Part 3: Amines: preparation, basicity and reactions

Follow the nitrogen lone pair through proton acceptance, nucleophilic substitution, acylation and copper complex formation, using butylamine as the Edexcel example.

Classify the bonds to nitrogen

A primary amine is RNH₂, a secondary amine is R₂NH and a tertiary amine is R₃N; classification counts carbon groups bonded directly to nitrogen. Butylamine, CH₃CH₂CH₂CH₂NH₂, is a primary aliphatic amine even though it has four carbons. Phenylamine, C₆H₅NH₂, is a primary aromatic amine because N is directly attached to the ring.

An amide has N bonded directly to a carbonyl carbon, RCONH₂ or RCONHR′. The nitrogen lone pair interacts with that C=O group, so an amide cannot simply be treated as an ordinary amine. Mark the N–C(=O) connection before predicting properties.

The amine nitrogen has three covalent bonds and a lone pair. It can use that pair to bond to H⁺, an electron-deficient carbon, or a metal ion. The partner being attacked determines whether the process is acid–base reaction, nucleophilic reaction or coordination.

Compare availability of the nitrogen lone pair

Butylamine accepts H⁺ from water to give butylammonium and OH⁻, so its aqueous solution is alkaline. The equilibrium lies mainly on the unprotonated side: it is a weak base, not a source of stoichiometrically complete OH⁻ release. Addition of acid drives salt formation; addition of strong alkali to the salt regenerates the amine.

Primary aliphatic amines are generally stronger bases than ammonia in the specified aqueous comparisons because an alkyl group donates electron density towards N, making its lone pair more available to accept H⁺. In phenylamine, the nitrogen lone pair is partly delocalised into the aromatic ring, so it is less available and phenylamine is a weaker base than ammonia. It is still a weak base; “not basic” is incorrect.

Use supplied data to make the actual order. Larger Kb means stronger base; larger pKa of the conjugate acid means the corresponding base is stronger. Illustrative conjugate-acid pKa values 10.6, 9.3 and 4.6 for an alkylammonium, ammonium and phenylammonium species imply alkylamine > ammonia > phenylamine. Do not extrapolate an unlimited alkyl-count rule to every secondary/tertiary amine: solvation also matters.

Nucleophile and base roles can occur in the same preparation

Butylamine attacks a halogenoalkane through the N lone pair. With bromoethane it can form the secondary amine C₄H₉NHC₂H₅ after proton removal. A second amine molecule can accept the proton to give an alkylammonium salt. Further alkylation can produce tertiary amine and eventually a quaternary ammonium salt because the intermediate amines still possess a lone pair.

With ethanoyl chloride, butylamine gives CH₃CONHC₄H₉, N-butylethanamide. Nitrogen bonds to carbonyl carbon, and excess amine neutralises the released acid. This is acylation, not simple protonation; the product contains a new covalent C–N bond.

Compare the products by connectivity: a hydrochloride salt has C₄H₉NH₃⁺ and Cl⁻ with no newly attached carbon group; an alkylation product has a new N–alkyl group; an acylation product has N–C(=O). These distinctions prevent the common error of drawing an amide after any amine reaction.

The same lone pair can form a coordinate bond

A primary amine can donate its N lone pair to Cu²⁺ and act as a ligand. Aqueous Cu(II) is initially pale blue. Small additions of a basic amine may first produce a pale-blue hydroxide precipitate because OH⁻ is generated in water; excess ligand can replace water molecules and produce a deeper blue soluble amine complex.

For the syllabus butylamine example, the usual simplified ligand-exchange equation replaces four water ligands: [Cu(H₂O)₆]²⁺ + 4C₄H₉NH₂ ⇌ [Cu(C₄H₉NH₂)₄(H₂O)₂]²⁺ + 4H₂O. The amine is a neutral ligand, so the complex remains 2+. Use the stoichiometry or structure supplied when a question specifies a different complex.

A coordinate-bond arrow starts at the N lone pair and points to the copper ion. This represents both bonding electrons supplied by the ligand. It is not an electron-transfer redox equation: copper remains in oxidation state +2.

Choose a route with the correct carbon count

Heat a halogenoalkane with excess ethanolic ammonia in an appropriate pressure-rated closed apparatus: nucleophilic substitution gives a primary amine. Excess NH₃ favours encounters with ammonia rather than the product amine and reduces further alkylation, although a mixture may still need separation. Do not heat an improvised sealed vessel.

Alternatively, reduce a nitrile using LiAlH₄ in dry ether followed by appropriate work-up, or catalytic hydrogenation under suitable supplied conditions. RCN + 4[H] → RCH₂NH₂. The nitrile carbon becomes the CH₂ next to nitrogen; it is not removed.

Worked route to butylamine: 1-bromopropane reacts with cyanide in ethanolic conditions under reflux to give butanenitrile; reduction gives butan-1-amine. The first step adds one carbon. Starting instead with 1-bromobutane plus NH₃ retains the original four-carbon chain.

Reduce nitrobenzene, then release the free amine

Heat an aromatic nitro compound with tin and concentrated hydrochloric acid, commonly under reflux. The NO₂ group is reduced while the aromatic ring is retained. The shorthand neutral-product equation is C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.

In the actual acidic mixture, phenylamine is protonated to phenylammonium, C₆H₅NH₃⁺. Addition of NaOH after the reduction releases free phenylamine for isolation. If a question asks for the species before work-up, show the salt; if it asks for the isolated amine, show NH₂.

This pairs naturally with benzene nitration: benzene → nitrobenzene → phenylammonium salt → phenylamine. Direct substitution of chlorobenzene with ammonia under ordinary halogenoalkane conditions is not an interchangeable simple route.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources