AQA A-Level Chemistry: Amines (7405)

This guide teaches 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. It covers every point in 3.3.11.1 Preparation, 3.3.11.2 Base properties and 3.3.11.3 Nucleophilic properties. The whole section is A-level only. The specification places organic chemistry (section 3.3) on A-level Paper 2, and Paper 3 can test any content.

For quick recall, use the amines revision notes. To test yourself, use the amines practice questions. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. The first substitution step, halogenoalkane plus ammonia, is taught in halogenoalkanes, and Brønsted–Lowry acids and bases are in acids and bases. Not sure where your gaps are? Try a free 10-minute diagnostic.

What this unit covers

Spec point	What you must be able to do	Status
3.3.11 introduction	Know that amines are based on ammonia with H atoms replaced by alkyl or aryl groups	A-level only
3.3.11.1 Preparation	Prepare primary aliphatic amines from ammonia and halogenoalkanes, and by reducing nitriles; know that aromatic amines are made by reducing nitro compounds and are used to make dyes	A-level only
3.3.11.2 Base properties	Know that amines are weak bases; explain the difference in base strength between ammonia, primary aliphatic and primary aromatic amines using lone pair availability on N	A-level only
3.3.11.3 Nucleophilic properties	Outline nucleophilic substitution with halogenoalkanes to give primary, secondary and tertiary amines and quaternary ammonium salts; know the use of quaternary ammonium salts as cationic surfactants; outline nucleophilic addition–elimination with acyl chlorides; know the reactions with acid anhydrides	A-level only

No required practical is listed under 3.3.11. The reactions of acyl chlorides and acid anhydrides with ammonia and primary amines also appear in 3.3.9.2 Acylation.

What an amine is

An amine is a compound based on ammonia, NH₃, in which one or more hydrogen atoms are replaced by alkyl or aryl groups. Amines are classified by the number of carbon atoms bonded to the nitrogen:

Type	General formula	Example
Primary	RNH₂	CH₃CH₂NH₂, ethylamine
Secondary	R₂NH	(CH₃CH₂)₂NH, diethylamine
Tertiary	R₃N	(CH₃)₃N, trimethylamine
Quaternary ammonium ion	R₄N⁺	(CH₃CH₂)₄N⁺, tetraethylammonium

Unlike alcohols, you count carbons on the nitrogen. (CH₃)₂CHNH₂ is primary: N carries one carbon.

An aromatic amine has N bonded directly to a benzene ring, as in phenylamine, C₆H₅NH₂; an aliphatic amine has N bonded to an alkyl chain.

You will also see substitutive names such as ethanamine for ethylamine.

The lone pair on N explains this whole section: it accepts a proton (base) and attacks a δ+ carbon (nucleophile). A quaternary ion has used its lone pair, so it is neither.

3.3.11.1 Preparation

Route 1: ammonia and a halogenoalkane

Reagent and conditions: excess ammonia in ethanol, heated in a sealed container.

CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br

The mechanism is in the halogenoalkanes guide. The weakness: the product amine is also a nucleophile, so it reacts with more halogenoalkane, giving a mixture (see 3.3.11.3). Excess ammonia makes the primary amine the main product, but the mixture still needs separating.

Route 2: reduction of nitriles

A nitrile, R–C≡N, is reduced to a primary amine, RCH₂NH₂. The specification names the reaction type but not a reagent. Two common choices are LiAlH₄ in dry ether, or hydrogen with a nickel catalyst.

CH3CN + 4[H] → CH3CH2NH2          (reducing agent written as [H])
CH3CN + 2H2  → CH3CH2NH2          (hydrogen, Ni catalyst)

The C≡N carbon becomes the CH₂ on N, so propanenitrile gives propylamine.

Advantages over route 1: only the primary amine forms. After the halogenoalkane-plus-cyanide reaction, it also gives an amine with one more carbon than the starting halogenoalkane.

Worked example 1 (atom economy of the two routes to ethylamine). Use Ar: H = 1.0, C = 12.0, N = 14.0, Br = 79.9. Compare route 1 (bromoethane + ammonia) with route 2 (ethanenitrile + hydrogen).

