Intext Questions
1. Classify the following amines as primary, secondary or tertiary: (i) 1-naphthylamine, C₁₀H₇–NH₂ (ii) C₁₀H₇–N(CH₃)₂ (iii) (C₂H₅)₂CHNH₂ (iv) (C₂H₅)₂NH
Answer
(i) Primary (ii) Tertiary (iii) Primary (–NH₂ attached to a carbon; N carries one carbon group) (iv) Secondary
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) and (ii) Eight isomers:
1. CH₃CH₂CH₂CH₂NH₂ – Butan-1-amine (1°)
2. CH₃CH₂CH(NH₂)CH₃ – Butan-2-amine (1°, chiral)
3. (CH₃)₂CHCH₂NH₂ – 2-Methylpropan-1-amine (1°)
4. (CH₃)₃CNH₂ – 2-Methylpropan-2-amine (1°)
5. CH₃CH₂CH₂NHCH₃ – N-Methylpropan-1-amine (2°)
6. (CH₃)₂CHNHCH₃ – N-Methylpropan-2-amine (2°)
7. CH₃CH₂NHCH₂CH₃ – N-Ethylethanamine (2°)
CH₃CH₂N(CH₃)₂ – N,N-Dimethylethanamine (3°)
Answer
(iii) Chain isomerism (1 and 3; 2 and 4), position isomerism (1 and 2; 3 and 4; 5 and 6), metamerism (5 and 7, or 6 and 7 – different groups on N), functional isomerism (primary vs secondary vs tertiary, e.g. 1, 5 and 8). Butan-2-amine also shows optical isomerism.
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) C₆H₆ —conc. HNO₃/conc. H₂SO₄
⇒ C₆H₅NO₂ —Sn/HCl (or Fe/HCl), then NaOH
⇒ C₆H₅NH₂
(ii) C₆H₆
⇒ C₆H₅NO₂
⇒ C₆H₅NH₂ (as above); C₆H₅NH₂ + 2CH₃I —Na₂CO₃
⇒ C₆H₅N(CH₃)₂ + 2HI (neutralised)
(iii) Cl–(CH₂)₄–Cl + 2KCN (ethanolic)
⇒ NC–(CH₂)₄–CN + 2KCl; NC–(CH₂)₄–CN —H₂/Ni (or LiAlH₄)
⇒ H₂N–CH₂–(CH₂)₄–CH₂–NH₂ (hexane-1,6-diamine)
4. Arrange the following in increasing order of their basic strength: (i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH (ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂ (iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂.
Answer
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
In aniline the N lone pair is delocalised into the ring, so it is the weakest base. In water, basicity of alkylamines depends on +I effect, solvation of the ammonium ion and steric hindrance together, giving the orders shown.
5. Complete the following acid-base reactions and name the products: (i) CH₃CH₂CH₂NH₂ + HCl → (ii) (C₂H₅)₃N + HCl →
Answer
(i) CH₃CH₂CH₂NH₃⁺Cl⁻ – propylammonium chloride (propan-1-aminium chloride)
(ii) (C₂H₅)₃NH⁺Cl⁻ – triethylammonium chloride
6. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Answer
Aniline is methylated stepwise; Na₂CO₃ neutralises the HI formed. The final product is the quaternary salt:
C₆H₅NH₂ + CH₃I
⇒ C₆H₅NHCH₃ + HI; C₆H₅NHCH₃ + CH₃I
⇒ C₆H₅N(CH₃)₂ + HI; C₆H₅N(CH₃)₂ + CH₃I
⇒ C₆H₅N⁺(CH₃)₃I⁻ (N,N,N-trimethylanilinium iodide)
2HI + Na₂CO₃ → 2NaI + H₂O + CO₂
7. Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer
C₆H₅NH₂ + C₆H₅COCl —pyridine
⇒ C₆H₅NHCOC₆H₅ + HCl. Product: N-phenylbenzamide (benzanilide). This is benzoylation.
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.
Answer
Isomers: CH₃CH₂CH₂NH₂ (propan-1-amine), CH₃CH(NH₂)CH₃ (propan-2-amine), CH₃CH₂NHCH₃ (N-methylethanamine), (CH₃)₃N (N,N-dimethylmethanamine).
