OxfordAQA A-Level Chemistry: Aromatic chemistry (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 question set is on section 3.3.10 (Aromatic chemistry) from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020: 3.3.10.1 Bonding and 3.3.10.2 Electrophilic substitution. All of it is International A-level only. It is assessed in Unit 4 (Organic 2 and Physical 2), and Unit 5 (Practical and synoptic) can test any content.

Learn the content first in the study guide and the revision notes. Course-wide links: the hub, the printable checklist and the free diagnostics help you plan what to do next. Use Ar: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5.

Questions

1. Which conditions convert methylbenzene into (chloromethyl)benzene?

• A chlorine with aluminium chloride, in the dark
• B chlorine with ultraviolet light
• C hydrogen chloride at room temperature
• D aqueous chlorine with sodium hydroxide

[1]

2. A C–C single bond is about 154 pm long and a C=C double bond about 134 pm. All six carbon–carbon bonds in benzene are about 139 pm. Explain how these data support a delocalised model of benzene rather than cyclohexa-1,3,5-triene. [3]

3. Use these enthalpies of hydrogenation, each forming cyclohexane: cyclohexa-1,3-diene (two C=C bonds), −230 kJ mol⁻¹; benzene, −206 kJ mol⁻¹.

(a) Use the diene value to estimate ΔH for hydrogenating one C=C bond. [1] (b) Predict ΔH for hydrogenating cyclohexa-1,3,5-triene. [1] (c) Calculate how much more stable benzene is than the triene, using these data. [2] (d) Calculate the heat released when 1.95 g of benzene is fully hydrogenated. [2]

4. In the nitration of benzene, the positive intermediate loses H⁺. It could instead, in principle, gain a hydrogensulfate ion and form an addition product. Explain why the H⁺ is lost instead. [3]

5. Methylbenzene is nitrated with a mixture of concentrated nitric and sulfuric acids.

(a) Give an equation showing how the electrophile forms. [1] (b) Draw an outline mechanism to show how 4-nitromethylbenzene forms. [3] (c) Name the other main organic product. [1] (d) 4.60 g of methylbenzene gives 5.48 g of a mixture of mononitro isomers, C₇H₇NO₂. Calculate the percentage yield of mononitro product. [3]

6. Benzene is warmed with fuming sulfuric acid.

(a) Identify the electrophile and explain why it acts as an electrophile. [2] (b) Using curly arrows, outline how benzene and this electrophile react to give benzenesulfonic acid. [3] (c) Name the main product when methylbenzene is sulfonated. [1]

7. Give one reason why nitration and one reason why sulfonation are important steps in synthesis. [2]

8. Benzene is heated with 2-chloro-2-methylpropane, (CH₃)₃CCl, and aluminium chloride.

(a) Give an equation showing how the carbocation forms. [1] (b) Name the organic product. [1] (c) Show, with an equation, that aluminium chloride is re-formed. [1] (d) Calculate the mass of product made from 5.46 g of benzene if the yield is 85.0%. [3] (e) Suggest why benzene is used in excess. [1]

9. A chemist wants to make 2-methyl-1-phenylpropan-1-one, C₆H₅COCH(CH₃)₂, in one step from benzene.

(a) Name the organic reagent needed. [1] (b) Give the formula of the electrophile. [1] (c) Suggest why the aluminium chloride must be kept dry. [1] (d) State the functional group in the product. [1]

10. Chlorine reacts with methylbenzene by a free-radical mechanism.

(a) State the condition needed. [1] (b) Write the equation for the initiation step. [1] (c) Write the two propagation steps that form (chloromethyl)benzene. [2] (d) A termination step forms a hydrocarbon, C₁₄H₁₄. Write its structural formula. [1] (e) With more chlorine, a product with Mr = 161.0 forms. Identify it. [1]

11. Benzene reacts with chlorine in ultraviolet light.

(a) Give the balanced equation. [1] (b) Name the type of reaction. [1] (c) Calculate the percentage by mass of chlorine in the organic product. [2]

12. Samples of chlorobenzene and (chloromethyl)benzene are each warmed with aqueous sodium hydroxide. Each mixture is then made acidic with dilute HNO₃ before aqueous AgNO₃ is added.

