OCR A Level Chemistry A: Amines (H432) – 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.

Here you practise section 6.2.1, Amines, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 6.2.1(a), (b)(i) and (b)(ii), from Module 6, Organic chemistry and analysis. H432 is the full A level. Questions 9 to 11 are longer problems; two of them draw on nitration (6.1.1) and nitrile reduction (6.2.4).

Revise first with the Amines study guide and the Amines revision notes. Other Module 6 topics are on the OCR chemistry hub; tick each secure outcome on the printable checklist. For a wider check, try the free diagnostics.

Use these A_r values throughout: H 1.0, C 12.0, N 14.0, O 16.0, S 32.1, Cl 35.5.

Questions

1. Explain why amines can act as bases. [2]

2. Three amines are shown: P (CH₃)₂CHNHCH₃, R CH₃CH₂C(CH₃)₂NH₂ and T (CH₃CH₂)₂NCH₃.

(a) Classify each amine as primary, secondary or tertiary. [3] (b) Write an equation for P reacting with dilute hydrochloric acid. [1]

3. Hexane-1,6-diamine, H₂N(CH₂)₆NH₂, is shaken with excess dilute hydrochloric acid.

(a) Write an equation for the reaction. [2] (b) Write an ionic equation to show how the free diamine can be recovered from the salt. [1]

4. 4-Methylpentan-1-amine is made from 1-chloro-4-methylpentane, (CH₃)₂CHCH₂CH₂CH₂Cl.

(a) State the reagent and conditions. [2] (b) Give the balanced equation for this preparation. [1] (c) Explain why the solvent is ethanol rather than water. [1] (d) Explain why the reagent is used in excess, and give the structural formula of one by-product that forms if it is not. [3]

5. N-methylbutan-1-amine, CH₃CH₂CH₂CH₂NHCH₃, is to be made from methylamine.

(a) Identify the organic reagent needed. [1] (b) Write an equation for the reaction, using two moles of methylamine, and explain why two moles are needed. [2] (c) The product can react further with the same organic reagent. Give the structural formula and class of the amine formed. [2]

6. 2-Fluorophenylamine is made from 1-fluoro-2-nitrobenzene, FC₆H₄NO₂.

(a) State the reagents and conditions for the reduction. [2] (b) Using [H] to represent the reducing agent, give the equation for this change. [1] (c) Explain why aqueous sodium hydroxide is added at the end. [1]

7. A solution of a primary aliphatic amine, CₙH₂ₙ₊₁NH₂, has a concentration of 7.52 g dm⁻³. A 25.0 cm³ portion is neutralised by 21.60 cm³ of 0.100 mol dm⁻³ hydrochloric acid.

(a) Write a general equation for the reaction, using RNH₂. [1] (b) Calculate the molar mass of the amine and deduce its molecular formula. [4] (c) Suggest a structural formula for the amine. [1]

8. A 6.39 g sample of 1,3,5-trinitrobenzene, C₆H₃(NO₂)₃, is fully reduced with tin and concentrated HCl. After the alkali step, 2.46 g of benzene-1,3,5-triamine, C₆H₃(NH₂)₃, is collected.

Give the equation, using [H], and calculate the percentage yield. [4]

9. 2,5-Dimethylphenylamine is made from 1,4-dimethylbenzene in two stages. Stage 1 gives 1,4-dimethyl-2-nitrobenzene, (CH₃)₂C₆H₃NO₂.

(a) State the reagents for stage 1. [1] (b) State the reagents and conditions for stage 2, including the final step. [2] (c) Write an equation for stage 2 using [H]. [1] (d) In stage 2, 7.55 g of the nitro compound gives 4.84 g of 2,5-dimethylphenylamine. Calculate the percentage yield. [3] (e) All 4.84 g of the amine is reacted with dilute sulfuric acid. Write an equation and calculate the mass of salt formed. [3]

10. Two routes to hexan-1-amine are proposed.

Route A: 1-bromohexane is heated with excess ethanolic ammonia. Route B: 1-bromopentane is converted into a nitrile with cyanide ions, then the nitrile is reduced with hydrogen and a nickel catalyst.

