OxfordAQA A-Level Chemistry: Amines (9620)

This study guide teaches section 3.3.11, Amines, of the OxfordAQA 9620 specification (International AS and A-level Chemistry, Version 5.3), which applies to exams from the May/June 2020 series onwards. All three subsections are covered, 3.3.11.1 to 3.3.11.3: preparation, base properties and nucleophilic properties. The specification lists this as International A2 content, which makes all of it International A-level only, and the Unit 4 paper (Organic 2 and Physical 2) examines it along with the rest of 3.3.7 to 3.3.16; Unit 5, the practical and synoptic paper, may test any content.

Pair it with the amines revision notes and the amines practice questions. Background you need: the ammonia and cyanide substitutions in halogenoalkanes, acids and bases for proton transfer, and carboxylic acids and derivatives for acylation. Course-wide pages: the OxfordAQA chemistry hub, a checklist you can print and our diagnostics.

Coverage table

Spec ref	Learning outcome	Stage
3.3.11	Amines are based on ammonia, with hydrogen atoms replaced by alkyl or aryl groups	International A-level only
3.3.11.1	Primary aliphatic amines are made from ammonia and halogenoalkanes, and by reducing nitriles	International A-level only
3.3.11.1	Aromatic amines come from reduction of nitro compounds; dye manufacture uses them	International A-level only
3.3.11.2	Amines are weak bases; explain the base strength of ammonia, primary aliphatic and primary aromatic amines from N lone-pair availability	International A-level only
3.3.11.3	Amines are nucleophiles; ammonia and amines substitute halogenoalkanes, giving primary, secondary and tertiary amines, then quaternary ammonium salts (outline the mechanisms)	International A-level only
3.3.11.3	Quaternary ammonium salts are used as cationic surfactants	International A-level only
3.3.11.3	Ammonia and primary amines react with acyl chlorides and acid anhydrides by nucleophilic addition–elimination (outline the mechanism with acyl chlorides)	International A-level only

Section 3.3.11 lists no required practical.

Amines: structure and classes

Every amine has a nitrogen lone pair. Classify by the carbons joined directly to N:

Class	N carries	Example
Primary	one C (–NH₂)	C₆H₁₁NH₂, cyclohexylamine
Secondary	two C (N–H left)	(CH₃CH₂CH₂CH₂)₂NH, dibutylamine
Tertiary	three C (no N–H)	(CH₃)₂NCH₂CH₂CH₂CH₃, N,N-dimethylbutylamine
Quaternary ammonium ion	four C, positive charge	(CH₃CH₂CH₂)₄N⁺, tetrapropylammonium

Classify by nitrogen, not by the carbon it sits on: (CH₃)₃CNH₂ is a primary amine, because N carries one carbon.

An amine is aromatic when a ring carbon of benzene itself carries the N, as in phenylamine (C₆H₅NH₂). If N is bonded to a chain or a non-aromatic ring carbon, the amine is aliphatic, even when a benzene ring sits elsewhere in the molecule. Names such as cyclohexanamine or propan-1-amine are also acceptable.

3.3.11.1 Preparation

Primary aliphatic amines from halogenoalkanes

Heat the halogenoalkane in a sealed vessel with excess ammonia in ethanol:

CH3CH2CH2CH2CH2Cl + 2NH3 → CH3CH2CH2CH2CH2NH2 + NH4Cl
1-chloropentane                pentan-1-amine

The mechanism is taught on the halogenoalkanes page. The drawback: the amine made still has a lone pair, so it can attack the halogenoalkane too. Excess ammonia keeps the primary amine as the main product, but some further-substituted amine still forms and must be separated.

Primary aliphatic amines by reducing nitriles

Reduction converts a nitrile, R–C≡N, into a primary amine, R–CH₂–NH₂. The 9620 text gives no reagent; hydrogen with a nickel catalyst, or LiAlH₄ in dry ether, both work. Write the reducing agent as [H] if you like:

R–C≡N + 4[H] → R–CH2–NH2        or        R–C≡N + 2H2 → R–CH2–NH2

This route gives the primary amine alone. And because the nitrile comes from a halogenoalkane and cyanide ions, the chain grows by one carbon compared with the halogenoalkane.

Worked example 1: choosing a route

Make 2-methylbutan-1-amine, CH₃CH₂CH(CH₃)CH₂NH₂, from 2-chlorobutane in two steps. Explain why ammonia alone would not work.

