2 Sn/HCl p-ClC6H4NH2.
7. (vii) Aniline → p-bromoaniline. Direct bromination of aniline gives 2,4,6-tribromoaniline (over-bromination). Protect -NH2 first by acetylation; then brominate (acetanilide gives mainly p-product); finally hydrolyse the amide back: aligned C6H5NH2 &(CH3CO)2O C6H5NHCOCH3 Br2/CH3COOH p-BrC6H4NHCOCH3
   &H+/H2O p-BrC6H4NH2. aligned
8. (viii) Benzamide → toluene. The standard NCERT route uses Hofmann to make aniline, the diazonium switchboard to strip the nitrogen, and Friedel–Crafts alkylation to add the methyl group:
   1. [label=., leftmargin=*]
   2. C6H5CONH2 → C6H5NH2 (Br2, NaOH; Hofmann, -1 C)
   3. C6H5NH2 → C6H5N2+Cl- (NaNO2/HCl, 273 K)
   4. C6H5N2+Cl- + H3PO2 + H2O → C6H6 + N2 + H3PO3 + HCl (reductive deamination)
   5. C6H6 + CH3Cl anhyd. AlCl3 C6H5CH3 + HCl (Friedel–Crafts alkylation).
   Net: benzamide (7 C) → aniline (6 C) → benzene (6 C) → toluene (7 C). The first step trims the amide carbon with Hofmann; the last step reinstalls a methyl on the ring.
9. (ix) Aniline → benzyl alcohol. Diazotise, treat with CuCN (Sandmeyer) to put a -CN on the ring, hydrolyse to benzoic acid, reduce to benzyl alcohol:
   1. [label=., leftmargin=*]
   2. C6H5NH2 → C6H5N2+Cl- (NaNO2/HCl, 273 K)
   3. C6H5N2+Cl- → C6H5CN (CuCN, Sandmeyer)
   4. C6H5CN → C6H5COOH (H2O/H+)
   5. C6H5COOH → C6H5CH2OH (LiAlH4).

Each conversion uses one or two of: diazonium replacement (Sandmeyer / Gattermann / Balz–Schiemann / H3PO2 / coupling), -NH2 protection by acetylation, or Hofmann bromamide for -1 C chain change.

Concept used. Each sequence chains together standard transformations: nucleophilic substitution by cyanide (+1 C), nitrile hydrolysis to acid or amide, amide to amine (Hofmann), nitro to amine (reduction), amine to diazonium (cold NaNO2/HCl), diazonium to phenol / coupling product.

Read each step's reagent and ``unfold'' the structure of the intermediate.

1. (i) Start: CH3CH2I. With NaCN (SN2): A = CH3CH2CN (propanenitrile). Partial alkaline hydrolysis (one equivalent of water on C#N): the nitrile becomes an amide, B = CH3CH2CONH2 (propanamide). Treatment with NaOBr (i.e. Br2/NaOH, Hofmann bromamide): C = CH3CH2NH2 (ethanamine). Net: 2 C → 3 C → 3 C → 2 C (the Hofmann step removes the carbonyl carbon). A = CH3CH2CN, B = CH3CH2CONH2, C = CH3CH2NH2
2. (ii) Start: C6H5N2+Cl-. With CuCN (Sandmeyer): A = C6H5CN (benzonitrile). Hydrolysis with H2O/H+: B = C6H5COOH (benzoic acid). With NH3 and Δ: ammonium salt → amide, C = C6H5CONH2 (benzamide). A = C6H5CN, B = C6H5COOH, C = C6H5CONH2
3. (iii) Start: CH3CH2Br. With KCN: A = CH3CH2CN. Reduction with LiAlH4 adds two hydrogens to the C of C#N: B = CH3CH2CH2NH2 (propan-1-amine). Cold HNO2 (i.e. NaNO2/HCl) on an aliphatic 1^∘ amine gives the alcohol with loss of N2: C = CH3CH2CH2OH (propan-1-ol). A = CH3CH2CN, B = CH3CH2CH2NH2, C = CH3CH2CH2OH
4. (iv) Start: C6H5NO2. With Fe/HCl: A = C6H5NH2 (aniline). Cold NaNO2/HCl at 273 K: B = C6H5N2+Cl- (benzenediazonium chloride). H2O/H+ (warm): C = C6H5OH (phenol). A = C6H5NH2, B = C6H5N2+Cl-, C = C6H5OH
5. (v) Start: CH3COOH. With NH3 and Δ: A = CH3CONH2 (acetamide). With NaOBr (Hofmann): B = CH3NH2 (methanamine). NaNO2/HCl on a 1^∘ aliphatic amine: C = CH3OH (methanol). A = CH3CONH2, B = CH3NH2, C = CH3OH
6. (vi) Start: C6H5NO2. With Fe/HCl: A = C6H5NH2. Cold HNO2 at 273 K: B = C6H5N2+Cl-. With phenol C6H5OH in mildly alkaline solution: coupling reaction, giving the p-hydroxyazobenzene dye, C = p-HOC6H4-N=N-C6H5 (an orange dye). A = C6H5NH2, B = C6H5N2+Cl-, C = p-HO-C6H4-N=N-C6H5