Only primary aliphatic amines liberate N₂ with HNO₂: propan-1-amine and propan-2-amine.
CH₃CH₂CH₂NH₂ + HNO₂
⇒ CH₃CH₂CH₂OH + N₂ + H₂O; (CH₃)₂CHNH₂ + HNO₂
⇒ (CH₃)₂CHOH + N₂ + H₂O
9. Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5-tribromobenzene.
Answer
(i) m-CH₃C₆H₄NH₂ —NaNO₂/HCl, 273–278 K
⇒ m-CH₃C₆H₄N₂⁺Cl⁻ —HBF₄
⇒ m-CH₃C₆H₄N₂⁺BF₄⁻ —NaNO₂/Cu, heat
⇒ m-CH₃C₆H₄NO₂ + N₂ + NaBF₄
(ii) C₆H₅NH₂ —Br₂(aq)
⇒ 2,4,6-tribromoaniline —NaNO₂/HCl, 273–278 K
⇒ 2,4,6-tribromobenzenediazonium chloride —H₃PO₂/H₂O
⇒ 1,3,5-tribromobenzene + N₂
Exercises
1. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH₃)₂CHNH₂ (ii) CH₃(CH₂)₂NH₂ (iii) CH₃NHCH(CH₃)₂ (iv) (CH₃)₃CNH₂ (v) C₆H₅NHCH₃ (vi) (CH₃CH₂)₂NCH₃ (vii) m–BrC₆H₄NH₂
Answer
(i) Propan-2-amine – primary
(ii) Propan-1-amine – primary
(iii) N-Methylpropan-2-amine – secondary
(iv) 2-Methylpropan-2-amine – primary
(v) N-Methylbenzenamine (N-methylaniline) – secondary
(vi) N-Ethyl-N-methylethanamine – tertiary
(vii) 3-Bromobenzenamine (3-bromoaniline) – primary
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) Carbylamine test: warm with CHCl₃ and alcoholic KOH. Methylamine (1°) gives the foul-smelling methyl isocyanide; dimethylamine does not. CH₃NH₂ + CHCl₃ + 3KOH
⇒ CH₃NC + 3KCl + 3H₂O
(ii) Hinsberg's test (benzenesulphonyl chloride): a secondary amine gives an N,N-dialkylbenzenesulphonamide, insoluble in alkali; a tertiary amine does not react.
(iii) Azo-dye test: treat with NaNO₂/HCl at 273–278 K, then alkaline β-naphthol. Aniline gives an orange-red azo dye; ethylamine gives N₂ gas (brisk effervescence) and no dye.
(iv) Nitrous acid at low temperature: benzylamine (aliphatic-type amine) liberates N₂ immediately (C₆H₅CH₂NH₂ + HNO₂
⇒ C₆H₅CH₂OH + N₂ + H₂O); aniline forms a diazonium salt that couples with β-naphthol to give an orange dye.
(v) Carbylamine test: aniline (1°) gives the offensive smell of phenyl isocyanide; N-methylaniline (2°) does not.
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 N is delocalised into the benzene ring by resonance, so it is less available for protonation; also the anilinium ion has no such resonance stabilisation. In methylamine the +I effect of CH₃ raises electron density on N. So aniline is a weaker base (higher pKb, 9.38 vs 3.38).
(ii) Ethylamine forms hydrogen bonds with water. In aniline the large hydrophobic phenyl group outweighs the hydrogen bonding of –NH₂, so it is only sparingly soluble.
(iii) Methylamine is a stronger base than water and produces OH⁻ ions: CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻. These precipitate Fe³⁺: FeCl₃ + 3OH⁻
⇒ Fe(OH)₃ (hydrated ferric oxide, Fe₂O₃·xH₂O) + 3Cl⁻.
(iv) Nitration is done in strongly acidic medium, where much of the aniline is protonated to the anilinium ion, C₆H₅NH₃⁺. The –NH₃⁺ group is deactivating and meta-directing, so about 47% m-nitroaniline forms (with 51% para and 2% ortho).
