Conversions and distinguishing tests for amines
Amine conversions use key moves: Hoffmann to lose a carbon, nitrile reduction to gain one, and diazonium salts to change the ring. Carbylamine and azo dye tests identify amines.
Conversions the board likes
| From → to | Steps, with reagents |
|---|---|
| Nitrobenzene → benzoic acid | Sn + HCl → aniline; NaNO₂ + HCl, 273 to 278 K → C₆H₅N₂⁺Cl⁻; CuCN/KCN → C₆H₅CN; H₃O⁺, heat → C₆H₅COOH |
| Benzene → m-bromophenol | conc. HNO₃ + conc. H₂SO₄ → nitrobenzene; Br₂/FeBr₃ → m-bromonitrobenzene (-NO₂ is meta-directing); Sn + HCl → m-bromoaniline; NaNO₂ + HCl, 273 to 278 K, then warm water → m-bromophenol |
| Benzoic acid → aniline | NH₃, heat → benzamide; Br₂ + NaOH (Hoffmann) → aniline |
| Aniline → 1,3,5-tribromobenzene | Br₂ water → 2,4,6-tribromoaniline; NaNO₂ + HCl, 273 to 278 K → diazonium salt; H₃PO₂ → 1,3,5-tribromobenzene |
| Aniline → 2,4,6-tribromofluorobenzene | Br₂ water → 2,4,6-tribromoaniline; NaNO₂ + HCl, 273 to 278 K; HBF₄, then heat |
| Benzyl chloride → 2-phenylethanamine | KCN → C₆H₅CH₂CN; LiAlH₄ (or H₂/Ni) → C₆H₅CH₂CH₂NH₂ |
| Chlorobenzene → p-chloroaniline | conc. HNO₃ + conc. H₂SO₄ → p-chloronitrobenzene (major; Cl is ortho-para directing); Sn + HCl → p-chloroaniline |
| Ethanoic acid → methanamine (descent) | NH₃, heat → ethanamide; Br₂ + NaOH → CH₃NH₂ |
| Methanamine → ethanamine (ascent) | HNO₂ (NaNO₂ + HCl) → CH₃OH; PCl₅ → CH₃Cl; KCN → CH₃CN; LiAlH₄ → CH₃CH₂NH₂ |
| Hexanenitrile → pentan-1-amine | Partial hydrolysis (H₂O, H⁺, mild) → hexanamide; Br₂ + NaOH → CH₃(CH₂)₄NH₂ |
| Nitromethane → N-methylmethanamine | Sn + HCl → CH₃NH₂; CHCl₃ + alcoholic KOH → CH₃NC; H₂/Pt → CH₃NHCH₃ |
| Propanoic acid → ethanoic acid | NH₃, heat → propanamide; Br₂ + NaOH → ethanamine; HNO₂ → ethanol; acidified KMnO₄ → ethanoic acid |
Distinguishing tests
| Pair | Test | What you see |
|---|---|---|
| Methanamine and N-methylmethanamine | Carbylamine: CHCl₃ + alcoholic KOH, heat | Methanamine: foul smell; N-methylmethanamine: none |
| N-Ethylethanamine and N,N-diethylethanamine | Hinsberg reagent, then KOH | Secondary: sulphonamide insoluble in alkali; tertiary: no reaction |
| Ethanamine and aniline | NaNO₂ + HCl at 273 to 278 K, then alkaline 2-naphthol | Aniline: orange-red dye; ethanamine: N₂ bubbles, no dye |
| Aniline and benzylamine | Same azo dye test | Aniline: dye; benzylamine: N₂ bubbles, no dye |
| Aniline and N-methylaniline | Carbylamine test | Aniline: foul smell; N-methylaniline: none |
| Aniline and ethanamine | Bromine water | Aniline: white precipitate of 2,4,6-tribromoaniline; ethanamine: no precipitate |
Activity: tests for the primary amino group
- 1★Carbylamine test: to 2 to 3 drops of the amine add 2 to 3 drops of chloroform and an equal volume of 0.5 M alcoholic KOH. Heat gently, in a fume cupboard.A very offensive smell of isocyanide shows a primary amine, aliphatic or aromatic. Afterwards, add conc. HCl to destroy the poisonous isocyanide
- 2★Azo dye test: dissolve about 0.2 g of the amine in 2 mL of dilute HCl and cool in ice. Add 2 mL of cold 2.5% NaNO₂ solution, then pour this slowly into a cold solution of 0.2 g of 2-naphthol in dilute NaOH.A scarlet red (orange-red) dye shows a primary aromatic amine
- 3Solubility test: shake the amine with dilute HCl, then add NaOH solution.The amine dissolves in acid as a salt and separates again with alkali, showing a basic group
- 4Record each result in a table with three columns: experiment, observation, inference.