See the six boxed answers above.

Concept used. The clue is the final product's molecular formula, C6H7N. Degree of unsaturation: DBE = 2(6)+2-7+12 = 4, which is exactly the DBE of a benzene ring with no other unsaturations. C6H7N with one ring is aniline, C6H5NH2.

Working backwards: the last step is Br2/KOH, a Hofmann bromamide reaction. So B must be a benzamide, C6H5CONH2, which on Hofmann gives aniline (chain shortens by one C). The first step, ``aromatic + aqueous NH3 + heat'' → amide, points to a benzoic-acid derivative. The simplest fit is benzoic acid itself.

1. Identify C. Molecular formula C6H7N with one ring and one nitrogen ⇒ aniline (C6H5NH2). IUPAC name: aniline (or benzenamine).
2. Identify B. The reaction B + Br2 + KOH → C is Hofmann bromamide. C has 6 carbons; the amide precursor has 7 carbons (C_n + carbonyl C). So B is C6H5CONH2, benzamide. IUPAC name: benzamide. C6H5CONH2 + Br2 + 4 KOH Δ C6H5NH2 + K2CO3 + 2 KBr + 2 H2O.
3. Identify A. ``A + aqueous NH3 + heat → benzamide'' is the standard route from an acid (or acid chloride / ester) to an amide. The simplest aromatic precursor is benzoic acid: C6H5COOH + NH3 Δ C6H5CONH2 + H2O. IUPAC name of A: benzoic acid.

A = C6H5COOH (benzoic acid); B = C6H5CONH2 (benzamide); C = C6H5NH2 (aniline).

Concept used. Each reaction is one of the named amine / diazonium transformations. The required products follow directly from the reagent set.