(v) Aniline, a Lewis base, reacts with the Lewis acid catalyst AlCl₃ to form a salt. The N then carries a positive charge, which strongly deactivates the ring, so the reaction does not occur.
(vi) In arenediazonium ions the positive charge is delocalised over the benzene ring by resonance, stabilising them (at 273–278 K). Alkyldiazonium ions lack this resonance and decompose at once, releasing N₂.
(vii) It gives pure primary amines with no contamination by secondary or tertiary amines (which ammonolysis of alkyl halides produces), because the phthalimide nitrogen can be alkylated only once.
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) C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH
(iii) (a) p-Nitroaniline < Aniline < p-Toluidine (–NO₂ withdraws, –CH₃ donates electrons) (b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂ (in benzylamine the lone pair is not delocalised into the ring)
(iv) (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ (in gas phase there is no solvation, so basicity follows the +I effect)
(v) (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH (the 2° amine has one N–H for H-bonding, the 1° amine two; O–H bonds are more polar than N–H)
(vi) C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂ (solubility falls as the hydrocarbon part grows)
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?
Answer
(i) CH₃COOH —NH₃, heat
⇒ CH₃CONH₂ —Br₂/NaOH
⇒ CH₃NH₂
(ii) CH₃(CH₂)₄CN —H₂O/H⁺ (partial hydrolysis)
⇒ CH₃(CH₂)₄CONH₂ —Br₂/KOH
⇒ CH₃(CH₂)₄NH₂ (pentan-1-amine)
(iii) CH₃OH —SOCl₂ (or PCl₅)
⇒ CH₃Cl —KCN
⇒ CH₃CN —H₃O⁺, heat
⇒ CH₃COOH
(iv) CH₃CH₂NH₂ —HNO₂
⇒ CH₃CH₂OH —KMnO₄/H⁺
⇒ CH₃COOH —NH₃, heat
⇒ CH₃CONH₂ —Br₂/KOH
⇒ CH₃NH₂
(v) CH₃COOH —LiAlH₄
⇒ CH₃CH₂OH —SOCl₂
⇒ CH₃CH₂Cl —KCN
⇒ CH₃CH₂CN —H₃O⁺
⇒ CH₃CH₂COOH
(vi) CH₃NH₂ —HNO₂
⇒ CH₃OH —SOCl₂
⇒ CH₃Cl —KCN
⇒ CH₃CN —LiAlH₄ (or Na/C₂H₅OH)
⇒ CH₃CH₂NH₂
(vii) CH₃NO₂ —Sn/HCl
⇒ CH₃NH₂ —CHCl₃/alc. KOH
⇒ CH₃NC —H₂/Pt (or Na/C₂H₅OH)
⇒ CH₃NHCH₃
(viii) CH₃CH₂COOH —NH₃, heat
⇒ CH₃CH₂CONH₂ —Br₂/KOH
⇒ CH₃CH₂NH₂ —HNO₂
⇒ CH₃CH₂OH —KMnO₄/H⁺
⇒ CH₃COOH
6. Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Answer
Hinsberg's test: the amine is shaken with benzenesulphonyl chloride (Hinsberg's reagent, C₆H₅SO₂Cl) and excess aqueous KOH.
Primary amine: forms an N-alkylbenzenesulphonamide whose N–H is acidic (strong electron-withdrawing –SO₂–), so it dissolves in KOH; acidifying gives a precipitate. C₆H₅SO₂Cl + C₂H₅NH₂
⇒ C₆H₅SO₂NHC₂H₅ + HCl; C₆H₅SO₂NHC₂H₅ + KOH
⇒ C₆H₅SO₂N⁻(C₂H₅)K⁺ (soluble) + H₂O
Secondary amine: forms an N,N-dialkylbenzenesulphonamide with no H on N, insoluble in alkali (precipitate remains). C₆H₅SO₂Cl + (C₂H₅)₂NH
⇒ C₆H₅SO₂N(C₂H₅)₂ + HCl
Tertiary amine: no H on N, so no reaction; the amine stays as an insoluble layer that dissolves on adding acid.
