OCR A Level Chemistry A: Aromatic compounds (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.

These questions test section 6.1.1, Aromatic compounds, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They reach all twelve outcomes, 6.1.1(a) to (l); question 9 uses the PAG7 phenol test. Module 6 is assessed in components 02 and 03.

Revise first with the aromatic compounds study guide and the aromatic compounds revision notes. Other H432 units are listed on the OCR chemistry hub; mark progress against the H432 checklist, or try the 10-minute diagnostics.

Relative atomic masses: N 14.0; H 1.0; O 16.0; C 12.0. Any further data is given in the question itself.

Questions

1. Describe the bonding in benzene according to the delocalised model, and state one way it differs from the Kekulé model. [3]

2. Hydrogenating one mole of cyclohexene releases 119 kJ. Fully hydrogenating 0.0500 mol of benzene releases 10.4 kJ.

(a) Calculate the enthalpy change of hydrogenation of benzene in kJ mol⁻¹. [1] (b) Use both values to show that benzene is more stable than the Kekulé structure predicts. Give the size of the difference. [2] (c) State one other piece of experimental evidence for delocalisation and explain how it rules out the Kekulé model. [2]

3. Name each compound.

(a) Phenol with an NO₂ group on both carbons next to the C–OH and on the carbon opposite it. [1] (b) Phenylamine with a Cl atom on the carbon opposite the C–NH₂. [1] (c) Benzene with Cl on one carbon and NO₂ on a carbon two positions away. [1]

4. Benzene is converted into nitrobenzene.

(a) State the reagents and conditions. [2] (b) Give an equation showing how the nitronium ion forms. [1] (c) Explain why the temperature is controlled. [1]

5. Benzene reacts with bromine in the presence of aluminium bromide, AlBr₃.

(a) Show, with an equation, how AlBr₃ produces the electrophile. [1] (b) Describe the mechanism. Show every curly arrow and the intermediate. [3] (c) Show how the AlBr₃ is regenerated. [1]

6. Cyclohexene decolourises bromine at room temperature. Benzene does not react unless a halogen carrier is present. Explain this difference. [3]

7. Benzene reacts with phenylethanoyl chloride, C₆H₅CH₂COCl, in the presence of AlCl₃ to form 1,2-diphenylethanone, C₆H₅COCH₂C₆H₅.

(a) Name the type of reaction and write an equation to show the role of AlCl₃. [2] (b) Write the overall equation for the reaction. [1] (c) 3.12 g of benzene reacts with excess acyl chloride. 6.04 g of the ketone is obtained. Calculate the percentage yield. [2] (d) Explain why reactions of this type are important in organic synthesis. [1]

8. In the presence of a strong acid, benzene reacts with cyclohexene to form cyclohexylbenzene, C₆H₅C₆H₁₁. You are told that H⁺ first adds to cyclohexene to form a carbocation.

(a) Give an equation for the carbocation formation. [1] (b) Outline the mechanism for its reaction with benzene. [2] (c) Explain why the acid acts as a catalyst. [1]

9. Three unlabelled samples are ethanol, phenol and propanoic acid, each in aqueous solution.

(a) Describe how aqueous sodium carbonate and a pH test would identify all three. [2] (b) Write an equation for the reaction of phenol with aqueous sodium hydroxide. [1] (c) Explain why phenol gives no gas with sodium carbonate. [1]

10. This question is about the reactions of phenol.

(a) State what you would see when bromine water is added to aqueous phenol, and name the organic product. [2] (b) State the reagent and conditions for nitrating phenol, and name the two organic products. [1] (c) Explain why phenol undergoes these substitutions more readily than benzene. [2]

11. This question is about directing effects.

(a) Name the main organic product when nitrobenzene is nitrated further. [1] (b) Write an equation for the reaction of phenylamine with excess bromine water. [2] (c) A chemist wants to make 1-chloro-3-nitrobenzene from benzene. You are told that a Cl substituent directs to the 2- and 4- positions. Give the two steps in the correct order, with reagents, and explain the order. [3] (d) You are told that the –CHO group directs to the 3- position. Name the main product of mononitration of benzaldehyde, C₆H₅CHO. [1]

12. 4.70 g of phenol reacts with dilute nitric acid. After separation, 3.13 g of 2-nitrophenol and 2.78 g of 4-nitrophenol are collected.