Step 1: Mr values. CH₃CH₂NH₂ = 45.0; CH₃CH₂Br = 108.9; NH₃ = 17.0; CH₃CN = 41.0; H₂ = 2.0. Step 2: route 1 uses CH₃CH₂Br + 2NH₃, total Mr = 108.9 + 34.0 = 142.9. Atom economy = 45.0 ÷ 142.9 × 100 = 31.5%. Step 3: route 2 uses CH₃CN + 2H₂, total Mr = 41.0 + 4.0 = 45.0. Atom economy = 45.0 ÷ 45.0 × 100 = 100%. Step 4: route 1 also loses yield to further substitution, which atom economy does not show.

Aromatic amines: reduction of nitro compounds

Benzene is nitrated (3.3.10.2) to nitrobenzene, and the nitro group is then reduced. A common method is tin and concentrated hydrochloric acid, heated under reflux. In acid the product forms as its salt, so sodium hydroxide is then added to release the free amine.

C6H5NO2 + 6[H] → C6H5NH2 + 2H2O

Count the [H]: two O atoms leave as 2H₂O (four H) and two H end up on N, so six [H] in total.

The specification says aromatic amines made this way are used in the manufacture of dyes.

Worked example 2 (percentage yield). 12.3 g of nitrobenzene gives 6.51 g of phenylamine. Use Ar: H = 1.0, C = 12.0, N = 14.0, O = 16.0. Calculate the percentage yield.

Step 1: Mr C₆H₅NO₂ = 123.0; Mr C₆H₅NH₂ = 93.0. Step 2: moles of nitrobenzene = 12.3 ÷ 123.0 = 0.100 mol. Step 3: the mole ratio is 1 : 1, so the theoretical mass = 0.100 × 93.0 = 9.30 g. Step 4: percentage yield = 6.51 ÷ 9.30 × 100 = 70.0%.

3.3.11.2 Base properties

Amines are weak bases

A Brønsted–Lowry base is a proton acceptor. The nitrogen lone pair forms a dative bond to H⁺. In water only a small fraction of the amine reacts, so an amine is a weak base:

CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH−

The OH⁻ ions make the solution alkaline. With a strong acid the reaction goes to completion and forms a salt:

CH3CH2NH2 + HCl → CH3CH2NH3+Cl−      (ethylammonium chloride)

Adding NaOH to the salt removes H⁺ from RNH₃⁺ and gives back the amine (the same step that releases phenylamine after the tin and acid reduction).

Comparing base strength

The specification asks you to compare three species. The order is:

primary aliphatic amine > ammonia > primary aromatic amine

for example, ethylamine > ammonia > phenylamine.

The rule is: the more available the lone pair on N, the more readily it accepts H⁺, so the stronger the base.

• Primary aliphatic amine. Alkyl groups are electron-releasing (they have a positive inductive effect). They push electron density onto N, so the lone pair is more available than in ammonia. Stronger base than NH₃.
• Ammonia. No electron-releasing group on N. It is the reference point.
• Primary aromatic amine. The lone pair on N overlaps with the delocalised π system of the benzene ring and is partly delocalised into it. Less electron density is available on N, so it is a weaker base than NH₃.

Worked example 3 (explain an order). Explain why propylamine is a stronger base than phenylamine.

Step 1: both accept H⁺ using the lone pair on N. Step 2: the propyl group is electron-releasing, increasing electron density on N. Step 3: in phenylamine the N lone pair is delocalised into the benzene ring, decreasing electron density on N. Step 4: the lone pair is more available in propylamine, so it accepts a proton more readily.

3.3.11.3 Nucleophilic properties

A nucleophile is an electron pair donor. The amine lone pair attacks an electron-deficient carbon, so amines take part in the same nucleophilic substitution as ammonia.

Substitution with halogenoalkanes: a ladder of products

Each product still has a lone pair on N, so it can attack another halogenoalkane:

NH3      + RX → RNH2 (primary)      + HX
RNH2     + RX → R2NH (secondary)    + HX
R2NH     + RX → R3N  (tertiary)     + HX
R3N      + RX → R4N+ X− (quaternary ammonium salt)

In practice the HX is taken up by more base as an ammonium salt.