1. (i) Carbylamine reaction. Aniline + CHCl3 + alc. KOH on heating gives phenyl isocyanide (a foul-smelling liquid): C6H5NH2 + CHCl3 + 3 KOH Δ C6H5NC + 3 KCl + 3 H2O.
2. (ii) Reduction of diazonium with hypophosphorous acid. H3PO2 replaces the -N2+ group by -H, liberating N2: C6H5N2+Cl- + H3PO2 + H2O -> C6H6 + N2 + H3PO3 + HCl.
3. (iii) Aniline + conc. H2SO4 → anilinium bisulphate, then sulphonation. On warming, the salt rearranges and sulphonation occurs at the para position. The final product is sulphanilic acid (p-aminobenzenesulphonic acid), which exists as a zwitterion: C6H5NH2 + H2SO4 (conc.) Δ p-NH2C6H4SO3H (sulphanilic acid).
4. (iv) Diazonium + ethanol. Ethanol acts as a reducing agent (like H3PO2): replaces -N2+ by -H and is itself oxidised to acetaldehyde: C6H5N2+Cl- + C2H5OH -> C6H6 + N2 + CH3CHO + HCl.
5. (v) Aniline + aqueous Br2. The -NH2 group is so strongly activating that all three free o/p-positions get brominated, giving the white precipitate 2,4,6-tribromoaniline: C6H5NH2 + 3 Br2 aq 2,4,6-Br3C6H2NH2 + 3 HBr.
6. (vi) Acetylation. Aniline + acetic anhydride gives acetanilide (N-phenylethanamide): C6H5NH2 + (CH3CO)2O -> C6H5NHCOCH3 + CH3COOH.
7. (vii) Two-step sequence ((i) HBF4, then (ii) NaNO2/Cu, Δ). Step (i) converts benzenediazonium chloride to the stable, isolable benzenediazonium tetrafluoroborate C6H5N2+BF4- (HCl is released). Step (ii) is a Sandmeyer-type substitution: the warm NaNO2/Cu pair replaces the -N2+ group with -NO2, giving nitrobenzene with evolution of N2. Net transformation: C6H5N2+Cl- + HBF4 -> C6H5N2+BF4- + HCl, C6H5N2+BF4- + NaNO2 Cu, Δ C6H5NO2 + N2 + NaBF4. Overall final product: nitrobenzene (C6H5NO2).

(i) Phenyl isocyanide. (ii) Benzene. (iii) Sulphanilic acid. (iv) Benzene + acetaldehyde. (v) 2,4,6-tribromoaniline. (vi) Acetanilide. (vii) Nitrobenzene (C6H5NO2) + N2 + NaBF4 (the diazonium tetrafluoroborate intermediate is decomposed by warm NaNO2/Cu, substituting -N2+ by -NO2).

Concept used. The Gabriel phthalimide synthesis proceeds by an SN2 reaction between potassium phthalimide (the soft, stabilised nitrogen nucleophile) and an alkyl halide R-X. The mechanism is bimolecular, requiring backside attack of the nucleophile on the carbon bearing the leaving group.

1. Step 1: Identify the required mechanism. Gabriel needs the phthalimide anion to displace the halide from R-X in an SN2 step.
2. Step 2: Examine the aryl halide. In an aryl halide Ar-X, the halogen is bonded to an sp²-hybridised carbon of the benzene ring. The C–X bond has partial double-bond character because of resonance: the lone pair on the halogen donates into the ring. This makes the C–X bond shorter, stronger and harder to break.
3. Step 3: SN2 on sp² carbon is forbidden. The nucleophile must approach from the back of the C–X bond, but on a benzene ring the back side is occupied by the ring's π-electron cloud, which repels any incoming nucleophile. Hence aryl halides do not undergo SN2 substitution with phthalimide.
4. Step 4: Conclusion. Since Gabriel needs SN2 and aryl halides cannot give SN2, the synthesis is not applicable to aryl-1^∘ amines (like aniline). They are prepared by other routes (reduction of nitrobenzene with Fe/HCl, or Hofmann bromamide from benzamide).

Aryl halides do not undergo SN2 substitution (the C–X bond is shortened/strengthened by resonance, and the back side is blocked by the ring's π-cloud), so the phthalimide anion cannot displace the halide. Hence aniline and other aryl-1^∘ amines are not accessible by Gabriel synthesis.

Concept used. Nitrous acid is generated in situ from sodium nitrite and a mineral acid (HCl or H2SO4) at 0–5 ^∘C: NaNO2 + HCl -> HNO2 + NaCl. The reaction proceeds by formation of the nitrosonium ion NO+, which attacks the lone pair of the primary amine.