(The carbylamine test can also pick out primary amines: only they give the foul-smelling isocyanide.)
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 or aromatic) heated with chloroform and alcoholic KOH give isocyanides (carbylamines) with a very unpleasant smell; secondary and tertiary amines do not. It is a test for 1° amines. RNH₂ + CHCl₃ + 3KOH
⇒ RNC + 3KCl + 3H₂O
(ii) Diazotisation: conversion of a primary aromatic amine into a diazonium salt with nitrous acid (NaNO₂ + HCl) at 273–278 K. C₆H₅NH₂ + NaNO₂ + 2HCl
⇒ C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(iii) Hofmann bromamide degradation: an amide heated with Br₂ and aqueous/ethanolic NaOH gives a primary amine with one carbon atom less (the alkyl/aryl group migrates from carbonyl carbon to N). RCONH₂ + Br₂ + 4NaOH
⇒ RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O
(iv) Coupling reaction: arenediazonium salts react with phenol (mildly alkaline medium) or aniline (mildly acidic medium) at the para position to give coloured azo compounds (–N=N– linking two rings). C₆H₅N₂⁺Cl⁻ + C₆H₅OH —OH⁻
⇒ p-HOC₆H₄N=NC₆H₅ (p-hydroxyazobenzene, orange) + HCl; with aniline: p-aminoazobenzene (yellow).
(v) Ammonolysis: cleavage of the C–X bond of an alkyl halide by ammonia (ethanolic NH₃, sealed tube, 373 K); a mixture of 1°, 2°, 3° amines and quaternary salt results. RX + NH₃
⇒ RNH₂ + HX
⇒ R₂NH
⇒ R₃N
⇒ R₄N⁺X⁻
(vi) Acetylation: replacement of an H of –NH₂ or –NH by an acetyl group using acetyl chloride or acetic anhydride, in presence of pyridine (which removes HCl). C₆H₅NH₂ + (CH₃CO)₂O
⇒ C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH
(vii) Gabriel phthalimide synthesis: phthalimide is converted to its potassium salt with ethanolic KOH, which is heated with an alkyl halide to give N-alkylphthalimide; alkaline hydrolysis then gives a pure primary amine. C₆H₄(CO)₂NH + KOH
⇒ C₆H₄(CO)₂N⁻K⁺; + RX
⇒ C₆H₄(CO)₂NR —NaOH(aq)
⇒ RNH₂ + C₆H₄(COONa)₂
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.
Answer
(i) C₆H₅NO₂ —Sn/HCl
⇒ C₆H₅NH₂ —NaNO₂/HCl, 273 K
⇒ C₆H₅N₂⁺Cl⁻ —CuCN/KCN
⇒ C₆H₅CN —H₃O⁺, heat
⇒ C₆H₅COOH
(ii) C₆H₆ —HNO₃/H₂SO₄
⇒ C₆H₅NO₂ —Br₂/FeBr₃
⇒ m-BrC₆H₄NO₂ —Sn/HCl
⇒ m-BrC₆H₄NH₂ —NaNO₂/HCl, 273 K
⇒ m-BrC₆H₄N₂⁺Cl⁻ —H₂O, warm
⇒ m-BrC₆H₄OH
(iii) C₆H₅COOH —SOCl₂
⇒ C₆H₅COCl —NH₃
⇒ C₆H₅CONH₂ —Br₂/KOH
⇒ C₆H₅NH₂
(iv) C₆H₅NH₂ —Br₂(aq)
⇒ 2,4,6-Br₃C₆H₂NH₂ —NaNO₂/HCl, 273 K
⇒ 2,4,6-Br₃C₆H₂N₂⁺Cl⁻ —HBF₄
⇒ 2,4,6-Br₃C₆H₂N₂⁺BF₄⁻ —heat
