OxfordAQA A-Level Chemistry: Amines (9620) – 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.

This set tests section 3.3.11, Amines, of the OxfordAQA 9620 specification (International AS and A-level Chemistry, Version 5.3), used for exams from the May/June 2020 series onwards, across subsections 3.3.11.1 to 3.3.11.3 (preparation, base properties, nucleophilic properties). As International A2 material, it is assessed at International A-level only: Unit 4 examines 3.3.7 to 3.3.16, while Unit 5 (synoptic) may test anything. Question 3 also refers to nitration (section 3.3.10.2), and question 11 uses substitution by cyanide ions (section 3.3.3).

Before starting, work through the amines study guide and the revision notes on amines. Planning tools: the OxfordAQA chemistry hub, a checklist you can print and our diagnostics.

Ar values to use: C = 12.0, H = 1.0, N = 14.0, O = 16.0 and Cl = 35.5.

Questions

1. State whether each of these amines is primary, secondary or tertiary. (i) (CH₃)₂CHCH₂NHCH₃ (ii) C₆H₅N(CH₃)₂ (iii) CH₃CH₂C(CH₃)₂NH₂ [3]

2. Pentanedinitrile, NC–CH₂CH₂CH₂–CN, is fully reduced to pentane-1,5-diamine. Name a reagent that can carry out this reduction, then write its equation with [H] as the reducing agent. [3]

3. 1-Chloro-3-nitrobenzene, ClC₆H₄NO₂, is made by nitration and then reduced to 3-chlorophenylamine, ClC₆H₄NH₂, using tin and concentrated hydrochloric acid.

(a) Write an equation for the reduction, using [H]. [1] (b) Explain why sodium hydroxide is added once the reduction is complete. [2] (c) 11.03 g of 1-chloro-3-nitrobenzene gave 6.43 g of 3-chlorophenylamine. Calculate the percentage yield. [3]

4. Benzylamine, C₆H₅CH₂NH₂, and phenylamine, C₆H₅NH₂, both contain a benzene ring.

(a) Explain why benzylamine is the stronger base. [4] (b) State where ammonia fits in base strength relative to these two amines. [1]

5. 3-Methylbutan-1-amine, (CH₃)₂CHCH₂CH₂NH₂, dissolves in water.

(a) Give the equation for its reaction with water and explain the term weak base for this amine. [2] (b) Give the equation for its complete reaction with sulfuric acid. [1] (c) Calculate the volume of 0.0500 mol dm⁻³ sulfuric acid that reacts exactly with 0.435 g of 3-methylbutan-1-amine. [3]

6. Methylamine reacts with 2-bromopropane to give the secondary amine N-methylpropan-2-amine, (CH₃)₂CHNHCH₃.

(a) Outline the mechanism. Describe each curly arrow and give the structure of the intermediate. [4] (b) Give the structural formula of the tertiary amine formed if the product reacts with more 2-bromopropane. [1]

7. Phenylamine is heated with a large excess of iodomethane.

(a) Give the formulae of the three nitrogen-containing products formed in turn. [3] (b) Explain why the third product does not react with more iodomethane. [1] (c) State how the conditions should change to make the first product the main one. [1]

8. N,N-Dimethyldodecylamine, CH₃(CH₂)₁₁N(CH₃)₂, reacts with chloromethane.

(a) Give the formula of the organic product. [1] (b) Explain why the product can act as a cationic surfactant. [2] (c) State one use of quaternary ammonium salts of this type. [1]

9. Benzoyl chloride, C₆H₅COCl, reacts with excess propan-1-amine, CH₃CH₂CH₂NH₂.

(a) Outline the mechanism, saying where each curly arrow starts and ends and showing the intermediate. [4] (b) Give the name of the amide formed. [1] (c) Write an equation for the reaction of cyclohexanecarbonyl chloride, C₆H₁₁COCl, with excess ammonia. [2]

10. 3-Chlorophenylamine, ClC₆H₄NH₂, reacts with ethanoic anhydride in a 1 : 1 ratio.