1. (i) Aromatic primary amines + HNO2. Aniline (and other aryl-1^∘ amines) with cold NaNO2/HCl at 273–278 K give a stable arenediazonium chloride. The -N2+ ion in C6H5N2+ is stabilised by resonance delocalisation of the + charge into the ring: C6H5NH2 + HNO2 + HCl 273--278 K C6H5N2+ Cl- + 2 H2O. The salt is isolable and is the key intermediate for Sandmeyer, Gattermann, Balz–Schiemann and coupling reactions.
2. (ii) Aliphatic primary amines + HNO2. An alkyl-1^∘ amine reacts with HNO2 to form the same type of diazonium ion, but the aliphatic R-N2+ is not stabilised by any π-system and falls apart at once, releasing N2 and giving an alcohol (with rearranged / minor olefin / halide side-products): R-NH2 + HNO2 -> R-OH + N2 + H2O. For example, ethanamine gives ethanol with vigorous effervescence of N2: C2H5NH2 + HNO2 -> C2H5OH + N2 + H2O. The brisk effervescence of N2 is a positive test for aliphatic 1^∘ amines.

(i) C6H5NH2 + HNO2 + HCl 273--278 K C6H5N2+Cl- + 2 H2O (stable salt). (ii) R-NH2 + HNO2 -> R-OH + N2 + H2O (effervescence of N2).

Concept used. Three different acid/base/physical comparisons, each resting on a single electronic or H-bonding argument.

1. (i) Acidity of amines vs alcohols. The acidity of R-Z-H depends on the stability of the conjugate base R-Z-. For an alcohol the conjugate base is R-O- (alkoxide); for an amine it is R-N-H- (amide ion). Oxygen is more electronegative than nitrogen (3.5 vs 3.0 on the Pauling scale), so O- holds the negative charge much more comfortably than N-:
   • pK_a of ethanol ≈ 16 (mildly acidic).
   • pK_a of ethylamine ≈ 35 (essentially not acidic in water).
   Comparing molecules of similar molar mass (e.g. ethanol vs propylamine, both around 60 g/mol), the alcohol is far more acidic because the alkoxide is much more stable than the corresponding aminide. Hence amines are less acidic.
2. (ii) Boiling point: 1^∘ vs 3^∘ amines. Boiling point is largely set by intermolecular forces, the strongest of which (for similar molar mass) is hydrogen bonding. Hydrogen bonding requires an N–H (donor) and an N lone pair (acceptor).
   • A primary amine has two N–H bonds per molecule: it can act as a donor twice, forming a 3D H-bond network. Boiling point is high.
   • A tertiary amine has zero N–H bonds (the N is fully substituted): it can only accept H-bonds, not donate them. So tertiary amines form no H-bond network and rely only on dipole-dipole and London forces. Boiling point is much lower.
   Numerical example: propan-1-amine (CH3CH2CH2NH2, M = 59) b.p. ≈ 49 ^∘C; trimethylamine ((CH3)3N, same M = 59) b.p. ≈ 3 ^∘C. A 46-^∘C difference at identical mass shows the dominant role of N–H hydrogen bonding.
3. (iii) Aliphatic amines > aromatic amines in basicity. Basicity reflects the availability of the nitrogen lone pair for protonation, plus stabilisation of the conjugate acid R-NH3+.
   • In aliphatic amines (e.g. CH3NH2), alkyl groups push electrons toward N by the +I effect, making the lone pair more available, and the conjugate acid CH3NH3+ is well solvated by water (three N–H donor bonds).
   • In aromatic amines (e.g. aniline, C6H5NH2), the lone pair on N is conjugated into the benzene ring and delocalised over the o and p carbons. So the lone pair is much less available for protonation. Furthermore, on protonation the aniline loses this resonance stabilisation (the N is now sp³ with no free lone pair), so the conjugate acid is destabilised relative to the free base. Both effects make aniline a weaker base than methylamine (pK_b 9.38 vs 3.38, a factor of 10⁶).

(i) Alkoxide RO- is more stable than amide RNH- because O is more electronegative than N, so alcohols are far more acidic. (ii) Primary amines form a 3D H-bond network via two N–H donors each; tertiary amines have no N–H donors and rely only on weaker forces. (iii) The aromatic ring delocalises the N lone pair in arylamines, lowering basicity by orders of magnitude.