⇒ 2,4,6-Br₃C₆H₂F + BF₃ + N₂
(v) C₆H₅CH₂Cl —KCN (ethanolic)
⇒ C₆H₅CH₂CN —LiAlH₄ (or H₂/Ni)
⇒ C₆H₅CH₂CH₂NH₂
(vi) C₆H₅Cl —conc. HNO₃/conc. H₂SO₄
⇒ p-ClC₆H₄NO₂ (separated from ortho isomer) —Sn/HCl
⇒ p-ClC₆H₄NH₂
(vii) C₆H₅NH₂ —(CH₃CO)₂O, pyridine
⇒ C₆H₅NHCOCH₃ —Br₂/CH₃COOH
⇒ p-BrC₆H₄NHCOCH₃ —H₂O/H⁺ (or OH⁻), heat
⇒ p-BrC₆H₄NH₂ (acetylation reduces the activating power of –NH₂, so only monobromination occurs)
(viii) C₆H₅CONH₂ —Br₂/KOH
⇒ C₆H₅NH₂ —NaNO₂/HCl, 273 K
⇒ C₆H₅N₂⁺Cl⁻ —H₃PO₂/H₂O
⇒ C₆H₆ —CH₃Cl, anhyd. AlCl₃
⇒ C₆H₅CH₃
(ix) C₆H₅NH₂ —NaNO₂/HCl, 273 K
⇒ C₆H₅N₂⁺Cl⁻ —CuCN/KCN
⇒ C₆H₅CN —LiAlH₄
⇒ C₆H₅CH₂NH₂ —HNO₂ (NaNO₂/HCl)
⇒ C₆H₅CH₂OH + N₂
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, heat→ C (iii) CH₃CH₂Br —KCN→ A —LiAlH₄→ B —HNO₂, 0 °C→ C (iv) C₆H₅NO₂ —Fe/HCl→ A —NaNO₂ + HCl, 273 K→ B —H₂O/H⁺, heat→ C (v) CH₃COOH —NH₃, heat→ A —NaOBr→ B —NaNO₂/HCl→ C (vi) C₆H₅NO₂ —Fe/HCl→ A —HNO₂, 273 K→ B —C₆H₅OH→ C
Answer
(i) A = CH₃CH₂CN (propanenitrile); B = CH₃CH₂CONH₂ (propanamide); C = CH₃CH₂NH₂ (ethanamine)
(ii) A = C₆H₅CN (benzonitrile); B = C₆H₅COOH (benzoic acid); C = C₆H₅CONH₂ (benzamide)
(iii) A = CH₃CH₂CN; B = CH₃CH₂CH₂NH₂ (propan-1-amine); C = CH₃CH₂CH₂OH (propan-1-ol)
(iv) A = C₆H₅NH₂ (aniline); B = C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride); C = C₆H₅OH (phenol)
(v) A = CH₃CONH₂ (ethanamide); B = CH₃NH₂ (methanamine); C = CH₃OH (methanol)
(vi) A = C₆H₅NH₂; B = C₆H₅N₂⁺Cl⁻; C = C₆H₅–N=N–C₆H₄–OH (p) (p-hydroxyazobenzene, an orange dye)
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.
Answer
C₆H₇N = C₆H₅NH₂, formed by Hofmann bromamide reaction, so B is the amide C₆H₅CONH₂ and A is the acid C₆H₅COOH.
A: C₆H₅COOH – benzoic acid (benzenecarboxylic acid)
B: C₆H₅CONH₂ – benzamide (benzenecarboxamide)
C: C₆H₅NH₂ – aniline (benzenamine)
C₆H₅COOH + NH₃
⇒ C₆H₅COONH₄ —heat
⇒ C₆H₅CONH₂ + H₂O; C₆H₅CONH₂ + Br₂ + 4KOH
⇒ C₆H₅NH₂ + K₂CO₃ + 2KBr + 2H₂O
11. Complete the following reactions: (i) C₆H₅NH₂ + CHCl₃ + alc. KOH → (ii) C₆H₅N₂Cl + H₃PO₂ + H₂O → (iii) C₆H₅NH₂ + H₂SO₄ (conc.) → (iv) C₆H₅N₂Cl + C₂H₅OH → (v) C₆H₅NH₂ + Br₂(aq) → (vi) C₆H₅NH₂ + (CH₃CO)₂O → (vii) C₆H₅N₂Cl —(i) HBF₄ (ii) NaNO₂/Cu, Δ→
Answer
(i) C₆H₅NH₂ + CHCl₃ + 3KOH
⇒ C₆H₅NC + 3KCl + 3H₂O
(ii) C₆H₅N₂Cl + H₃PO₂ + H₂O
⇒ C₆H₆ + N₂ + H₃PO₃ + HCl
(iii) C₆H₅NH₂ + H₂SO₄
⇒ C₆H₅NH₃⁺HSO₄⁻ (anilinium hydrogensulphate); on heating at 453–473 K it gives p-aminobenzenesulphonic acid (sulphanilic acid), which exists as a zwitterion.