(a) Write an equation for route A. [1] (b) Give the structural formula of the nitrile in route B and write an equation for its reduction. [2] (c) Explain why route B gives hexan-1-amine from a five-carbon haloalkane. [1] (d) Suggest one advantage of route B over route A for obtaining a pure primary amine, and explain it. [2]

11. Amine X is a saturated, non-cyclic secondary amine. X reacts with dilute hydrochloric acid to give a salt that contains 25.8% chlorine by mass.

(a) Calculate the molar mass of the salt and of X, and deduce the molecular formula of X. [3] (b) X is made by heating butan-1-amine with a chloroalkane in ethanol. Identify the chloroalkane and write an equation using two moles of butan-1-amine. [2] (c) Give the structural formula of a tertiary amine that could also form in this reaction. [1] (d) State how X could be released from its hydrochloride salt. [1]

Answers

1. Nitrogen in an amine carries a lone pair [1], which can bond datively to an H⁺ ion, so amines are proton acceptors [1] Examiner insight: Both ideas are needed: “lone pair” alone or “accepts H⁺” alone is half an answer, and the lone pair must be placed on nitrogen.

2. (a) P secondary [1]; R primary [1]; T tertiary [1] (b) (CH₃)₂CHNHCH₃ + HCl → (CH₃)₂CHNH₂⁺CH₃ Cl⁻ [1] Examiner insight: R is primary: only one carbon is bonded to its nitrogen, even though that carbon carries three others. Count carbons on N.

3. (a) H₂N(CH₂)₆NH₂ + 2HCl [1] → Cl⁻ ⁺H₃N(CH₂)₆NH₃⁺ Cl⁻ [1] (b) ⁺H₃N(CH₂)₆NH₃⁺ + 2OH⁻ → H₂N(CH₂)₆NH₂ + 2H₂O [1] Examiner insight: With excess acid both nitrogens are protonated; an equation with 1 HCl leaves one lone pair unreacted and does not match “excess”.

4. (a) Excess ammonia dissolved in ethanol [1]; heated in a sealed vessel [1] (b) (CH₃)₂CHCH₂CH₂CH₂Cl + 2NH₃ → (CH₃)₂CHCH₂CH₂CH₂NH₂ + NH₄Cl [1] (c) In water, hydroxide ions or water would substitute the chlorine and give an alcohol [1] (d) The amine product has a lone pair on N, so it is also a nucleophile [1]; it can react with more haloalkane, and excess ammonia makes reaction with ammonia much more likely [1]; by-product: ((CH₃)₂CHCH₂CH₂CH₂)₂NH [1] Examiner insight: “Heat” on its own, or “aqueous ammonia”, misses the conditions; the words “excess” and “ethanol” each carry meaning here.

5. (a) 1-bromobutane (or another 1-halobutane) [1] (b) 2CH₃NH₂ + CH₃CH₂CH₂CH₂Br → CH₃CH₂CH₂CH₂NHCH₃ + CH₃NH₃⁺Br⁻ [1]; the HBr formed is an acid and protonates a second methylamine molecule [1] (c) (CH₃CH₂CH₂CH₂)₂NCH₃ [1]; a tertiary amine [1] Examiner insight: An equation ending in HBr is acceptable chemistry only if you then show where the HBr goes; when two moles are asked for, the salt must appear as a product.

6. (a) Reagents: Sn with concentrated HCl [1]; conditions: heat, under reflux [1] (b) FC₆H₄NO₂ + 6[H] → FC₆H₄NH₂ + 2H₂O [1] (c) In the acid the amine is present as its protonated salt, FC₆H₄NH₃⁺; OH⁻ removes the proton to release the free amine [1] Examiner insight: “Dilute HCl” or “HCl” without “concentrated” is a common way to lose the reagent point; check the [H] count by balancing H atoms on both sides.

7. (a) RNH₂ + HCl → RNH₃⁺Cl⁻ [1] (b) n(HCl) = 0.100 × 21.60/1000 = 0.00216 mol, so n(amine) = 0.00216 mol [1]; concentration = 0.00216 ÷ 0.0250 = 0.0864 mol dm⁻³ [1]; M = 7.52 ÷ 0.0864 = 87.0 g mol⁻¹ [1]; 14n + 17 = 87.0 gives n = 5, so C₅H₁₃N (C₅H₁₁NH₂) [1] (c) Any primary C₅ amine, e.g. CH₃CH₂CH₂CH₂CH₂NH₂ [1] Examiner insight: Show the 1 : 1 ratio and each conversion; a molar mass with no working gives nothing to credit if the arithmetic slips.

