OxfordAQA A-Level Chemistry: Amines (9620) – Revision Notes

For full explanations and longer worked examples, read the amines study guide first.

In these notes, section 3.3.11 (Amines) of the OxfordAQA 9620 specification (International AS and A-level Chemistry, Version 5.3) is cut down to essentials for exams from the May/June 2020 series onwards. They span subsections 3.3.11.1 to 3.3.11.3 (preparation, base properties, nucleophilic properties). Being International A2 material, the whole section is International A-level only, with Unit 4 examining 3.3.7 to 3.3.16 and Unit 5 (synoptic) able to test anything. The section lists no required practical.

Test yourself afterwards with the amines practice questions. Related pages: halogenoalkanes for the first ammonia substitution, and carboxylic acids and derivatives for acylation. Planning tools: the OxfordAQA chemistry hub, a checklist you can print and our diagnostics.

Key terms

Term	Meaning
Amine	Compound based on ammonia with one or more H atoms replaced by alkyl or aryl groups
Primary / secondary / tertiary	One / two / three carbon atoms bonded directly to N
Quaternary ammonium ion	R₄N⁺: four carbons on N, no lone pair, no N–H
Aliphatic amine	N bonded to a chain or non-aromatic ring carbon
Aromatic amine	N bonded directly to a benzene ring
Base	Proton (H⁺) acceptor
Nucleophile	Electron-pair donor that attacks an electron-deficient carbon
Cationic surfactant	Surfactant whose water-attracting head is positively charged

The lone pair on N explains the whole section. It accepts H⁺ (base) and attacks δ+ carbon (nucleophile).

3.3.11.1 Preparation at a glance

Product	Starting material	Reagent and conditions	Equation pattern
Primary aliphatic amine	Halogenoalkane RX	Excess NH₃ in ethanol, heated in a sealed vessel	RX + 2NH₃ → RNH₂ + NH₄X
Primary aliphatic amine	Nitrile RCN	Reduce: H₂ with Ni catalyst, or LiAlH₄ in dry ether (no reagent named in the specification)	RCN + 4[H] → RCH₂NH₂
Aromatic amine	Nitroarene ArNO₂	Tin and concentrated HCl, heat under reflux; then NaOH to free the amine	ArNO₂ + 6[H] → ArNH₂ + 2H₂O

• The ammonia route gives a mixture (the amine attacks more RX). Excess NH₃ limits this.
• The nitrile route gives only the primary amine and adds one carbon (RX → RCN → RCH₂NH₂).
• Aromatic amines made this way are used to manufacture dyes.

Method in steps: counting [H]

1. Write the starting group and the product group.
2. Each O removed leaves as H₂O, using 2[H].
3. Add the H atoms that end up on N or C.
4. For C≡N → CH₂–NH₂: 0 O removed, 4 H added, so 4[H].
5. For NO₂ → NH₂: 2 O removed (4[H]), 2 H added on N, so 6[H] and 2H₂O.
6. Multiply for molecules with two groups.

Worked reminder. Make heptan-1-amine, CH₃(CH₂)₆NH₂, from 1-bromohexane. Six carbons become seven, so a C–C bond must form. Step 1: KCN in ethanol and water, heat under reflux, giving heptanenitrile, CH₃(CH₂)₅CN. Step 2: reduce with H₂/Ni: CH₃(CH₂)₅CN + 4[H] → CH₃(CH₂)₆NH₂.

3.3.11.2 Base properties

In water, an amine is a weak base: only a small fraction is protonated, so use ⇌.

RNH2 + H2O ⇌ RNH3+ + OH−          (solution is alkaline)
RNH2 + HCl → RNH3+Cl−             (salt: alkylammonium chloride)
RNH3+ + OH− → RNH2 + H2O          (NaOH frees the amine)

Order of base strength: primary aliphatic amine > ammonia > primary aromatic amine.

Method in steps: explaining base strength

1. Say that the lone pair on N is what accepts H⁺.
2. Aliphatic: alkyl group is electron-releasing, so electron density on N rises; lone pair more available.
3. Ammonia: no alkyl or aryl group on N; reference point.
4. Aromatic: the ring delocalises the lone pair, lowering the electron density on N; lone pair less available.
5. Conclude: greater lone-pair availability means a stronger base.

3.3.11.3 Nucleophilic properties

Repeated substitution by halogenoalkanes

Nucleophile	+ RX gives	Class
NH₃	RNH₂	primary amine
RNH₂	R₂NH	secondary amine
R₂NH	R₃N	tertiary amine
R₃N	R₄N⁺X⁻	quaternary ammonium salt (end of the ladder)

• Excess ammonia → mainly primary amine.
• Excess halogenoalkane → mainly quaternary salt.
• R₄N⁺ has no lone pair, so it is neither a base nor a nucleophile.

Method in steps: substitution mechanism

1. Show δ+ on the carbon bonded to the halogen, δ− on the halogen.
2. Arrow from the N lone pair to the δ+ carbon.
3. Arrow from the C–X bond to X; X⁻ leaves.
4. Draw the intermediate with + on N.
5. Arrow from the lone pair of a second NH₃ or amine molecule to an H on N⁺; arrow from that N–H bond to N⁺.
6. Products: the new amine and an ammonium salt.

Exception: a tertiary amine forming a quaternary salt stops at step 4. There is no H on N⁺ to remove.

Worked reminder. Propan-2-amine with a large excess of bromomethane ends at (CH₃)₂CHN⁺(CH₃)₃ Br⁻, after passing through (CH₃)₂CHNHCH₃ and (CH₃)₂CHN(CH₃)₂.

