AQA A-Level Chemistry: Aromatic chemistry (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 3.3.10 Aromatic chemistry of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.3.10.1 Bonding and 3.3.10.2 Electrophilic substitution. The whole section is A-level only. It is examined on A-level Paper 2, and A-level Paper 3 can test any content.

Learn the ideas first in the aromatic chemistry study guide and the aromatic chemistry revision notes. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are. For a mechanism to compare with, see the alkenes practice questions. Questions 3, 8, 9 and 11 need a calculator. Use Ar: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5.

Questions

1. Which statement about the carbon–carbon bonds in benzene is correct?

• A There are three short bonds and three long bonds.
• B All six bonds are the same length as a C=C bond.
• C All six bonds are the same length, between that of a C–C bond and a C=C bond.
• D All six bonds are the same length as a C–C bond.

[1]

2. Describe the bonding in a benzene molecule. Refer to the shape of the molecule and the p electrons. [3]

3. The enthalpy of hydrogenation of cyclohexene to cyclohexane is −120 kJ mol⁻¹. Benzene is 152 kJ mol⁻¹ more stable than the theoretical molecule cyclohexa-1,3,5-triene.

(a) Calculate the enthalpy of hydrogenation of benzene to cyclohexane. [2] (b) Explain, in terms of bonding, why benzene is more stable than cyclohexa-1,3,5-triene. [1] (c) Calculate the heat energy released when 0.250 mol of benzene is fully hydrogenated. [1]

4. Explain why benzene reacts by substitution rather than by addition. [2]

5. Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid.

(a) Give the formula of the electrophile. [1] (b) Write an equation for the formation of this electrophile. [1] (c) Outline the mechanism for the reaction of benzene with this electrophile. [3] (d) Write an equation to show that sulfuric acid acts as a catalyst. [1]

6. Benzene reacts with an acyl chloride, RCOCl, in the presence of aluminium chloride. The organic product, C₆H₅COR, has Mr = 148.0. R is an unbranched alkyl group.

(a) Deduce the formula of R. Show your working. [2] (b) Name the acyl chloride. [1] (c) Write an equation for the formation of the electrophile. [1] (d) Outline the mechanism for the reaction of benzene with this electrophile. [3]

7. State two reasons why nitration of aromatic compounds is important in synthesis. [2]

8. A student nitrates 4.08 g of methyl benzoate, C₈H₈O₂, and obtains 3.85 g of recrystallised methyl 3-nitrobenzoate, C₈H₇NO₄. One molecule of product forms from each molecule of methyl benzoate.

(a) Calculate the percentage yield. [3] (b) Describe how the student could use a melting point to show that the product is pure. [2]

9. 2.34 g of benzene reacts with excess ethanoyl chloride in the presence of aluminium chloride.

(a) Calculate the mass of phenylethanone, C₆H₅COCH₃, formed if the yield is 80.0%. [3] (b) Name the inorganic by-product of the reaction. [1] (c) Suggest why the reaction is carried out in dry conditions. [1]

10. Cyclohexene reacts with bromine water at room temperature, but benzene does not. Benzene does react with the nitronium ion. Explain these observations. In your answer, compare the electron density in the two molecules, the type of electrophile needed and the type of reaction each compound undergoes. [6]

11. A chemist makes 1-phenylpropan-1-one, C₆H₅COCH₂CH₃, from benzene and propanoyl chloride, CH₃CH₂COCl, using an aluminium chloride catalyst.

(a) Write an equation for the overall reaction. [1] (b) Calculate the percentage atom economy for making 1-phenylpropan-1-one. [2] (c) Write two equations to show how aluminium chloride acts as a catalyst. [2] (d) The chemist needs 5.00 g of product. The yield is 60.0%. Calculate the minimum mass of benzene needed. [2]

Answers

1. C [1] [1] Examiner insight: A one-mark multiple-choice item has no partial credit; B and D are the common traps because they describe one of the two bond types in the triene model.

2. Planar molecule with bond angles of 120° [1]. Each carbon has one p electron in a p orbital at right angles to the ring, and these overlap sideways [1]. The six p electrons are delocalised over all six carbons, in a ring of electron density above and below the plane [1]. [3] Examiner insight: “Delocalised” on its own rarely earns all the credit; say which electrons (p) and where (above and below the plane, over all six carbons).

3. (a) Predicted for triene = 3 × (−120) = −360 kJ mol⁻¹ [1]. Benzene = −360 + 152 = −208 kJ mol⁻¹ [1]. (b) Delocalisation of the p electrons (ring of delocalised electrons) [1]. (c) 0.250 × 208 = 52.0 kJ [1]. [4] Examiner insight: In (a) the sign matters: benzene releases less energy than the triene, so its value is less negative; adding 152 to −360 is the right direction.

4. Addition would break up the delocalised ring (π system) [1]. So the product would lose the extra stability; substitution keeps the delocalised ring intact [1]. [2] Examiner insight: “Benzene has no double bonds” scores nothing; the mark depends on linking addition to loss of delocalisation and stability.

5. (a) NO₂⁺ [1] (b) HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ (or HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O) [1] (c) Curly arrow from inside the ring to the N of NO₂⁺ [1]. Intermediate: attacked carbon carrying H and NO₂, horseshoe over the other five carbons with + inside, gap facing the attacked carbon [1]. Curly arrow from the C–H bond back into the ring, giving nitrobenzene and H⁺ [1]. (d) H⁺ + HSO₄⁻ → H₂SO₄ [1] [6] Examiner insight: An equation for the electrophile must balance for charge as well as atoms; check that the positive and negative charges on the right cancel.