★ key step
Worked example: identify A, B and C
Compound A (C₇H₇NO) is heated with Br₂ and NaOH to give B (C₆H₇N). B gives an offensive smell when heated with CHCl₃ and alcoholic KOH. B with NaNO₂ and HCl at 273 to 278 K gives C, which forms an orange dye with phenol in alkaline solution. Identify A, B and C.
- Br₂ + NaOH on A removes one carbon and gives an amine: this is Hoffmann bromamide degradation, so A is an amide.
- B gives the carbylamine test, so B is a primary amine. C₆H₇N with a ring is aniline, C₆H₅NH₂.
- So A, with one carbon more, is benzamide, C₆H₅CONH₂ (C₇H₇NO). C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.
- Aniline at 273 to 278 K with NaNO₂ + HCl gives C, benzenediazonium chloride, C₆H₅N₂⁺Cl⁻.
- C couples with phenol in alkali to give orange p-hydroxyazobenzene, which confirms C.
A: benzamide; B: aniline; C: benzenediazonium chloride
How to plan an amine conversion
- Count carbons first. One fewer: make the amide and use Hoffmann degradation. One more: make the nitrile (KCN) and reduce it.
- To get an arylamine, reduce a nitro compound (Sn + HCl); do not use Gabriel synthesis, which fails for aryl halides.
- To put F, I, CN, OH or H on a ring, go through the diazonium salt at 273 to 278 K.
- To control substitution in aniline, protect -NH₂ by acetylation, then hydrolyse.
- Check directing effects: -NO₂ is meta-directing, so brominate nitrobenzene before reducing it if you want a meta product.
Many medicines contain amine groups: sulpha drugs carry a primary aromatic amine, while many antihistamines are tertiary amines. Tests like these tell chemists which kind of amine they are handling.
2 marks: “Give one chemical test to distinguish: (i) propan-1-amine and N-methylethanamine (ii) 4-methylaniline and 4-methylbenzylamine.” (i) Carbylamine test (CHCl₃ + alcoholic KOH, heat): foul smell with propan-1-amine only. (ii) Azo dye test (NaNO₂ + HCl at 273 to 278 K, then alkaline 2-naphthol): orange-red dye with 4-methylaniline; 4-methylbenzylamine gives N₂ and no dye.
3 marks: “How will you convert: (i) 4-nitrotoluene to 4-methylbenzoic acid (ii) benzene to 3-chlorophenol (iii) propanoic acid to ethanamine?” (i) Sn/HCl → 4-methylaniline → NaNO₂/HCl, 273 to 278 K → CuCN/KCN → 4-CH₃C₆H₄CN → H₃O⁺ → 4-CH₃C₆H₄COOH. (ii) HNO₃/H₂SO₄ → nitrobenzene → Cl₂/FeCl₃ → 3-chloronitrobenzene → Sn/HCl → diazotise → warm water. (iii) NH₃, heat → CH₃CH₂CONH₂ → Br₂/NaOH → CH₃CH₂NH₂.
3 marks: “Bring about: (i) 1-bromobutane to pentan-1-amine (ii) 4-methylaniline to 3,5-dibromotoluene (iii) ethanamine to propan-1-amine.” (i) KCN → CH₃(CH₂)₃CN → LiAlH₄ → CH₃(CH₂)₄NH₂. (ii) Br₂ water → 2,6-dibromo-4-methylaniline → NaNO₂/HCl, 273 to 278 K → H₃PO₂ → 3,5-dibromotoluene. (iii) HNO₂ → C₂H₅OH → PCl₅ → C₂H₅Cl → KCN → C₂H₅CN → LiAlH₄ → CH₃CH₂CH₂NH₂.
Down one: amide + Br₂/NaOH. Up one: CN, then reduce. Off the ring: diazotise, then H₃PO₂. A 1° amine stinks with CHCl₃; a 1° aromatic amine gives a dye.