(a) Write an equation for the reaction. [2] (b) 5.10 g of 3-chlorophenylamine gave 5.42 g of the amide, C₈H₈ClNO. Calculate the percentage yield. [3]

11. A chemist needs cyclopentylmethanamine, C₅H₉CH₂NH₂, and has bromocyclopentane, C₅H₉Br.

(a) Give a two-step route. For each step give the reagent and conditions, name the intermediate, and write an equation for step 2 using [H]. [4] (b) A colleague suggests heating bromocyclopentane with excess ammonia instead. Give two reasons why this would not give a pure sample of the target. [2] (c) A 0.500 g sample of the product, impure, was dissolved in water and titrated with 0.100 mol dm⁻³ hydrochloric acid. The impurities do not react with acid. The titre was 48.60 cm³.

(i) Write an equation for the reaction. [1] (ii) Calculate the percentage by mass of cyclopentylmethanamine in the sample. [3]

Answers

1. (i) secondary [1]; (ii) tertiary [1]; (iii) primary [1] [3] Examiner insight: In (iii) only one carbon is bonded to N, even though that carbon carries three others.

2. H₂ with a nickel catalyst, or LiAlH₄ in dry ether [1]; NC(CH₂)₃CN + 8[H] → H₂N(CH₂)₅NH₂: correct diamine [1]; 8[H] [1] [3] Examiner insight: Two nitrile groups need twice the hydrogen; an equation balanced for one group only is unbalanced, so count H atoms on both sides before moving on.

3. (a) ClC₆H₄NO₂ + 6[H] → ClC₆H₄NH₂ + 2H₂O [1] (b) In the acid, the amine is protonated and is present as its salt, ClC₆H₄NH₃⁺ [1]; NaOH removes H⁺ from the –NH₃⁺ group and releases the free amine [1]. (c) n(ClC₆H₄NO₂) = 11.03 ÷ 157.5 = 0.0700 mol [1]; theoretical mass = 0.0700 × 127.5 = 8.93 g [1]; yield = 6.43 ÷ 8.93 × 100 = 72.0% [1] [6] Examiner insight: Show both Mr values; a percentage worked from masses alone (6.43 ÷ 11.03) skips the mole step and is not a valid method.

4. (a) Both bases accept H⁺ using the lone pair on N [1]. In benzylamine, N is bonded to a CH₂ group, not the ring, so its lone pair is not delocalised and the alkyl carbon releases electrons onto N [1]. In phenylamine, N is bonded directly to the ring and its lone pair is partly delocalised into the ring, reducing electron density on N [1]. So the lone pair is more available in benzylamine, which accepts H⁺ more readily [1]. (b) Benzylamine > ammonia > phenylamine [1] [5] Examiner insight: The key fact is that the CH₂ group separates N from the ring; an answer that treats both as aromatic amines cannot explain the difference.

5. (a) (CH₃)₂CHCH₂CH₂NH₂ + H₂O ⇌ (CH₃)₂CHCH₂CH₂NH₃⁺ + OH⁻ [1]; only a small proportion is protonated / the equilibrium lies to the left [1]. (b) 2(CH₃)₂CHCH₂CH₂NH₂ + H₂SO₄ → ((CH₃)₂CHCH₂CH₂NH₃)₂SO₄ [1] (c) n(amine) = 0.435 ÷ 87.0 = 5.00 × 10⁻³ mol [1]; n(H₂SO₄) = 5.00 × 10⁻³ ÷ 2 = 2.50 × 10⁻³ mol [1]; V = 2.50 × 10⁻³ ÷ 0.0500 = 0.0500 dm³ = 50.0 cm³ [1] [6] Examiner insight: Sulfuric acid has two protons to give, so it reacts with two amine molecules; using 1 : 1 doubles the volume to 100 cm³.

6. (a) Methylamine’s N lone pair attacks the δ+ carbon carrying Br [1]; the C–Br bonding pair moves onto Br, so bromide is expelled [1]; intermediate (CH₃)₂CH–N⁺H₂CH₃, with the positive charge on N [1]; a second methylamine uses its lone pair to remove H⁺ from N⁺, and the N–H electrons return to N [1]. (b) ((CH₃)₂CH)₂NCH₃ [1] [5] Examiner insight: Begin the first arrow at the lone pair itself, not at the N symbol or a hydrogen, so the electron pair donor is clear.