• Target: five carbons; start: four. A C–C bond must form, so use the nitrile route.
• Step 1: heat under reflux with potassium cyanide in ethanol and water: CH₃CHClCH₂CH₃ + CN⁻ → CH₃CH(CN)CH₂CH₃ + Cl⁻. Counting from the nitrile carbon, the intermediate is 2-methylbutanenitrile.
• Step 2: reduce with H₂ and a nickel catalyst (or LiAlH₄): CH₃CH(CN)CH₂CH₃ + 4[H] → CH₃CH₂CH(CH₃)CH₂NH₂.
• Ammonia would give butan-2-amine, CH₃CH(NH₂)CH₂CH₃: four carbons, N on the wrong carbon, plus further-substituted products.

Aromatic amines from nitro compounds

Nitration (section 3.3.10.2) puts an –NO₂ group on the ring; reducing it gives an aromatic amine. A usual laboratory method is to heat the nitro compound under reflux with tin and concentrated hydrochloric acid. The acid protonates the amine as it forms, so sodium hydroxide is then added to free it.

CH3C6H4NO2 + 6[H] → CH3C6H4NH2 + 2H2O
1-methyl-4-nitrobenzene    4-methylphenylamine

The specification notes that dye manufacture uses aromatic amines made this way.

Worked example 2: two nitro groups

1,3-Dinitrobenzene, C₆H₄(NO₂)₂, is reduced to benzene-1,3-diamine, C₆H₄(NH₂)₂. Write the equation using [H], then find the mass of diamine from 8.40 g of the nitro compound at 65.0% yield. Use H = 1.0, C = 12.0, N = 14.0 and O = 16.0.

• Each –NO₂ group loses two O atoms as 2H₂O (four H) and gains two H on N, so it needs 6[H]. Two groups need 12[H].
• C₆H₄(NO₂)₂ + 12[H] → C₆H₄(NH₂)₂ + 4H₂O (H: 16 on each side; O: 4 on each side).
• Mr: C₆H₄N₂O₄ = 168.0 and C₆H₈N₂ = 108.0.
• Amount = 8.40 ÷ 168.0 = 0.0500 mol; ratio 1 : 1, so theoretical mass = 0.0500 × 108.0 = 5.40 g.
• Actual mass = 5.40 × 0.650 = 3.51 g.

3.3.11.2 Base properties

Why amines are weak bases

A base accepts a proton; the lone pair on N forms a dative covalent bond to H⁺. In water the equilibrium lies well to the left, so only a small proportion of amine molecules are protonated:

C6H11NH2 + H2O ⇌ C6H11NH3+ + OH−

The OH⁻ ions make the solution alkaline. A strong acid protonates the amine completely to give a salt; sulfuric acid needs two amine molecules per acid molecule:

2C6H11NH2 + H2SO4 → (C6H11NH3)2SO4      cyclohexylammonium sulfate

Sodium hydroxide takes the H⁺ back and releases the free amine.

Comparing ammonia, aliphatic and aromatic amines

Base strength depends on how available the nitrogen lone pair is to accept a proton. More electron density on N means a stronger base:

primary aliphatic amine > ammonia > primary aromatic amine

• Primary aliphatic amines. An alkyl group releases electrons towards N (a positive inductive effect), making its lone pair more available than ammonia’s.
• Ammonia. Only H atoms on N, so no electron release: the reference point.
• Primary aromatic amines. The N lone pair overlaps with the delocalised p electrons of the ring and is partly drawn into it, so it is less available than in ammonia.

Worked example 3: same carbon count, different strength

Cyclohexylamine, C₆H₁₁NH₂, and phenylamine, C₆H₅NH₂, both have six ring carbons. Explain which is the stronger base and where ammonia fits.

• Both use the lone pair on N to accept H⁺.
• In cyclohexylamine, N sits on a non-aromatic ring carbon. The ring releases electrons onto N, so the lone pair is more available.
• In phenylamine, the lone pair is partly delocalised into the benzene ring, so less electron density stays on N.
• So cyclohexylamine is stronger, and ammonia, with neither effect, lies between: cyclohexylamine > ammonia > phenylamine.

3.3.11.3 Nucleophilic properties

The lone pair also lets amines act as nucleophiles, attacking an electron-deficient carbon atom.

