AQA A-Level Chemistry: Amines (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

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

Learn the ideas first in the amines study guide and the amines revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. For the first substitution step, see the halogenoalkanes practice questions. Use Ar values H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5, Br = 79.9 wherever they are needed.

Questions

1. Classify each amine as primary, secondary or tertiary. (i) CH₃CH(NH₂)CH₂CH₃ (ii) CH₃NHCH₂CH₂CH₃ (iii) (CH₃CH₂)₃N [3]

2. Propanenitrile, CH₃CH₂CN, can be reduced to a primary amine. Suggest a suitable reducing agent and write an equation for the reaction, using [H] for the reducing agent. [3]

3. Phenylamine is made from nitrobenzene.

(a) State the reagents used to reduce nitrobenzene, and the substance added afterwards to release free phenylamine. [2] (b) Write an equation for the reduction, using [H]. [1] (c) 18.45 g of nitrobenzene gives 10.32 g of phenylamine. Calculate the percentage yield. [3] (d) State one use of aromatic amines given in the specification. [1]

4. Methylamine, ammonia and phenylamine all act as bases.

(a) Write the three in order of decreasing base strength. [1] (b) Explain the order. [4]

5. Butylamine, CH₃CH₂CH₂CH₂NH₂, dissolves in water to give an alkaline solution.

(a) Write an equation for butylamine reacting with water, and explain why butylamine is described as a weak base. [2] (b) Write an equation for butylamine reacting with hydrochloric acid and name the salt formed. [2] (c) State how butylamine can be recovered from this salt. [1]

6. Propylamine reacts with bromoethane to form a secondary amine.

(a) Outline the mechanism. Describe each curly arrow and the intermediate. [4] (b) Name the type of mechanism. [1]

7. Ammonia reacts with excess chloromethane.

(a) Give the formulae of the three amines formed in sequence, and of the final product. [2] (b) Explain why the sequence stops at the final product. [2] (c) Trimethylamine reacts with 1-bromohexadecane, CH₃(CH₂)₁₅Br. Give the formula of the product and explain why it can act as a cationic surfactant. [3]

8. 2-Methylpropanoyl chloride, (CH₃)₂CHCOCl, reacts with excess ethylamine.

(a) Write an equation for the reaction. [2] (b) Name the organic product. [1] (c) Outline the mechanism. Describe each curly arrow. [4]

9. Phenylamine reacts with propanoic anhydride, (CH₃CH₂CO)₂O.

(a) Write an equation for the reaction (1 : 1). [2] (b) Name the organic product containing nitrogen. [1] (c) State one difference between this reaction and the reaction of phenylamine with propanoyl chloride. [1]

10. 1.18 g of a primary aliphatic amine was dissolved in water and made up to 250 cm³. A 25.0 cm³ portion needed 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid for neutralisation. One mole of amine reacts with one mole of HCl.

(a) Calculate the Mr of the amine. [3] (b) Deduce the molecular formula. [1] (c) Draw or write the structures, and give the names, of the two primary amines with this formula. [2]

11. A chemist wants to make hexylamine, CH₃(CH₂)₅NH₂, from 1-bromopentane.

(a) Give a two-step route. For each step state the reagent and conditions, and name the intermediate. [4] (b) Step 1 has an 85% yield and step 2 a 60% yield. Calculate the overall percentage yield. [1] (c) A second chemist heats 1-bromopentane with excess ammonia instead. Name the main organic product and calculate the percentage atom economy for its formation. [3] (d) Explain why the product mixture in (c) is harder to purify than the product of the nitrile route. [2]

Answers

1. (i) primary [1] (one carbon on N); (ii) secondary [1]; (iii) tertiary [1] [3] Examiner insight: For (i) the carbon bearing N is secondary, but the amine is primary; count carbons on N, not on the carbon next to it.

2. LiAlH₄ (in dry ether), or H₂ with a nickel catalyst [1]; CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂: correct amine [1]; 4[H] to balance [1] [3] Examiner insight: An equation with 2[H] is unbalanced and loses the balancing mark, even with the correct product.

3. (a) Tin and concentrated hydrochloric acid [1]; then sodium hydroxide [1]. (b) C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O [1] (c) n(C₆H₅NO₂) = 18.45 ÷ 123.0 = 0.150 mol [1]; theoretical mass = 0.150 × 93.0 = 13.95 g [1]; yield = 10.32 ÷ 13.95 × 100 = 74.0% [1] (d) Manufacture of dyes [1]. [7] Examiner insight: An error in an Mr in (c) can usually be carried forward into the later steps, so show each value.

4. (a) methylamine > ammonia > phenylamine [1] (b) Each base accepts H⁺ using the lone pair on N [1]. The methyl group is electron-releasing and increases electron density on N [1]. In phenylamine the N lone pair is delocalised into the benzene ring, decreasing electron density on N [1]. So lone-pair availability, and the ability to accept H⁺, is greatest in methylamine and least in phenylamine [1]. [5] Examiner insight: “Benzene is electron-withdrawing” without saying the lone pair is delocalised into the ring is too vague for the phenylamine mark.