(a) Give the equation for this nitration. [1] (b) Calculate the percentage yield of each isomer and the overall percentage yield. [3] (c) Ka for phenol is 1.3 × 10⁻¹⁰ mol dm⁻³. Calculate the pH of a 0.20 mol dm⁻³ solution of phenol. [3] (d) Explain why no 3-nitrophenol forms. [1]

Answers

1. Each carbon has one electron in a p-orbital, above and below the ring plane [1]; the p-orbitals overlap sideways all round the ring, forming a π-system [1]; the six π-electrons are delocalised over all six carbons, whereas Kekulé has three localised C=C bonds alternating with C–C bonds [1] [3] Examiner insight: “Delocalised electrons” alone describes the result; the answer needs the sideways p-orbital overlap that produces it.

2. (a) 10.4 ÷ 0.0500 = 208, so ΔH = −208 kJ mol⁻¹ [1] (b) Kekulé prediction = 3 × (−119) = −357 kJ mol⁻¹ [1]; benzene releases 149 kJ mol⁻¹ less energy, so it is 149 kJ mol⁻¹ more stable than predicted [1] (c) X-ray data show six equal ring bonds, intermediate in length between C–C and C=C [1]; the Kekulé structure needs two bond lengths, alternating round the ring [1] [5] Examiner insight: State the conclusion in words, “more stable than predicted”, rather than leaving a bare number such as 149.

3. (a) 2,4,6-trinitrophenol [1] (b) 4-chlorophenylamine [1] (c) 1-chloro-3-nitrobenzene [1] [3] Examiner insight: In (c), the locants 1 and 3 could go either way; chloro comes first alphabetically, so it takes locant 1.

4. (a) Concentrated nitric acid and concentrated sulfuric acid [1]; warm to about 50 °C [1] (b) HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O [1] (c) At higher temperatures further substitution occurs, giving dinitrobenzene [1] [4] Examiner insight: “Concentrated” must appear for both acids; “nitric and sulfuric acid” alone describes a different reaction.

5. (a) Br₂ + AlBr₃ → Br⁺ + AlBr₄⁻ [1] (b) Curly arrow from the π-system to Br⁺ [1]; intermediate with a broken horseshoe over five carbons, a + charge, and H and Br on the sixth carbon [1]; the C–H bond’s electrons return to the ring by a curly arrow, giving C₆H₅Br and H⁺ [1] (c) H⁺ + AlBr₄⁻ → AlBr₃ + HBr [1] [5] Examiner insight: The first arrow must start inside the ring at the π-system, not at a single carbon or an H atom.

6. The C=C π-bond of cyclohexene holds its two electrons between just two atoms: a region of high electron density [1]; benzene’s six π-electrons spread round the whole ring, so electron density at any bond is lower [1]; benzene cannot induce a large enough dipole in Br₂, so a halogen carrier is needed to form Br⁺ [1] [3] Examiner insight: Compare electron density in both molecules; saying only that benzene is “stable” does not explain why Br₂ fails to react.

7. (a) Friedel–Crafts acylation (electrophilic substitution) [1]; C₆H₅CH₂COCl + AlCl₃ → C₆H₅CH₂CO⁺ + AlCl₄⁻ [1] (b) C₆H₆ + C₆H₅CH₂COCl → C₆H₅COCH₂C₆H₅ + HCl [1] (c) n(benzene) = 3.12 ÷ 78.0 = 0.0400 mol; Mr(C₁₄H₁₂O) = 196.0, theoretical mass = 7.84 g [1]; yield = 6.04 ÷ 7.84 × 100 = 77.0% [1] (d) It forms a new C–C bond to the aromatic ring, extending the carbon skeleton [1] [6] Examiner insight: Keep 7.84 g unrounded; rounding to 7.8 g early gives 77.4%.

