A Level Chemistry: Arenes and Halogenoarenes — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Arenes and Halogenoarenes study guide

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Section A

1. State the reagents and conditions needed for the nitration of benzene. [2]

2. State the two conditions that determine whether Cl₂ reacts with methylbenzene at the ring or at the side-chain. [2]

3. State whether the –NO₂ group is ring-activating or ring-deactivating, and which position(s) it directs to. [2]

4. State one reason why chlorobenzene resists nucleophilic substitution under conditions that readily hydrolyse chloroethane. [2]

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Section B

5. Describe, using curly arrows, the mechanism for the nitration of benzene using concentrated HNO₃ and concentrated H₂SO₄, including the equation for generating the electrophile. [5]

6. Explain why benzene reacts with electrophiles by substitution rather than addition, even though addition would seem to be the more obvious analogy with an alkene. [3]

7. Methylbenzene is reacted with Br₂ under two different sets of conditions.

(a) State the reagents/conditions and the organic product when ring substitution is required. [2]

(b) State the reagents/conditions, the type of mechanism, and the organic product when side-chain substitution is required. [3]

8. Chlorobenzene is made by chlorinating benzene with a halogen carrier (e.g. AlCl₃ or FeCl₃). When chlorobenzene is heated under reflux with aqueous sodium hydroxide, no reaction occurs, whereas chloroethane is hydrolysed by aqueous sodium hydroxide only on heating under reflux (and even then only slowly, being the least reactive of the common halogenoalkanes) — never at room temperature.

(a) Explain, in terms of bonding, why the C–Cl bond in chlorobenzene is shorter and stronger than in chloroethane. [2]

(b) Explain the second reason chlorobenzene resists nucleophilic substitution, referring to the ring system. [2]

9. Predict and explain the position(s) at which nitration would occur on benzoic acid (C₆H₅COOH). [3]

10. State the reagents and conditions for the Friedel-Crafts acylation of benzene with CH₃COCl, and name the organic product. [2]

11. State the reagents and conditions needed to fully hydrogenate benzene, and explain why this reaction (unlike halogenation or nitration) proceeds by addition rather than substitution. [3]

12. Methylbenzene is oxidised using hot alkaline KMnO₄ followed by dilute acid.

(a) Name the organic product. [1] (b) Predict and explain the position(s) at which further nitration of this product would occur. [2]

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Answers

1. Concentrated HNO₃ and concentrated H₂SO₄ [1], at 25–60 °C [1].

2. Cl₂ with an AlCl₃ catalyst, in the dark/cold, gives ring substitution; Cl₂ with UV light and no catalyst gives side-chain substitution [2 — one mark for each correct pairing of conditions with outcome].

3. Ring-deactivating [1]; directs to the 3- (and 5-) position(s) [1].

4. Either: the C–Cl bond has partial double-bond character from lone-pair delocalisation into the ring, making it shorter and stronger; or: the carbon bonded to Cl is part of the delocalised ring, so nucleophilic attack would disrupt the ring’s aromatic stabilisation [2, either reason in full].

5. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ [1] generates the electrophile, NO₂⁺. Step 1: a curly arrow from the ring’s delocalised π system attacks NO₂⁺ [1], breaking delocalisation at that carbon and forming a positively charged arenium-ion intermediate with the remaining electrons delocalised over the other five carbons [1]. Step 2: a curly arrow from the C–H bond at that carbon reforms a bond within the ring’s π system [1], and H⁺ is lost (to HSO₄⁻, regenerating H₂SO₄), restoring full delocalisation and giving nitrobenzene [1].

6. Addition would permanently destroy the ring’s delocalised π system, converting it into isolated, higher-energy double bonds [1]. Substitution only briefly disrupts delocalisation, in the arenium-ion intermediate, before it is immediately restored in the second step [1]. The aromatic stabilisation retained by substitution (and lost by addition) makes substitution far more energetically favourable [1].

7. (a) Br₂ with an AlBr₃ catalyst, in the dark/cold [1]; product is a mixture of 2-bromomethylbenzene and 4-bromomethylbenzene [1].

(b) Br₂ with UV light, no catalyst [1]; free-radical substitution mechanism [1]; product is (bromomethyl)benzene, i.e. bromination of the side-chain CH₃ group [1].

8. (a) One of chlorine’s lone pairs partially delocalises into the ring’s π system [1], giving the C–Cl bond partial double-bond character, which shortens and strengthens it compared with the single C–Cl bond in chloroethane [1].

(b) The carbon bonded to Cl is part of the delocalised aromatic ring itself [1], so a nucleophile attacking that carbon would have to disrupt the ring’s aromatic stabilisation, which is energetically very costly [1].

9. Nitration would occur mainly at the 3-position [1]. The –COOH group withdraws electron density from the ring (by induction and its own delocalisation pulling electron density away) [1], deactivating the ring and leaving the 3-position comparatively the most electron-rich of the available positions for the electrophile to attack [1].

10. CH₃COCl with an AlCl₃ catalyst, heat [1]; the organic product is phenylethanone [1].

11. H₂ gas with a platinum or nickel catalyst, and heat [1]. This proceeds by addition because it fully saturates the ring, converting it into cyclohexane; this needs a strong catalyst and heat precisely because disrupting the ring’s delocalised π system is energetically costly, unlike halogenation or nitration, where substitution allows the ring’s aromatic stabilisation to be restored after only briefly being disrupted [2].

12. (a) Benzoic acid [1]. (b) Nitration would occur mainly at the 3-position [1], since the –COOH group already present withdraws electron density from the ring, deactivating it and leaving the 3-position comparatively the most electron-rich of the available positions [1].

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Where marks are usually lost

• Drawing benzene reacting with Br₂ alone (no catalyst) as if it were an alkene undergoing addition.
• Omitting the second step of the electrophilic substitution mechanism (loss of H⁺ to restore aromaticity) — this step is not optional and is specifically credited.
• Mixing up which conditions give ring substitution versus side-chain substitution on an alkylbenzene side-chain.
• Assuming halogenoarenes and halogenoalkanes behave the same way in nucleophilic substitution simply because both contain a C–X bond.