Substitution with halogenoalkanes

Each new amine keeps its lone pair, so substitution repeats. Ammonia and 1-iodopropane give, in turn:

CH3CH2CH2NH2          propylamine (primary)
(CH3CH2CH2)2NH        dipropylamine (secondary)
(CH3CH2CH2)3N         tripropylamine (tertiary)
(CH3CH2CH2)4N+ I−     tetrapropylammonium iodide (quaternary ammonium salt)

Excess ammonia favours the primary amine; excess halogenoalkane pushes through to the quaternary salt. The sequence stops there because N in R₄N⁺ has no lone pair left.

Worked example 4: outline a substitution mechanism

Outline the mechanism for dimethylamine, (CH₃)₂NH, reacting with 1-chlorobutane.

• Curly arrow 1: from the lone pair on N of dimethylamine to the δ+ carbon of C–Cl.
• Curly arrow 2: from the C–Cl bond to Cl. The bond breaks heterolytically and Cl⁻ leaves.
• Intermediate: (CH₃)₂N⁺H–CH₂CH₂CH₂CH₃, with the positive charge on N.
• Curly arrow 3: from the lone pair of a second dimethylamine molecule to the H on N⁺. Curly arrow 4: from that N–H bond back to N⁺.
• Products: the tertiary amine (CH₃)₂NCH₂CH₂CH₂CH₃ (N,N-dimethylbutylamine) and (CH₃)₂NH₂⁺Cl⁻.

Quaternary ammonium salts as cationic surfactants

A tertiary amine and a long-chain halogenoalkane give a quaternary salt with a long hydrocarbon tail, such as CH₃(CH₂)₁₃N⁺(CH₂CH₃)₃ Cl⁻ from triethylamine and 1-chlorotetradecane. The charged N⁺ head interacts with water; the non-polar tail mixes with oils and grease. A surfactant with a positive head is cationic. Quaternary ammonium salts of this kind are used in liquid fabric softeners and hair conditioners.

Addition–elimination with acyl chlorides

The carbonyl carbon of an acyl chloride is strongly δ+, because O and Cl both pull electron density from it. Ammonia gives an amide and a primary amine gives an N-substituted amide. A second mole of the nitrogen compound removes the HCl:

(CH3)3CCOCl + 2NH3 → (CH3)3CCONH2 + NH4Cl            2,2-dimethylpropanamide

Worked example 5: outline the addition–elimination mechanism

Outline the mechanism for propanoyl chloride, CH₃CH₂COCl, with cyclohexylamine.

• Addition. Curly arrow from the lone pair on N of C₆H₁₁NH₂ to the δ+ carbonyl carbon; a second arrow moves the C=O double-bond pair onto O, giving O⁻.
• Intermediate: a tetrahedral carbon carrying O⁻, Cl, CH₂CH₃ and N⁺H₂C₆H₁₁.
• Elimination. Curly arrow from a lone pair on O⁻ back to reform C=O; the C–Cl bonding pair moves onto Cl, expelling Cl⁻.
• Loss of H⁺. A second cyclohexylamine molecule (or Cl⁻) takes H⁺ from N⁺; the N–H electrons return to nitrogen.
• Overall: CH₃CH₂COCl + 2C₆H₁₁NH₂ → CH₃CH₂CONHC₆H₁₁ + C₆H₁₁NH₃⁺Cl⁻. The amide is N-cyclohexylpropanamide.

Acid anhydrides

Acid anhydrides react in the same way, more slowly, and the by-product is a carboxylic acid, not HCl. The specification asks for the mechanism only with acyl chlorides.

(CH3CO)2O + CH3C6H4NH2 → CH3CONHC6H4CH3 + CH3COOH
                         N-(4-methylphenyl)ethanamide

A second mole of amine would neutralise the ethanoic acid.

Common errors

• Calling (CH₃)₃CNH₂ tertiary. Count carbons on nitrogen only.
• Treating any amine with a benzene ring as aromatic. N must be bonded to the ring.
• Under-counting hydrogen. Each C≡N needs 4[H] and each NO₂ needs 6[H].
• Explaining phenylamine’s weakness without saying the N lone pair is delocalised into the ring.
• Leaving the positive charge off nitrogen in either intermediate, or stopping before the H⁺ is removed.
• Giving HCl as a by-product of an anhydride reaction. It is a carboxylic acid.
• Dropping the N- from names such as N-cyclohexylpropanamide.

Next steps

Move on to the amines revision notes and, after those, the amines practice set. Amide links return in polymers, and amine routes in organic synthesis. The free diagnostics show where to focus.

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.