Part 2: Nucleophilic substitution, acylation and ammonium salts

Follow further alkylation to a quaternary salt and distinguish amine nucleophilicity from proton acceptance.

First form a C–N bond, then remove H⁺

For bromoethane plus NH₃, draw the nitrogen lone-pair arrow to the δ⁺ carbon bonded to Br, and a second arrow from C–Br to Br. The product of this step is CH₃CH₂NH₃⁺ and Br⁻. A second ammonia acts as a base: its lone pair attacks an N–H proton and the N–H bond electrons return to the alkylammonium nitrogen.

The same ammonia molecule can act as a nucleophile in one step and another ammonia molecule as a base in the next. A nucleophile donates a pair to an electrophilic atom; a Brønsted base specifically accepts a proton. Explain the actual step rather than treating the labels as mutually exclusive.

Primary, secondary, tertiary, then quaternary

Primary and secondary amines undergo the same attack and deprotonation sequence with a halogenoalkane, adding a carbon group to N each time. A tertiary amine can still attack because it has a lone pair, but the quaternary ammonium product has no N–H bond to deprotonate. It stays positively charged, with the halide as counter-ion.

For mixed groups, retain every original N substituent. Excess halogenoalkane favours further alkylation and can produce a quaternary ammonium salt. The equations below use a second amine as the proton acceptor where deprotonation is needed; other appropriate bases can fulfil that role.

A permanent ionic head and a hydrocarbon tail

A quaternary ammonium salt with a sufficiently long hydrocarbon group can act as a cationic surfactant. The positively charged head interacts with water and can be attracted to negatively charged surfaces; the hydrocarbon tail associates with non-polar material. Such structures can be used in conditioning and related surface-treatment products.

Not every quaternary salt is an effective surfactant: a small tetramethylammonium ion lacks the long hydrophobic tail of the standard surfactant example. Quaternary nitrogen has no lone pair to use in the ordinary proton-acceptance mechanism of an amine.

An acyl group makes an amide, not a more substituted amine

A primary amine reacts with an acyl chloride through nucleophilic addition–elimination. The N lone pair attacks carbonyl carbon, C=O electrons move to O, the tetrahedral intermediate collapses with chloride loss, and deprotonation leaves an N-substituted amide. With excess amine, a second molecule captures the acid to form an alkylammonium salt.

Ethanoic anhydride also acylates primary amines; its by-product is ethanoic acid or ethanoate if neutralised by excess amine. Amide nitrogen is next to a carbonyl and its lone pair is delocalised, so do not treat the product as another ordinary strongly nucleophilic alkylamine. The full four-nucleophile mechanism set is in Carboxylic Acids and Derivatives.

Choose a route by selectivity and carbon count

To make a primary amine with the same carbon skeleton as a nitrile, reduction is selective for that conversion under the stated method. To make an aromatic amine, nitration then reduction provides a common two-step route from benzene. To attach an alkyl group to an existing amine, use halogenoalkane substitution while considering over-alkylation.

Original quantity check: 0.0200 mol nitrile requires 0.0400 mol H₂ for full reduction, since RCN + 2H₂ → RCH₂NH₂. At 80.0% isolated yield, the amount of desired amine is 0.0160 mol. The hydrogen stoichiometry is not the same as the two [H] equivalents used to reduce one aldehyde carbonyl.

Quick checks

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

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