Give one test for each pair: methylamine/dimethylamine; secondary/tertiary amine; ethylamine/aniline; aniline/benzylamine; aniline/N-methylaniline.
Steps only
Use the carbylamine test for primary versus secondary amine: only methylamine, ethylamine, aniline and benzylamine give foul-smelling isocyanides. Use Hinsberg test (benzenesulphonyl chloride/NaOH) for secondary versus tertiary: secondary forms an insoluble sulphonamide in alkali; tertiary does not react and dissolves in acid. Ethylamine gives an aliphatic diazonium product that releases N2 with nitrous acid; aniline forms a stable diazonium salt at 273–278 K. Aniline forms a diazonium salt with nitrous acid; benzylamine releases nitrogen. Aniline is primary and gives carbylamine; N-methylaniline is secondary and does not.
Answer
Use the carbylamine test for primary versus secondary amine: only methylamine, ethylamine, aniline and benzylamine give foul-smelling isocyanides. Use Hinsberg test (benzenesulphonyl chloride/NaOH) for secondary versus tertiary: secondary forms an insoluble sulphonamide in alkali; tertiary does not react and dissolves in acid. Ethylamine gives an aliphatic diazonium product that releases N2 with nitrous acid; aniline forms a stable diazonium salt at 273–278 K. Aniline forms a diazonium salt with nitrous acid; benzylamine releases nitrogen. Aniline is primary and gives carbylamine; N-methylaniline is secondary and does not.
Question 9.3
Explain the seven observations in Exercise 9.3.
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(i) Aniline is less basic because its N lone pair is delocalised into the benzene ring; its pKb is therefore higher than methylamine’s. (ii) Small ethylamine molecules hydrogen-bond with water; the large hydrophobic phenyl group makes aniline poorly soluble. (iii) Methylamine makes water basic, so Fe3+ forms hydrated Fe(OH)3 precipitate. (iv) Nitration uses strongly acidic medium, which protonates aniline to anilinium; this group directs meta, while some unprotonated aniline gives ortho/para products. (v) Aniline’s lone pair binds the AlCl3 catalyst, forming a salt and deactivating the ring. (vi) Aromatic diazonium ions are resonance-stabilised; aliphatic ones quickly lose N2. (vii) Gabriel synthesis gives primary alkyl amines cleanly and avoids over-alkylation; it does not make aryl amines.
Answer
(i) Aniline is less basic because its N lone pair is delocalised into the benzene ring; its pKb is therefore higher than methylamine’s.
(ii) Small ethylamine molecules hydrogen-bond with water; the large hydrophobic phenyl group makes aniline poorly soluble.
(iii) Methylamine makes water basic, so Fe3+ forms hydrated Fe(OH)3 precipitate.
(iv) Nitration uses strongly acidic medium, which protonates aniline to anilinium; this group directs meta, while some unprotonated aniline gives ortho/para products.
(v) Aniline’s lone pair binds the AlCl3 catalyst, forming a salt and deactivating the ring.
How can primary, secondary and tertiary amines be identified? Give the reaction idea.
Steps only
Treat with benzenesulphonyl chloride in aqueous NaOH. A primary amine forms R–NH–SO2Ph, which has an acidic N–H and dissolves in excess alkali as a salt; acidification precipitates it. A secondary amine forms R2N–SO2Ph with no N–H, so it is insoluble in alkali. A tertiary amine does not form a sulphonamide; it dissolves in dilute acid as an ammonium salt.
Answer
Treat with benzenesulphonyl chloride in aqueous NaOH. A primary amine forms R–NH–SO2Ph, which has an acidic N–H and dissolves in excess alkali as a salt; acidification precipitates it. A secondary amine forms R2N–SO2Ph with no N–H, so it is insoluble in alkali. A tertiary amine does not form a sulphonamide; it dissolves in dilute acid as an ammonium salt.
Question 9.7
Write short notes on carbylamine, diazotisation, Hofmann bromamide, coupling, ammonolysis, acetylation and Gabriel synthesis.