8. (a) C₆H₁₀ + H⁺ → C₆H₁₁⁺ [1] (b) Curly arrow from the π-system to the positive carbon of C₆H₁₁⁺, forming the charged intermediate [1]; C–H bond electrons move into the ring (curly arrow) and H⁺ leaves [1] (c) H⁺ is used in forming the electrophile and regenerated in the final step [1] [4] Examiner insight: Unfamiliar substitutions follow the same pattern as nitration; reuse the three-arrow sequence rather than inventing a new one.

9. (a) Only propanoic acid effervesces with sodium carbonate [1]; of the other two, phenol gives a weakly acidic pH (below 7) while ethanol is neutral [1] (b) C₆H₅OH + NaOH → C₆H₅ONa + H₂O [1] (c) Phenol is too weak an acid to release CO₂ from carbonate [1] [4] Examiner insight: Give the reagent and the result for each substance; NaOH alone cannot separate phenol from the acid.

10. (a) Bromine water is decolourised and a white precipitate forms [1]; 2,4,6-tribromophenol [1] (b) Dilute nitric acid at room temperature; 2-nitrophenol and 4-nitrophenol [1] (c) A lone pair in an oxygen p-orbital is partly delocalised into the ring [1]; the ring’s electron density rises, so it is more susceptible to electrophilic attack and can polarise Br₂ without a carrier [1] [5] Examiner insight: Keep (c) to electron density; an argument based on the stability of the intermediate is outside what the specification asks for.

11. (a) 1,3-dinitrobenzene [1] (b) Product C₆H₂Br₃NH₂ (2,4,6-tribromophenylamine) [1]; C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr [1] (c) Step 1: nitrate with concentrated HNO₃ and concentrated H₂SO₄ [1]; step 2: Cl₂ with a halogen carrier such as AlCl₃ [1]; NO₂ directs the chlorine to the 3- position, whereas chlorinating first would give 2- and 4- isomers [1] (d) 3-nitrobenzaldehyde [1] [7] Examiner insight: In (c), a correct order with no reason gives only part of the credit; link the order to the directing group.

12. (a) C₆H₅OH + HNO₃ → C₆H₄(OH)NO₂ + H₂O [1] (b) n(phenol) = 4.70 ÷ 94.0 = 0.0500 mol [1]; Mr(C₆H₅NO₃) = 139.0, theoretical mass = 6.95 g [1]; 2-isomer 45.0%, 4-isomer 40.0%, overall 85.0% [1] (c) Ka = [H⁺]² ÷ [C₆H₅OH] [1]; [H⁺] = √(1.3 × 10⁻¹⁰ × 0.20) = 5.1 × 10⁻⁶ mol dm⁻³ [1]; pH = 5.29 [1] (d) OH is 2- and 4-directing, so substitution at position 3 does not happen [1] [8] Examiner insight: Both isomers share the formula C₆H₅NO₃, so each yield uses the same theoretical mass; a separate Mr for each is a sign of a slip.

Where marks are usually lost

• Describing p-orbital overlap without saying it is sideways and all round the ring.
• Giving a hydrogenation difference without the words “more stable”.
• Omitting “concentrated” for benzene nitration, or adding H₂SO₄ to phenol nitration.
• Leaving out the halogen carrier, or the equation that regenerates it.
• Starting the first curly arrow at a carbon atom instead of the π-system.
• Explaining phenol’s reactivity by intermediate stability.
• Saying phenol fizzes with carbonate.
• Ignoring directing effects when ordering synthesis steps.
• Forgetting the 3HBr in the bromination equations of phenol and phenylamine.

Next steps

• Recap with the aromatic compounds revision notes
• Reread the explanations in the aromatic compounds study guide
• Compare with addition in the OCR alkenes practice questions
• Browse H432 units from the OCR chemistry hub
• Mark progress on the printable H432 checklist
• Book a free trial class
• Find your weakest topic with the free diagnostics

Official syllabus

Questions set on OCR A Level Chemistry A (H432), specification Version 3.1 (May 2026), for first assessment in 2017, published by OCR: section 6.1.1 Aromatic compounds, learning outcomes (a) to (l).