7. (a) C₆H₅NHCH₃ [1]; C₆H₅N(CH₃)₂ [1]; C₆H₅N⁺(CH₃)₃ I⁻ [1] (b) In C₆H₅N⁺(CH₃)₃ the nitrogen has no lone pair left, so it cannot act as a nucleophile [1]. (c) Use a large excess of phenylamine (not of iodomethane) [1]. [5] Examiner insight: Show the quaternary product as a salt with its iodide ion; a neutral formula with five bonds to N is not a sensible structure.

8. (a) CH₃(CH₂)₁₁N⁺(CH₃)₃ Cl⁻ [1] (b) The N⁺ head is positively charged and interacts with water [1]; the long hydrocarbon tail has no charge and dissolves in grease or oil, so the ion collects at the oil–water boundary [1]. (c) In fabric softeners or hair conditioners [1]. [4] Examiner insight: “Cationic” refers to the charge on the head group, so name the positive N⁺; describing only the chain gives half the explanation.

9. (a) Arrow from the lone pair on N of propan-1-amine to the δ+ carbonyl carbon, and arrow from the C=O bond to O [1]; tetrahedral intermediate with O⁻ and N⁺H₂CH₂CH₂CH₃ on the same carbon [1]; the O⁻ lone pair re-forms C=O while the C–Cl bonding pair moves onto Cl, expelling chloride [1]; chloride or another propan-1-amine molecule takes H⁺ off N⁺ [1]. (b) N-propylbenzamide [1] (c) C₆H₁₁COCl + 2NH₃ → C₆H₁₁CONH₂ [1] + NH₄Cl, with 2NH₃ on the left [1] [7] Examiner insight: A direct one-step swap of Cl for N describes substitution, not addition–elimination; the O⁻ intermediate must appear.

10. (a) (CH₃CO)₂O + ClC₆H₄NH₂ → CH₃CONHC₆H₄Cl [1] + CH₃COOH [1] (b) n(amine) = 5.10 ÷ 127.5 = 0.0400 mol [1]; Mr(C₈H₈ClNO) = 169.5, so theoretical mass = 0.0400 × 169.5 = 6.78 g [1]; yield = 5.42 ÷ 6.78 × 100 = 79.9% [1] [5] Examiner insight: The by-product here is ethanoic acid; writing HCl, as for an acyl chloride, is wrong chemistry for an anhydride.

11. (a) Step 1: KCN in ethanol and water, heat under reflux [1]; intermediate cyclopentanecarbonitrile, C₅H₉CN [1]. Step 2: H₂ with a nickel catalyst, or LiAlH₄ in dry ether [1]; C₅H₉CN + 4[H] → C₅H₉CH₂NH₂ [1]. (b) Ammonia puts N straight onto the ring carbon, giving cyclopentylamine, C₅H₉NH₂, which has one carbon fewer than the target [1]; the amine formed can react further with bromocyclopentane, giving a mixture of secondary and tertiary amines and a quaternary salt [1]. (c) (i) C₅H₉CH₂NH₂ + HCl → C₅H₉CH₂NH₃⁺Cl⁻ [1] (ii) n(HCl) = 0.100 × 48.60 ÷ 1000 = 4.86 × 10⁻³ mol = n(amine) [1]; mass = 4.86 × 10⁻³ × 99.0 = 0.481 g [1]; percentage = 0.481 ÷ 0.500 × 100 = 96.2% [1] [10] Examiner insight: In (b), “it gives a mixture” is only one of the two reasons; the wrong carbon count is a separate point and needs its own sentence.

Where marks are usually lost

• Classifying an amine by the carbon next to N rather than by the carbons on N.
• Treating benzylamine-type compounds as aromatic amines.
• Unbalanced [H] equations, especially for molecules with two nitrile or nitro groups.
• Arrows that start at an atom rather than at its lone pair or a bond.
• No + charge on N in the intermediate, or no final H⁺ removal.
• An addition–elimination mechanism drawn as one step with no O⁻.
• HCl given as a by-product of an acid anhydride reaction.
• Dropping the N- from N-substituted amide names.

Next steps

• Recap with the revision notes for amines
• Re-read the study guide for amines
• More substitution work: halogenoalkanes practice
• All free 10-minute diagnostics
• The OxfordAQA chemistry course hub
• Book a free trial class
• A checklist you can print for the whole course

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.