Steps only
Carbylamine: primary amine + CHCl3 + alcoholic KOH → isocyanide. Diazotisation: aromatic primary amine + NaNO2/HCl at 273–278 K → diazonium salt. Hofmann bromamide: amide + Br2/KOH → primary amine with one fewer carbon. Coupling: aryl diazonium salt couples with an activated ring (phenol/aniline) to form an azo dye. Ammonolysis: NH3 replaces halogen in an alkyl halide to form amines. Acetylation: an amine reacts with an acetylating agent to form an amide, protecting the –NH2 group. Gabriel synthesis: potassium phthalimide alkylation then hydrolysis gives a primary alkyl amine.
Answer
Carbylamine: primary amine + CHCl3 + alcoholic KOH → isocyanide. Diazotisation: aromatic primary amine + NaNO2/HCl at 273–278 K → diazonium salt. Hofmann bromamide: amide + Br2/KOH → primary amine with one fewer carbon. Coupling: aryl diazonium salt couples with an activated ring (phenol/aniline) to form an azo dye. Ammonolysis: NH3 replaces halogen in an alkyl halide to form amines. Acetylation: an amine reacts with an acetylating agent to form an amide, protecting the –NH2 group. Gabriel synthesis: potassium phthalimide alkylation then hydrolysis gives a primary alkyl amine.
Question 9.8
Give reagent sequences for the nine aromatic conversions in Exercise 9.8.
Steps only
(i) Nitrobenzene → aniline (reduction) → diazonium salt (NaNO2/HCl, 273 K) → benzonitrile (CuCN) → benzoic acid (hydrolysis). (ii) Benzene → nitrobenzene → m-bromonitrobenzene (Br2/FeBr3; NO2 directs meta) → m-bromoaniline (reduction) → diazonium salt → m-bromophenol (warm water). (iii) Benzoic acid → benzamide → aniline (Br2/KOH; Hofmann). (iv) Aniline → 2,4,6-tribromoaniline (Br2 water) → diazonium salt → replace diazonium group by F (HBF4, heat). (v) Benzyl chloride → phenylacetonitrile (KCN) → 2-phenylethanamine (LiAlH4). (vi) Chlorobenzene → p-nitrochlorobenzene (nitration; isolate para) → p-chloroaniline (reduction). (vii) Protect aniline as acetanilide, brominate to the para product, then hydrolyse the protecting group. (viii) Benzamide → benzylamine (LiAlH4) → benzyl alcohol (HNO2) → toluene (Zn dust). (ix) Aniline → diazonium salt → benzonitrile (CuCN) → benzaldehyde (partial reduction) → benzyl alcohol (NaBH4).
Why can aromatic primary amines not be prepared by Gabriel phthalimide synthesis?
Steps only
Aryl halides do not undergo the required SN2 substitution with phthalimide ion: the C–X bond is on an sp2 carbon and has partial double-bond character.
Answer
Aryl halides do not undergo the required SN2 substitution with phthalimide ion: the C–X bond is on an sp2 carbon and has partial double-bond character.
Question 9.13
Write reactions of primary aromatic and primary aliphatic amines with nitrous acid.
Explain why (i) amines are less acidic than similar alcohols, (ii) primary amines boil higher than tertiary amines, and (iii) aliphatic amines are stronger bases than aromatic amines.
Steps only
(i) Nitrogen is less electronegative than oxygen, so an N–H proton is less easily released. (ii) Primary amines have N–H bonds and form intermolecular hydrogen bonds; tertiary amines cannot donate such bonds. (iii) Alkyl groups push electron density toward nitrogen, while an aromatic ring delocalises the N lone pair.
Answer
(i) Nitrogen is less electronegative than oxygen, so an N–H proton is less easily released.
(ii) Primary amines have N–H bonds and form intermolecular hydrogen bonds; tertiary amines cannot donate such bonds.
(iii) Alkyl groups push electron density toward nitrogen, while an aromatic ring delocalises the N lone pair.