5. (a) CH₃CH₂CH₂CH₂NH₂ + H₂O ⇌ CH₃CH₂CH₂CH₂NH₃⁺ + OH⁻ [1]; it is only partly protonated / the equilibrium lies to the left [1]. (b) CH₃CH₂CH₂CH₂NH₂ + HCl → CH₃CH₂CH₂CH₂NH₃⁺Cl⁻ [1]; butylammonium chloride [1]. (c) Add a strong base such as sodium hydroxide [1]. [5] Examiner insight: The equilibrium sign is needed in (a); a full arrow there contradicts “weak base”.

6. (a) Arrow from the lone pair on N of propylamine to the δ+ carbon of C–Br [1]; arrow from the C–Br bond to Br, Br⁻ leaves [1]; intermediate CH₃CH₂CH₂N⁺H₂CH₂CH₃ [1]; a second propylamine removes H⁺, with an arrow from its lone pair to H and from the N–H bond to N⁺, giving CH₃CH₂CH₂NHCH₂CH₃ [1]. (b) Nucleophilic substitution [1]. [5] Examiner insight: An arrow starting from the N atom rather than from its lone pair is a common way to lose the first mark.

7. (a) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N [1]; (CH₃)₄N⁺Cl⁻ [1] (b) In (CH₃)₄N⁺ the N has no lone pair [1], so it cannot act as a nucleophile and attack another chloromethane [1]. (c) CH₃(CH₂)₁₅N⁺(CH₃)₃ Br⁻ [1]; it has a positively charged ionic head that is attracted to water [1] and a long hydrocarbon chain that is not [1]. [7] Examiner insight: In (c) the word “cationic” needs the positive head to be named; a description of the chain alone earns only one mark.

8. (a) (CH₃)₂CHCOCl + 2CH₃CH₂NH₂ → (CH₃)₂CHCONHCH₂CH₃ [1] + CH₃CH₂NH₃⁺Cl⁻ [1] (b) N-ethyl-2-methylpropanamide [1] (c) Arrow from N lone pair to the δ+ carbonyl carbon [1]; arrow from C=O to O, giving O⁻ [1]; arrow from O⁻ reforming C=O with the C–Cl bond breaking and Cl⁻ leaving [1]; Cl⁻ or a second ethylamine removes H⁺ from N⁺ [1]. [7] Examiner insight: A one-step substitution with no O⁻ intermediate is not an addition–elimination mechanism and loses the second and third marks.

9. (a) (CH₃CH₂CO)₂O + C₆H₅NH₂ → CH₃CH₂CONHC₆H₅ [1] + CH₃CH₂COOH [1] (b) N-phenylpropanamide [1] (c) The by-product is propanoic acid rather than HCl, or the reaction is slower [1]. [4] Examiner insight: Writing HCl as a by-product of an anhydride reaction is wrong chemistry and loses the second mark in (a).

10. (a) n(HCl) = 0.0200 × 0.100 = 2.00 × 10⁻³ mol [1]; n(amine) in 250 cm³ = 2.00 × 10⁻³ × 10 = 0.0200 mol [1]; Mr = 1.18 ÷ 0.0200 = 59.0 [1] (b) C₃H₉N (C₃H₇NH₂) [1] (c) CH₃CH₂CH₂NH₂, propylamine (propan-1-amine) [1]; CH₃CH(NH₂)CH₃, 1-methylethylamine (propan-2-amine) [1]. [6] Examiner insight: Forgetting the ×10 scale-up from 25.0 to 250 cm³ loses one mark; the Mr can still earn credit as follow-through.

11. (a) Step 1: KCN in aqueous ethanol, heat under reflux [1]; intermediate hexanenitrile, CH₃(CH₂)₄CN [1]. Step 2: reduce with LiAlH₄ in dry ether, or H₂ with a Ni catalyst [1]; CH₃(CH₂)₄CN + 4[H] → CH₃(CH₂)₅NH₂ [1]. (b) 0.85 × 0.60 × 100 = 51% [1] (c) Pentylamine, CH₃(CH₂)₄NH₂ [1]; Mr of reactants C₅H₁₁Br + 2NH₃ = 150.9 + 34.0 = 184.9 [1]; atom economy = 87.0 ÷ 184.9 × 100 = 47.1% [1] (d) Pentylamine has a lone pair on N, so it reacts further with 1-bromopentane to give secondary and tertiary amines and a quaternary ammonium salt [1]; the nitrile route gives only the primary amine [1]. [10] Examiner insight: Naming the intermediate pentanenitrile in (a) is the usual slip; the C≡N carbon is the sixth carbon in the chain.

Where marks are usually lost

• Classifying amines by the carbon next to N instead of by the number of carbons on N.
• Using 2[H] for nitrile reduction or 4[H] for nitro reduction.
• Naming the nitrile without counting the C≡N carbon.
• Base-strength explanations with no mention of the lone pair or of accepting H⁺.
• Saying phenylamine’s ring “withdraws electrons” without saying the N lone pair is delocalised into it.
• A full arrow instead of ⇌ for an amine reacting with water.
• Leaving out the final deprotonation step in either mechanism.
• Drawing addition–elimination as one step, with no O⁻ intermediate.
• Writing HCl as the by-product of an acid anhydride reaction.
• Forgetting to scale titration moles from the portion to the whole solution.

Next steps

• Amines revision notes
• Amines study guide
• Halogenoalkanes study guide
• AQA A-Level Chemistry exam preparation
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

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.