AQA A-Level Chemistry: Amines (7405) – Revision Notes

These notes condense section 3.3.11 Amines of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. They cover 3.3.11.1 Preparation, 3.3.11.2 Base properties and 3.3.11.3 Nucleophilic properties. Everything here is A-level only. Organic chemistry is placed on A-level Paper 2, and Paper 3 can test any content.

For full explanations and worked examples, use the amines study guide. Test yourself with the amines practice questions. The first substitution step is also in the halogenoalkanes revision notes, and proton transfer is in acids and bases. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. To find your weak spots fast, take a free 10-minute diagnostic.

Definitions to learn

• Amine: a compound based on ammonia in which one or more H atoms are replaced by alkyl or aryl groups.
• Primary / secondary / tertiary amine: one / two / three carbons bonded to N (RNH₂, R₂NH, R₃N).
• Quaternary ammonium ion: four carbons bonded to N, R₄N⁺. Its salts are R₄N⁺X⁻.
• Aliphatic amine: N bonded to an alkyl group. Aromatic amine: N bonded directly to a benzene ring (phenylamine, C₆H₅NH₂).
• Base (Brønsted–Lowry): proton acceptor. Weak base: only partly reacts with water.
• Nucleophile: electron pair donor.
• N-substituted amide: an amide with an alkyl group on N, e.g. CH₃CONHCH₂CH₃, N-ethylethanamide.
• Cationic surfactant: a surfactant whose ionic head group is positive, e.g. a quaternary ammonium salt with one long hydrocarbon chain.

Classify amines by counting carbons on N. (CH₃)₃CNH₂ is a primary amine even though the carbon next to N is tertiary.

3.3.11.1 Preparation summary

Product	Starting material	Reagent and conditions	Equation (example)	Note
Primary aliphatic amine	Halogenoalkane	Excess NH₃ in ethanol, heat in a sealed container	C₃H₇Cl + 2NH₃ → C₃H₇NH₂ + NH₄Cl	Mixture of products by further substitution
Primary aliphatic amine	Nitrile	Reduction, e.g. LiAlH₄ in dry ether, or H₂ with Ni catalyst	C₂H₅CN + 4[H] → C₂H₅CH₂NH₂	Primary amine only; no mixture
Aromatic amine	Nitro compound	Reduction, e.g. Sn and conc. HCl under reflux, then NaOH	C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O	Used in the manufacture of dyes

The specification names the reaction types (reaction with ammonia; reduction) but not specific reducing agents. The reagents in the table are common choices.

[H] counting. Nitrile: C≡N → CH₂–NH₂ adds four H, so 4[H] (or 2H₂). Nitro: –NO₂ → –NH₂ removes two O as 2H₂O and adds two H to N, so 6[H].

Chain length. Halogenoalkane + NH₃ keeps the carbon count. Halogenoalkane + KCN, then reduction, adds one carbon: 1-bromopropane → butanenitrile → butylamine.

3.3.11.2 Base properties

Reactions as a base:

RNH2 + H2O ⇌ RNH3+ + OH−        (weak base, solution alkaline)
RNH2 + HCl → RNH3+Cl−            (salt, e.g. propylammonium chloride)
RNH3+ + OH− → RNH2 + H2O         (NaOH releases the amine from its salt)

Base strength order (learn it):

Strongest	Middle	Weakest
Primary aliphatic amine (e.g. CH₃NH₂)	Ammonia, NH₃	Primary aromatic amine (C₆H₅NH₂)

Method in steps: explaining a base-strength order

1. Say that the base accepts H⁺ using the lone pair on N.
2. For an aliphatic amine: alkyl group is electron-releasing (positive inductive effect), so electron density on N increases.
3. For an aromatic amine: the N lone pair is delocalised into the benzene ring, so electron density on N decreases.
4. Conclude: more available lone pair → accepts H⁺ more readily → stronger base.

Ammonia sits in the middle because it has neither effect.

3.3.11.3 Nucleophilic properties

Substitution ladder with halogenoalkanes

NH3   →  RNH2  →  R2NH  →  R3N  →  R4N+X−
        primary  secondary tertiary quaternary salt

Each step: amine + RX → next amine (+ HX, taken up by more base as a salt).

• Excess NH₃ → mainly primary amine.
• Excess halogenoalkane → mainly quaternary ammonium salt.
• The ladder stops at R₄N⁺: no lone pair and no H on N.

Method in steps: substitution mechanism (amine + halogenoalkane)

1. Curly arrow: lone pair on N → δ+ carbon of C–X.
2. Curly arrow: C–X bond → X. X⁻ leaves.
3. Intermediate: N⁺ with an extra alkyl group and at least one H.
4. Second amine (or NH₃) removes H⁺ from N⁺: arrow from its lone pair to H, arrow from N–H bond to N⁺.

Cationic surfactants. A quaternary ammonium salt with one long chain, e.g. CH₃(CH₂)₁₃N⁺(CH₃)₃ Cl⁻. The N⁺ head is attracted to water and to negatively charged surfaces; the long chain is not attracted to water.