(a) State what is seen for each compound. [2] (b) Write an equation for the reaction of (chloromethyl)benzene with hydroxide ions. [1] (c) Explain the difference in reactivity of the chlorine in the two compounds. [4]

Answers

1. B [1] [1] Examiner insight: Option A puts chlorine on the ring because a catalyst in the dark gives electrophilic attack, so read the conditions, not just the reagent.

2. The triene would have three shorter C=C and three longer C–C bonds, so two different lengths [1]. All benzene’s bonds are equal, so every bond shares the same delocalised p electrons [1]. Their length lies between single and double, so each bond has more electron density than a single bond but less than a double bond [1]. [3] Examiner insight: Each point should use the data: an answer that only says “benzene is delocalised” without comparing the lengths does not explain anything.

3. (a) −230 / 2 = −115 kJ mol⁻¹ [1] (b) 3 × (−115) = −345 kJ mol⁻¹ [1] (c) −206 − (−345) = +139, so benzene releases 139 kJ mol⁻¹ less energy [1]; benzene is 139 kJ mol⁻¹ more stable [1]. (d) n = 1.95 / 78.0 = 0.0250 mol [1]; 0.0250 × 206 = 5.15 kJ [1] [6] Examiner insight: In (c), the final mark needs the direction stated: say benzene is the more stable compound, not just “a difference of 139”.

4. Losing H⁺ restores the full delocalised ring of p electrons [1]. An addition product would not have a complete delocalised ring [1]. So it would lose the extra stability from delocalisation, making substitution energetically favoured [1]. [3] Examiner insight: Link each step: delocalisation, then stability, then the choice of substitution. “Benzene doesn’t do addition” restates the question.

5. (a) HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ [1] (b) Arrow from the ring to the N of NO₂⁺, attacking the carbon opposite CH₃ [1]. Intermediate: that carbon carries H and NO₂, with a partial ring (horseshoe) across the remaining five carbons and + in its centre [1]. The C–H bond electrons move back into the ring (arrow), releasing H⁺ and forming 4-nitromethylbenzene [1]. (c) 2-nitromethylbenzene (1-methyl-2-nitrobenzene) [1] (d) n(C₇H₈) = 4.60 / 92.0 = 0.0500 mol [1]. Mr(C₇H₇NO₂) = 137.0; theoretical mass = 0.0500 × 137.0 = 6.85 g [1]. Yield = 5.48 / 6.85 × 100 = 80.0% [1] [8] Examiner insight: In (b) the NO₂ must be drawn opposite the methyl group; a correct mechanism at the wrong position does not give the named product.

6. (a) Sulfur trioxide, SO₃ [1]. The three oxygen atoms withdraw electron density, so the sulfur atom is strongly δ+ and accepts an electron pair [1]. (b) Arrow from the ring to the S of SO₃ [1]. Intermediate: attacked carbon carries H and SO₃⁻, horseshoe with + inside [1]. Arrow from the C–H bond into the ring; H⁺ is lost and the SO₃⁻ group is protonated to give C₆H₅SO₃H [1]. (c) 4-methylbenzenesulfonic acid [1] [6] Examiner insight: SO₃ is neutral; giving it a positive charge, or pointing the arrow at an oxygen, costs the relevant mark.

7. Nitration: manufacture of explosives, or forming amines by reducing the nitro group [1]. Sulfonation: manufacture of surfactants (detergents) or of sulfonamides [1]. [2] Examiner insight: Each mark is for a use matched to the right reaction; listing uses with no link to a reaction gives no credit.

