Part 1: Amine basicity, salts and preparation

Use the nitrogen lone pair to explain proton acceptance and substitution, then choose a route to an aliphatic or aromatic amine.

Count carbon groups on nitrogen, not on the adjacent carbon

A primary amine has one carbon group attached to nitrogen, RNH₂; a secondary amine has two, R₂NH; and a tertiary amine has three, R₃N. This classification counts carbon groups bonded directly to N. It differs from classifying an alcohol by the carbon bearing OH.

For example (CH₃)₂CHNH₂ is a primary amine: nitrogen is bonded to only one carbon group, even though the carbon carrying NH₂ has two carbon neighbours. CH₃NHCH₂CH₃ is a secondary amine because N has two carbon groups.

Amines have a nitrogen lone pair. An amide contains nitrogen directly attached to a carbonyl carbon, RCONH₂ or RCONHR, and has different electronic behaviour. Do not treat every molecule containing NH₂ as an ordinary amine.

The lone pair accepts a proton

A Brønsted–Lowry base accepts H⁺. An amine uses the nitrogen lone pair to form a bond to the proton, giving an ammonium ion. Methylamine reacts with HCl to form methylammonium chloride, CH₃NH₃⁺Cl⁻. Nitrogen has four bonds and a positive charge in the product.

In water, RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. This equilibrium explains basic behaviour. Aromatic amines can also act as weak bases; it is incorrect to claim that they cannot form basic aqueous solutions at all. OCR does not require a memorised comparative-basicity ranking in this section.

Adding a strong base to an ammonium salt removes a proton and regenerates the neutral amine. This acid–base interconversion is useful in understanding work-up after amine preparation. Draw the species appropriate to the pH, rather than using NH₂ and NH₃⁺ interchangeably.

Excess ammonia favours the primary amine

Heating a haloalkane with excess ethanolic ammonia, in suitable apparatus that retains volatile ammonia, prepares an aliphatic amine by nucleophilic substitution. Nitrogen’s lone pair attacks the carbon bonded to the halogen and the C–halogen bond pair goes to the leaving halide.

The immediate substitution product is an alkylammonium ion, which loses H⁺ to another ammonia molecule. The overall equation for bromoethane is CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. One NH₃ supplies the new nitrogen group; the other accepts a proton.

The primary amine also has a lone pair and can react with more haloalkane, producing secondary and tertiary amines, and eventually a quaternary ammonium salt. Excess ammonia increases the chance that haloalkane encounters NH₃ rather than the amine product. It favours primary-amine formation but does not justify claiming a perfectly pure single product in every preparation.

Reduce nitrobenzene, then release the free amine

An aromatic nitro compound is reduced using tin and concentrated HCl with heating. The nitro group becomes NH₂ overall: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O. Both oxygen atoms are removed as water; the benzene ring and C–N connection remain.

Because the reaction mixture is acidic, the amine is protonated and present as an ammonium salt. Addition of alkali in the work-up releases the free aromatic amine. Distinguish the overall reduction equation from the actual acid–base form present before work-up.

For an unfamiliar substituted nitrobenzene, preserve every other group unless the stated reagents also affect it. Count six [H] per NO₂ group. Do not replace the nitro group with a completely new carbon chain or apply an ordinary haloalkane substitution route directly to chlorobenzene without justification.

Worked route choice and a mole calculation

To prepare propylamine while retaining a three-carbon haloalkane skeleton, use 1-bromopropane with excess ethanolic ammonia. To lengthen a two-carbon skeleton first, substitute bromoethane with CN⁻, then reduce the resulting three-carbon nitrile using H₂/Ni. These routes answer different carbon-counting problems.

For an original nitrobenzene reduction calculation, 6.15 g of nitrobenzene, Mr 123.0, is 0.0500 mol. The 1:1 organic ratio gives a theoretical 0.0500 mol phenylamine, or 4.65 g using Mr 93.0. At 72.0% isolated yield the mass is 3.35 g. The 6[H] coefficient does not make six moles of amine.

Before finalising any amine product, count N bonds, assign charge and check whether an acid or alkaline work-up is specified. Then count carbons to distinguish direct substitution from a nitrile-extension route.

Quick checks

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

Sources