Addition–elimination with acyl chlorides and acid anhydrides

Nucleophile	Acyl chloride, RCOCl	Acid anhydride, (RCO)₂O
NH₃	RCONH₂ (primary amide) + NH₄Cl	RCONH₂ + RCOOH
R′NH₂	RCONHR′ (N-substituted amide) + R′NH₃⁺Cl⁻	RCONHR′ + RCOOH
Speed	Fast at room temperature	Slower

Example: CH₃CH₂CH₂COCl + 2NH₃ → CH₃CH₂CH₂CONH₂ (butanamide) + NH₄Cl.

Method in steps: addition–elimination mechanism (acyl chloride)

1. Addition. Arrow from lone pair on N → δ+ carbonyl carbon. Arrow from C=O π bond → O, giving O⁻.
2. Tetrahedral intermediate: C bonded to O⁻, Cl, R and N⁺.
3. Elimination. Arrow from O⁻ lone pair → reforms C=O. Arrow from C–Cl bond → Cl, Cl⁻ leaves.
4. Cl⁻ or a second NH₃/amine removes H⁺ from N⁺.

You must be able to outline this mechanism for acyl chlorides. For acid anhydrides the specification asks only for the reactions.

Small worked reminders

Naming an N-substituted amide. Ethanoyl chloride + propylamine. The acyl part (CH₃CO–) names the amide, ethanamide; the amine’s alkyl group goes on N. Product: CH₃CONHCH₂CH₂CH₃, N-propylethanamide.

Choosing a route. To make CH₃CH₂CH₂CH₂NH₂ cleanly from a three-carbon halogenoalkane: 1-bromopropane + KCN (ethanol/water, reflux) → butanenitrile; then reduce → butylamine. Using NH₃ directly would give propylamine, with one carbon too few, and a mixture.

Salt formula. Phenylamine + HCl → C₆H₅NH₃⁺Cl⁻, phenylammonium chloride. The positive charge is on N.

Must-know distinctions

• Amine vs amide. Amine: N bonded to C and H only. Amide: N bonded to a C=O carbon. Amides are made from amines (or NH₃) plus acyl chlorides or anhydrides.
• Primary amine vs primary amide. CH₃CH₂NH₂ vs CH₃CONH₂.
• Substitution vs addition–elimination. Halogenoalkane: one-step replacement of X. Acyl chloride: adds across C=O, then eliminates Cl⁻.
• Ammonium salt vs quaternary ammonium salt. RNH₃⁺Cl⁻ gives back the amine with NaOH; R₄N⁺Cl⁻ cannot, because N has no H to lose.
• By-product. Acyl chloride gives HCl (as a salt with excess base); acid anhydride gives a carboxylic acid.

Quick self-test

1. Classify (CH₃CH₂)₂NCH₃.
2. Give the product of reducing butanenitrile.
3. How many [H] are needed to reduce one nitrobenzene molecule to phenylamine?
4. Write the equation for methylamine with water.
5. Put in order of increasing base strength: phenylamine, propylamine, ammonia.
6. Why is phenylamine a weaker base than ammonia?
7. What conditions favour a quaternary ammonium salt?
8. Calculate the Mr of tetramethylammonium chloride (Ar: H = 1.0, C = 12.0, N = 14.0, Cl = 35.5).
9. Name the organic product of butanoyl chloride with excess methylamine.
10. Give the by-product when ethanoic anhydride reacts with ammonia (1 : 1).
11. Why can a quaternary ammonium ion not act as a nucleophile?

Answers

1. Tertiary amine: three carbons on N.
2. Butylamine, CH₃CH₂CH₂CH₂NH₂ (the nitrile carbon becomes the CH₂ on N, so still four carbons).
3. 6[H], giving 2H₂O.
4. CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻
5. Phenylamine < ammonia < propylamine.
6. The N lone pair is delocalised into the benzene ring, so it is less available to accept H⁺.
7. Excess halogenoalkane.
8. (CH₃)₄N⁺Cl⁻ = 4(12.0 + 3.0) + 14.0 + 35.5 = 109.5.
9. N-methylbutanamide, CH₃CH₂CH₂CONHCH₃.
10. Ethanoic acid, CH₃COOH (as ammonium ethanoate if ammonia is in excess).
11. Its N has no lone pair: all four N electron pairs are used in bonds.

Where marks are usually lost

• Calling (CH₃)₃CNH₂ a tertiary amine because the carbon is tertiary.
• Unbalanced [H] in reductions: 2[H] for a nitrile or 4[H] for a nitro compound.
• Forgetting that nitrile reduction keeps the C≡N carbon in the chain.
• Base-strength answers that never mention the lone pair or proton acceptance.
• Saying the alkyl group “donates a lone pair” instead of “releases electron density”.
• Missing the deprotonation step at the end of either mechanism.
• Arrows starting from the N atom or the minus sign instead of from the lone pair.
• In addition–elimination, leaving out the arrow from C=O to O in the first step.
• Writing the amide product with an extra H on N (RCONH₂R′).
• Giving HCl as a free by-product when excess amine or ammonia is used.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.11 Amines (3.3.11.1 Preparation, 3.3.11.2 Base properties and 3.3.11.3 Nucleophilic properties), all A-level only.