8. (a) (CH₃)₃CCl + AlCl₃ → (CH₃)₃C⁺ + AlCl₄⁻ [1] (b) (1,1-dimethylethyl)benzene, C₆H₅C(CH₃)₃ [1] (c) H⁺ + AlCl₄⁻ → AlCl₃ + HCl [1] (d) n(benzene) = 5.46 / 78.0 = 0.0700 mol [1]. Mr(C₁₀H₁₄) = 134.0; theoretical mass = 9.38 g [1]. At 85.0%: 9.38 × 0.850 = 7.97 g [1]. (e) The alkyl group makes the product more reactive than benzene, so excess benzene makes further substitution less likely [1]. [7] Examiner insight: An electrophile equation must balance for charge; the + on the carbocation and the − on AlCl₄⁻ must both appear.

9. (a) 2-methylpropanoyl chloride, (CH₃)₂CHCOCl [1] (b) (CH₃)₂CHCO⁺ [1] (c) Water would react with (hydrolyse) AlCl₃ and remove the catalyst [1]. (d) Ketone (carbonyl group between the ring and the alkyl group) [1] [4] Examiner insight: For (a), work backwards: the C=O carbon joined to the ring comes from the acyl chloride, so all four carbons of the side chain must be in the reagent.

10. (a) Ultraviolet light [1] (b) Cl₂ → 2Cl• [1] (c) C₆H₅CH₃ + Cl• → C₆H₅CH₂• + HCl [1]; C₆H₅CH₂• + Cl₂ → C₆H₅CH₂Cl + Cl• [1] (d) C₆H₅CH₂CH₂C₆H₅ [1] (e) (Dichloromethyl)benzene, C₆H₅CHCl₂ [1] [6] Examiner insight: Propagation steps lose credit if the radical dot is on the wrong atom or if H• appears instead of HCl.

11. (a) C₆H₆ + 3Cl₂ → C₆H₆Cl₆ [1] (b) Free-radical addition [1] (c) Mr(C₆H₆Cl₆) = 291.0 [1]; 213.0 / 291.0 × 100 = 73.2% [1] [4] Examiner insight: Writing C₆H₅Cl here shows a mix-up with ring substitution; the specification describes ring addition for benzene.

12. (a) (Chloromethyl)benzene: white precipitate [1]. Chlorobenzene: no precipitate [1]. (b) C₆H₅CH₂Cl + OH⁻ → C₆H₅CH₂OH + Cl⁻ [1] (c) In (chloromethyl)benzene the Cl is on an sp³ side-chain carbon that is δ+, so the hydroxide nucleophile substitutes it and releases Cl⁻ [1]. In chlorobenzene one chlorine lone pair becomes partly delocalised into the ring’s p-electron system [1]. So the C–Cl bond is strengthened (partial π character) and harder to break [1]. The delocalised electron density also repels the approaching nucleophile [1]. [7] Examiner insight: Each explanation point should be tied to the compound it describes; one line on each compound earns less than a full comparison.

Where marks are usually lost

• Quoting a stability figure without saying benzene is the more stable compound.
• Starting the first arrow on a ring carbon instead of the delocalised ring.
• Drawing the intermediate without the + or with a full circle.
• Giving 3-nitromethylbenzene as a main product from methylbenzene.
• Treating SO₃ as a positive ion, or arrowing to its oxygen.
• Mixing up conditions: AlCl₃ for ring reactions, UV light for radical reactions.
• Writing chlorobenzene for the product of benzene with chlorine in UV light.
• Leaving HCl out of Friedel–Crafts equations.
• Explaining chlorobenzene’s low reactivity without the lone pair overlapping with the ring.

Next steps

• Revision notes for this topic
• Study guide for this topic
• OxfordAQA A-Level Chemistry hub
• Halogenoalkanes practice questions
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.10 Aromatic chemistry (International A2): subsections 3.3.10.1 (Bonding) and 3.3.10.2 (Electrophilic substitution).