(iv) C₆H₅N₂Cl + CH₃CH₂OH
⇒ C₆H₆ + N₂ + CH₃CHO + HCl
(v) C₆H₅NH₂ + 3Br₂
⇒ 2,4,6-tribromoaniline (white ppt) + 3HBr
(vi) C₆H₅NH₂ + (CH₃CO)₂O
⇒ C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH
(vii) C₆H₅N₂⁺Cl⁻ + HBF₄
⇒ C₆H₅N₂⁺BF₄⁻ + HCl; C₆H₅N₂⁺BF₄⁻ + NaNO₂ —Cu, heat
⇒ C₆H₅NO₂ + N₂ + NaBF₄
12. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer
The method needs the phthalimide anion to displace a halide from the organic halide by nucleophilic substitution. Aryl halides do not undergo nucleophilic substitution under these conditions (C–X bond has partial double-bond character), so N-arylphthalimide cannot be formed and aryl amines are not obtained.
13. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Answer
(i) Aromatic: form fairly stable diazonium salts at 273–278 K: C₆H₅NH₂ + NaNO₂ + 2HCl
⇒ C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(ii) Aliphatic: form unstable alkyldiazonium salts which decompose at once, giving alcohols and nitrogen gas: RNH₂ + HNO₂ —(NaNO₂ + HCl)
⇒ [RN₂⁺Cl⁻] —H₂O
⇒ ROH + N₂ + HCl. The quantitative evolution of N₂ is used to estimate amino acids and proteins.
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) Loss of a proton gives RNH⁻ from an amine and RO⁻ from an alcohol. Oxygen is more electronegative than nitrogen, so it holds the negative charge better; RO⁻ is more stable than RNH⁻ and alcohols are more acidic. Equivalently, the O–H bond is more polar than N–H.
(ii) Primary amines have two H atoms on N and form intermolecular hydrogen bonds; tertiary amines have no H on N and cannot, so they boil lower.
(iii) In aromatic amines the lone pair on N is delocalised into the benzene ring, making it less available for protonation; in aliphatic amines, alkyl groups push electrons (+I) toward N and the lone pair is fully available. So aliphatic amines are stronger bases.
Important Extra Questions
Why is the –NH₂ group of aniline acetylated before nitration or bromination?
Answer
Acetylation reduces the strong activating effect of –NH₂ (the lone pair also interacts with the C=O of the acetyl group), preventing polysubstitution and oxidation, and protects the group from protonation, giving mainly the para product.
Why is methylamine a stronger base than ammonia?
Answer
The +I effect of the CH₃ group increases electron density on N, making its lone pair more available for donating to a proton.
What is the Sandmeyer reaction?
Answer
Replacement of the diazonium group by Cl, Br or CN using Cu(I) salts: C₆H₅N₂⁺Cl⁻ —CuCl/HCl
⇒ C₆H₅Cl + N₂; —CuBr/HBr
⇒ C₆H₅Br; —CuCN/KCN
⇒ C₆H₅CN.
Why is benzene diazonium chloride not stored and used immediately after preparation?
Answer
It is unstable; it is highly soluble and decomposes on warming or in the dry state (explosive), so it is made at 273–278 K and used at once.
Name the reaction in which the Cu catalyst is used with HCl and diazonium salt, as distinct from Sandmeyer.
Answer
Gattermann reaction: C₆H₅N₂⁺Cl⁻ —Cu/HCl
⇒ C₆H₅Cl + N₂ (copper powder instead of cuprous halide).