AS Chemistry: Nitrogen Compounds — Revision Notes

Condensed for the final weeks. For the full explanation, use the Nitrogen Compounds study guide.

Amines

Classified by how many alkyl groups are attached to nitrogen: primary (one), secondary (two), tertiary (three) — useful terminology, but syllabus 19.1 states explicitly that classifying amines this way will not be tested at AS.

Basicity (A Level extension — Topic 34, not an AS 19.1 outcome): the nitrogen lone pair accepts a proton. Strength depends on how available that lone pair is:

secondary aliphatic > primary aliphatic > ammonia > phenylamine

• Alkyl groups donate electron density, increasing availability of the lone pair.
• In phenylamine the lone pair is delocalised into the ring, so it is much less available — making phenylamine a weaker base than ammonia.

This ranking presupposes arenes and is included here as forward context only; AS 19.1 is limited to the production of a primary amine and does not require basicity comparisons.

Preparation — the single AS 19.1 outcome:

halogenoalkane + EXCESS NH3, ethanolic, heat under pressure  ->  primary amine

Excess ammonia is essential because the primary amine is itself a nucleophile and reacts further, producing secondary, tertiary and quaternary products.

A Level extension (Topic 34) — further preparation routes and reactions, not required at AS:

nitrile + LiAlH4 (dry ether) or H2/Ni                        ->  primary amine
nitrobenzene + Sn/conc HCl, then NaOH                        ->  phenylamine

The NaOH step in the aromatic route liberates the free amine from the phenylammonium salt formed by the reduction. Amines also react with acids to form salts, with halogenoalkanes to give secondary amines, and with acyl chlorides to give N-substituted amides — all A Level Topic 34 content. See the Amines: Aliphatic and Aromatic study guide for the full treatment.

Nitriles

Contain the C≡N group.

Preparation:

halogenoalkane + KCN in ethanol, heat under reflux  ->  nitrile
aldehyde/ketone + HCN (with KCN catalyst)          ->  hydroxynitrile

Both routes add a carbon atom. This is the key synthetic use of nitriles and the reason they appear in multi-step route questions: if the product has one more carbon than the starting material, a nitrile step is almost certainly required — a pattern worth checking for before attempting a full multi-step synthesis route.

The solvent matters: KCN in ethanol gives substitution to the nitrile; KCN in aqueous solution favours hydrolysis to the alcohol instead.

Reactions:

nitrile + dilute HCl(aq), reflux    ->  carboxylic acid + ammonium salt

A Level extension (Topic 34): a nitrile can also be reduced to a primary amine (nitrile + LiAlH4 or H2/Ni), making it a branch point — hydrolyse it for an acid (AS), or reduce it for an amine (A Level). Either product has one more carbon than the original halogenoalkane.

Same nucleophile, different mechanism

Cyanide, CN⁻, is the nucleophile in both the nitrile and hydroxynitrile reactions — but the mechanism differs because the carbon attacked is different. Attacking a halogenoalkane’s sp³ carbon (bonded to a leaving group) is substitution — the halide leaves as CN⁻ arrives. Attacking a carbonyl’s sp² carbon (no leaving group, just a π bond) is addition — nothing leaves; the π bond breaks and the oxygen becomes an alkoxide. Recognising a C–X bond versus a C=O bond tells you immediately which mechanism is expected.

Worked example. Convert bromoethane into 2-hydroxypropanenitrile. Since a halogenoalkane cannot become a carbonyl compound directly, three steps are needed: (1) hydrolyse with NaOH(aq), heat, to give ethanol; (2) oxidise with acidified K₂Cr₂O₇ and distil, to give ethanal; (3) react the ethanal with HCN (KCN catalyst, heat) to give the hydroxynitrile, CH₃CH(OH)CN.

The hydroxynitrile mechanism

Nucleophilic addition of CN⁻ to the carbonyl carbon, then protonation of the resulting alkoxide.

The carbonyl carbon is planar, so CN⁻ attacks with equal probability from either face, producing equal amounts of both enantiomers — a racemic mixture (a term properly introduced at A Level; at AS it’s enough to say equal amounts of both enantiomers form).

That is direct evidence for the shape of the carbonyl compound (planar, trigonal) rather than the intermediate formed after attack — the alkoxide intermediate produced once CN⁻ has added is tetrahedral, not planar. Explaining the racemic outcome in terms of the planar carbonyl reactant, rather than just stating “a racemate forms,” is what distinguishes a top answer.

Exam traps

• Explaining basicity without lone-pair availability.
• Saying phenylamine is a stronger base than ammonia.
• Forgetting excess ammonia, or the NaOH step.
• Using aqueous rather than ethanolic KCN for nitrile formation.
• Using NaBH₄ to reduce a nitrile — it is not powerful enough.
• Stating that a racemate forms without explaining the planar carbonyl reactant (not the tetrahedral alkoxide intermediate formed after attack).

Self-test

1. Why is a secondary amine a stronger base than ammonia?
2. Why is phenylamine a weaker base than ammonia?
3. Why must ammonia be in excess when preparing a primary amine?
4. Why are nitriles important in synthesis?
5. Why does HCN addition to a ketone give a racemic mixture?
6. Why does CN⁻ give substitution with a halogenoalkane but addition with a carbonyl compound?
7. Why does converting a halogenoalkane into a hydroxynitrile take three steps rather than two?

Answers: 1. Two electron-donating alkyl groups increase the electron density on nitrogen, making the lone pair more available to accept a proton. 2. Its nitrogen lone pair is delocalised into the benzene ring, so it is much less available to accept a proton. 3. The primary amine formed is itself nucleophilic and reacts further to give secondary, tertiary and quaternary amines; excess ammonia reduces the chance of this. 4. They add a carbon atom to the chain, and can then be hydrolysed to a carboxylic acid or reduced to an amine. 5. The carbonyl carbon is planar, so the cyanide ion attacks with equal probability from either face, producing equal amounts of the two enantiomers. 6. Mechanism depends on the substrate, not the nucleophile: a halogenoalkane has a leaving group at an sp³ carbon (substitution), while a carbonyl compound has a π bond to add across at an sp² carbon with nothing to leave (addition). 7. A halogenoalkane cannot be converted directly into a carbonyl compound, so it must first be hydrolysed to an alcohol, then oxidised to an aldehyde or ketone, before HCN can add to it.