6. (a) Mr(C₆H₅CO) = 105.0, so Mr(R) = 148.0 − 105.0 = 43.0 [1]. CₙH₂ₙ₊₁: 14n + 1 = 43, n = 3, so R = C₃H₇ (CH₂CH₂CH₃) [1]. (b) Butanoyl chloride, CH₃CH₂CH₂COCl [1] (c) CH₃CH₂CH₂COCl + AlCl₃ → CH₃CH₂CH₂CO⁺ + AlCl₄⁻ [1] (d) Curly arrow from inside the ring to the C of CH₃CH₂CH₂CO⁺ [1]. Intermediate with horseshoe and +, attacked carbon carrying H and COCH₂CH₂CH₃ [1]. Curly arrow from the C–H bond into the ring, giving 1-phenylbutan-1-one and H⁺ [1]. [7] Examiner insight: Allow error carried forward in (b)–(d) from a wrong R in (a), but the arrow in (d) must point to the carbon of the acylium ion, not to the oxygen.

7. Manufacture of explosives [1]. Formation of aromatic amines by reducing the nitro compound; these amines are used to make dyes [1]. [2] Examiner insight: Two separate uses are needed for two marks; “making amines” and “making dyes from those amines” describe one route, so pair one of them with explosives.

8. (a) Mr(C₈H₈O₂) = 136.0; n = 4.08 / 136.0 = 0.0300 mol [1]. Mr(C₈H₇NO₄) = 181.0; theoretical mass = 0.0300 × 181.0 = 5.43 g [1]. Yield = 3.85 / 5.43 × 100 = 70.9% [1]. (b) Measure the melting point and compare it with the data-book value [1]. A pure sample melts sharply at (or very close to) that value; an impure sample melts lower and over a range [1]. [5] Examiner insight: Give the yield to 3 significant figures to match the data, and allow error carried forward from a wrong Mr.

9. (a) n(C₆H₆) = 2.34 / 78.0 = 0.0300 mol [1]. Theoretical mass of C₆H₅COCH₃ (Mr 120.0) = 0.0300 × 120.0 = 3.60 g [1]. At 80.0%: 3.60 × 0.800 = 2.88 g [1]. (b) Hydrogen chloride, HCl [1] (c) Aluminium chloride reacts with water, so the catalyst would be destroyed (or the acyl chloride would react with water) [1]. [5] Examiner insight: In (a) multiply by 0.800, not divide; dividing gives a mass larger than the theoretical maximum, which should prompt a check.

10.

• In cyclohexene the C=C bond is a localised region of high electron density between two carbons [1].
• In benzene the p electrons are delocalised over six carbons, so the electron density is lower in any one region [1].
• Cyclohexene can polarise Br₂ (induce a dipole) and react with this weak electrophile [1].
• Benzene cannot polarise Br₂ enough, so it needs a strong electrophile such as the positive NO₂⁺ ion [1].
• Cyclohexene undergoes electrophilic addition, giving a saturated product [1].
• Benzene undergoes electrophilic substitution, because addition would destroy the delocalised ring and its stability [1].

[6] Examiner insight: An extended “explain” question rewards a comparison: each point about cyclohexene should be matched by the corresponding point about benzene rather than describing one compound only.

11. (a) C₆H₆ + CH₃CH₂COCl → C₆H₅COCH₂CH₃ + HCl [1] (b) Mr: C₆H₅COCH₂CH₃ 134.0; C₆H₆ 78.0; CH₃CH₂COCl 92.5. Atom economy = 134.0 / (78.0 + 92.5) × 100 [1] = 78.6% [1]. (c) CH₃CH₂COCl + AlCl₃ → CH₃CH₂CO⁺ + AlCl₄⁻ [1]; H⁺ + AlCl₄⁻ → AlCl₃ + HCl [1]. (d) n(product) = 5.00 / 134.0 = 0.03731 mol; n(benzene) = 0.03731 / 0.600 = 0.06219 mol [1]. Mass = 0.06219 × 78.0 = 4.85 g [1]. [7] Examiner insight: Atom economy uses the total Mr of all reactants in the balanced equation, not the mass actually used, and it is independent of the yield in (d).

Where marks are usually lost

• Describing benzene’s bonds as “between single and double” without saying it is their length.
• Missing the sign in hydrogenation calculations, or not stating which compound is more stable.
• Explaining “no addition” without mentioning loss of delocalisation.
• Curly arrow 1 drawn from a carbon atom or a C–H bond instead of from the delocalised ring.
• Intermediates drawn with a full circle, without a +, or with the horseshoe gap facing the wrong way.
• Arrows pointing at oxygen in NO₂⁺ or RCO⁺ instead of N or C.
• Electrophile equations that do not balance for charge.
• Leaving out catalyst regeneration, or giving HCl as the catalyst in acylation.
• Rounding Mr or moles too early in yield calculations.
• Naming the acylation product as an aldehyde.

Next steps

• Aromatic chemistry revision notes
• Aromatic chemistry study guide
• Alkenes practice questions
• Mole practice questions
• 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 3.3.10 Aromatic chemistry (A-level only), covering 3.3.10.1 Bonding and 3.3.10.2 Electrophilic substitution.