8. C₆H₃(NO₂)₃ + 18[H] → C₆H₃(NH₂)₃ + 6H₂O [1]; n(trinitro) = 6.39 ÷ 213.0 = 0.0300 mol [1]; n(triamine) = 2.46 ÷ 123.0 = 0.0200 mol [1]; yield = 0.0200 ÷ 0.0300 × 100 = 66.7% [1] Examiner insight: Three nitro groups need 18[H] and give 6H₂O; check by counting H (21 on each side), since the 1 : 1 mole ratio hides an unbalanced equation.

9. (a) Concentrated nitric acid and concentrated sulfuric acid [1] (b) Tin and concentrated HCl, heated under reflux [1]; then excess NaOH(aq) [1] (c) (CH₃)₂C₆H₃NO₂ + 6[H] → (CH₃)₂C₆H₃NH₂ + 2H₂O [1] (d) n(C₈H₉NO₂) = 7.55 ÷ 151.0 = 0.0500 mol [1]; n(C₈H₁₁N) = 4.84 ÷ 121.0 = 0.0400 mol [1]; yield = 0.0400 ÷ 0.0500 × 100 = 80.0% [1] (e) 2(CH₃)₂C₆H₃NH₂ + H₂SO₄ → ((CH₃)₂C₆H₃NH₃⁺)₂SO₄²⁻ [1]; n(salt) = 0.0400 ÷ 2 = 0.0200 mol [1]; M = 2(122.0) + 96.1 = 340.1 g mol⁻¹, so mass = 0.0200 × 340.1 = 6.80 g [1] Examiner insight: In (e) the 2 : 1 ratio is the deciding step; a 1 : 1 ratio doubles the moles of salt and gives an answer twice too big.

10. (a) CH₃(CH₂)₅Br + 2NH₃ → CH₃(CH₂)₅NH₂ + NH₄Br [1] (b) Nitrile: CH₃(CH₂)₄CN (hexanenitrile) [1]; CH₃(CH₂)₄CN + 2H₂ → CH₃(CH₂)₅NH₂ [1] (c) The cyanide carbon adds a sixth carbon to the chain and becomes the CH₂ next to NH₂ [1] (d) Route B avoids secondary and tertiary amines from further substitution [1], because the haloalkane is used up making the nitrile before any amine forms, so less separation is needed [1] Examiner insight: In (d), “higher yield” without a reason linked to further substitution in route A does not answer the question.

11. (a) M(salt) = 35.5 ÷ 0.258 = 137.6 g mol⁻¹ [1]; M(X) = 137.6 − 36.5 = 101 g mol⁻¹ [1]; 14n + 17 = 101 gives n = 6, so C₆H₁₅N [1] (b) Chloroethane [1]; 2CH₃(CH₂)₃NH₂ + CH₃CH₂Cl → CH₃(CH₂)₃NHCH₂CH₃ + CH₃(CH₂)₃NH₃⁺Cl⁻ [1] (c) CH₃(CH₂)₃N(CH₂CH₃)₂ [1] (d) Add aqueous sodium hydroxide (or another strong alkali) [1] Examiner insight: Subtract HCl (36.5), not Cl (35.5), to get from the salt to the amine; the salt contains the extra proton as well as the chloride.

Where marks are usually lost

• Counting carbons on the wrong atom when deciding whether an amine is primary, secondary or tertiary (question 2).
• Protonating only one nitrogen in a diamine with excess acid (question 3).
• Writing “ammonia, heat” with no mention of excess or ethanol (question 4).
• Leaving HX as a product without showing it react with a second base molecule (question 5).
• Omitting “concentrated” with Sn/HCl, or leaving out the NaOH step (questions 6 and 9).
• Getting the 2 : 1 amine to sulfuric acid ratio wrong (question 9).
• Subtracting 35.5 instead of 36.5 when working back from a hydrochloride salt (question 11).

Next steps

• Recap the tables and self-test: Amines revision notes
• Full explanations and worked examples: Amines study guide
• Check the whole course: all free 10-minute diagnostics
• Mole, titration and yield methods: Amount of substance practice
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.2.1 Amines, learning outcomes (a) and (b)(i)–(ii). Published by OCR.