To favour the primary amine, use excess ammonia. To favour the quaternary ammonium salt, use excess halogenoalkane. The quaternary ion has no H on N to lose and no lone pair left, so the sequence stops there.

Worked example 4 (outline the mechanism for ethylamine + bromomethane). Describe each arrow.

Step 1: a curly arrow from the lone pair on N of CH₃CH₂NH₂ to the δ+ carbon of CH₃Br. Step 2: a curly arrow from the C–Br bond to Br. The bond breaks heterolytically and Br⁻ leaves. Step 3: the intermediate is CH₃CH₂N⁺H₂CH₃, with a positive charge on N. Step 4: a second ethylamine removes H⁺ from N⁺ (arrow from its lone pair to the H; arrow from the N–H bond to N⁺). Products: N-methylethylamine, CH₃CH₂NHCH₃ (a secondary amine), and CH₃CH₂NH₃⁺Br⁻.

Quaternary ammonium salts as cationic surfactants

A quaternary ammonium salt with one long hydrocarbon chain, such as CH₃(CH₂)₁₅N⁺(CH₃)₃ Br⁻, behaves as a surfactant. The ionic N⁺ head is attracted to water and the long chain is not. Because the head group is positive it is called a cationic surfactant, and it is attracted to negatively charged surfaces.

Addition–elimination with acyl chlorides

Ammonia and primary amines react with acyl chlorides, RCOCl, to give amides. The reaction is fast at room temperature.

CH3COCl + 2NH3       → CH3CONH2 + NH4Cl                   (ethanamide)
CH3COCl + 2CH3NH2    → CH3CONHCH3 + CH3NH3+Cl−            (N-methylethanamide)

A primary amine gives an N-substituted amide; the N- shows the alkyl group is on nitrogen. The second mole of base takes up HCl as a salt.

Worked example 5 (outline the mechanism for ethanoyl chloride + methylamine).

Step 1 (addition): a curly arrow from the lone pair on N of CH₃NH₂ to the δ+ carbon of C=O. A second arrow from the C=O double bond to O, giving O⁻. Step 2: the intermediate has a tetrahedral carbon bonded to O⁻, Cl, CH₃ and N⁺H₂CH₃. Step 3 (elimination): a curly arrow from the lone pair on O⁻ back to form C=O again, and an arrow from the C–Cl bond to Cl, which leaves as Cl⁻. Step 4: a base (Cl⁻, or a second CH₃NH₂) removes H⁺ from N⁺, with an arrow from the N–H bond to N. Product: CH₃CONHCH₃, N-methylethanamide.

The carbonyl carbon is strongly δ+ because both O and Cl are more electronegative than carbon.

Acid anhydrides

Acid anhydrides, (RCO)₂O, give the same amides, more slowly. The by-product is a carboxylic acid instead of HCl:

(CH3CO)2O + NH3     → CH3CONH2 + CH3COOH
(CH3CO)2O + CH3NH2  → CH3CONHCH3 + CH3COOH

With excess base the acid is then neutralised to its salt. The specification does not ask for the anhydride mechanism.

Common errors

• Classifying (CH₃)₂CHNH₂ as secondary. Count carbons on N, not on the carbon next to N.
• Writing the nitrile reduction with 2[H]. C≡N to CH₂–NH₂ needs 4[H].
• Using 3[H] or 4[H] for nitrobenzene. It needs 6[H] and gives 2H₂O.
• Explaining phenylamine’s weakness as “the benzene ring is electron-withdrawing” with no mention of the lone pair being delocalised into the ring.
• Stopping the substitution mechanism at the N⁺ intermediate without the deprotonation step.
• Naming CH₃CONHCH₃ “methylethanamide” without the N-, which places the methyl group on carbon.

Next steps

Next, use the revision notes and the practice questions. Infrared absorptions for N–H are in organic analysis. The exam preparation guide covers paper structure, and the free 10-minute diagnostics show where to focus.

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.