Cationic surfactants

Give a quaternary ammonium salt a single long hydrocarbon chain and it has a positive N⁺ head that interacts with water and a non-polar tail that mixes with grease and oils. Because the head is positive, it is a cationic surfactant. Such salts are used in fabric softeners and hair conditioners.

Addition–elimination: acyl chlorides and acid anhydrides

	Acyl chloride, RCOCl	Acid anhydride, (RCO)₂O
With NH₃	RCONH₂ + NH₄Cl (2 mol NH₃)	RCONH₂ + RCOOH (or RCOONH₄ with excess NH₃)
With primary amine R′NH₂	RCONHR′ + R′NH₃⁺Cl⁻ (2 mol amine)	RCONHR′ + RCOOH
Rate	fast, at room temperature	slower
Mechanism needed?	yes	no (products only)

The product from a primary amine is an N-substituted amide; the N- shows that the group is on nitrogen.

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

1. δ+ on the carbonyl carbon (O and Cl both draw electrons away).
2. Arrow from the N lone pair to the carbonyl carbon; arrow from C=O to O, giving O⁻.
3. Tetrahedral intermediate: O⁻, Cl, R and N⁺ on the same carbon.
4. Arrow from O⁻ lone pair to re-form C=O; arrow from C–Cl to Cl; Cl⁻ leaves.
5. Cl⁻ or a second NH₃/amine molecule removes H⁺ from N⁺.

Must-know distinctions

• Class of amine vs class of alcohol. Amines count carbons on N; alcohols count carbons on the C–OH carbon. (CH₃)₃CNH₂ is a primary amine, but (CH₃)₃COH is a tertiary alcohol.
• Aromatic vs aliphatic. C₆H₅NH₂ is aromatic; C₆H₅CH(CH₃)NH₂ is aliphatic because N is on a chain carbon.
• Nitrile route vs ammonia route. Pure primary amine with one extra carbon vs a mixture with the same carbon count.
• Base vs nucleophile. Same lone pair; a base gives it to H⁺, a nucleophile gives it to a carbon.
• Amine vs amide. An amide has C=O bonded to N (–CONH–); an amine has no C=O on N.
• Acyl chloride by-product vs anhydride by-product. HCl (taken up as a chloride salt) vs a carboxylic acid.

Quick self-test

1. Is (CH₃CH₂)₂NCH₃ primary, secondary or tertiary?
2. Is C₆H₅CH(CH₃)NH₂ an aromatic or an aliphatic amine? Give a reason.
3. How many [H] are needed to reduce butanedinitrile, NCCH₂CH₂CN, to butane-1,4-diamine?
4. Give the reagents used to reduce 1-ethyl-4-nitrobenzene, and name the amine formed.
5. Put in order of increasing base strength: butan-1-amine, phenylamine, ammonia.
6. Explain why the tetramethylammonium ion cannot act as a nucleophile.
7. Write an equation for hexan-1-amine reacting with nitric acid and name the salt.
8. Calculate the volume of 0.200 mol dm⁻³ hydrochloric acid that exactly reacts with 0.606 g of hexan-1-amine, C₆H₁₅N (Mr = 101.0).
9. Name the organic product when heptanoyl chloride reacts with excess methylamine, and name the other product.
10. Name the by-product when ethanoic anhydride, not ethanoyl chloride, reacts with ammonia (1 : 1).
11. State the two structural features that let a quaternary ammonium salt act as a cationic surfactant.
12. Name the mechanism for (a) an amine with a halogenoalkane and (b) an amine with an acyl chloride.

Answers

1. Tertiary: three carbon groups on N, no N–H.
2. Aliphatic. N is bonded to a chain carbon, not to the ring.
3. 8[H]. Each C≡N needs 4[H]: NCCH₂CH₂CN + 8[H] → H₂N(CH₂)₄NH₂.
4. Tin and concentrated hydrochloric acid under reflux, then sodium hydroxide; 4-ethylphenylamine, CH₃CH₂C₆H₄NH₂ (6[H] used).
5. Phenylamine < ammonia < butan-1-amine.
6. N has used its lone pair in four bonds, so it has no lone pair to donate.
7. CH₃(CH₂)₅NH₂ + HNO₃ → CH₃(CH₂)₅NH₃⁺NO₃⁻; hexylammonium nitrate.
8. n = 0.606 ÷ 101.0 = 6.00 × 10⁻³ mol; ratio 1 : 1; V = 6.00 × 10⁻³ ÷ 0.200 = 0.0300 dm³ = 30.0 cm³.
9. N-methylheptanamide, CH₃(CH₂)₅CONHCH₃; by-product methylammonium chloride, CH₃NH₃⁺Cl⁻.
10. Ethanoic acid, CH₃COOH.
11. A positively charged N⁺ head and a long hydrocarbon chain.
12. (a) Nucleophilic substitution; (b) nucleophilic addition–elimination.

Where marks are usually lost

• Using the carbon next to N, not N itself, to classify an amine.
• Calling an amine aromatic because the molecule has a benzene ring somewhere.
• Equations with 2[H] for nitriles or 3[H] for nitro groups, or no water shown for nitro reduction.
• Leaving out the NaOH step after reduction with tin and hydrochloric acid.
• A full arrow, not ⇌, for an amine reacting with water.
• Base-strength answers that never mention the lone pair on N or accepting H⁺.
• “Phenylamine is weaker because benzene withdraws electrons” with no mention of delocalisation into the ring.
• Missing the + charge on N in either intermediate, or omitting the final H⁺ removal.
• Writing HCl as a product of an anhydride reaction.

Official syllabus

Source document: the 9620 specification from OxfordAQA (International AS and A-level Chemistry), Version 5.3, first teaching 2019, for exams from May/June 2020. Section covered: 3.3.11 Amines, subsections 3.3.11.1 to 3.3.11.3, all International A2.