Edexcel International A Level Chemistry: Organic chemistry: arenes (YCH11) – 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 Topic 18: Organic Chemistry – Arenes of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 18.1 to 18.6. Topic 18 belongs to Unit 5: Transition Metals and Organic Nitrogen Chemistry, an International A2 (IA2) unit. Learn the content first from the arenes study guide and the arenes revision notes, and tick off outcomes on the printable checklist. More resources are on the course hub.

Data for this set: H = 1.0, C = 12.0, N = 14.0, O = 16.0, S = 32.1, Br = 79.9.

Questions

1. This question is about the bonding in benzene.

(a) Describe how the π-bonding in benzene forms from the p-orbitals on the carbon atoms. [2] (b) State the H–C–C bond angle in benzene. [1]

2. A crystallographer records these carbon–carbon distances: the C–C bond in cyclohexane 0.154 nm, the C=C bond in cyclohexene 0.134 nm, and all six ring bonds in benzene 0.139 nm.

(a) State the bond lengths the Kekulé structure would predict for the ring. [1] (b) Explain what the benzene measurement shows about its bonding. [2]

3. Use these values for the enthalpy change of hydrogenation: cyclohexene −120 kJ mol⁻¹; benzene (to cyclohexane) −205 kJ mol⁻¹.

(a) Write the equation for the complete hydrogenation of benzene. [1] (b) Calculate the enthalpy change of hydrogenation predicted for the Kekulé structure. [1] (c) Use your answer to comment on the stability of benzene. [2]

4. Explain how infrared spectroscopy could distinguish a sample of benzene from a sample of cyclohexene. [2]

5. Methylbenzene, C₇H₈, and heptane, C₇H₁₆, are burned in separate spirit burners.

(a) Give the equation for methylbenzene burning in excess oxygen. [1] (b) Calculate the percentage by mass of carbon in each compound. [2] (c) Explain why methylbenzene burns with the smokier flame. [1]

6. Cyclohexene decolourises bromine at room temperature in the dark, but benzene does not react unless a catalyst is added, and then it gives a substitution product. Explain both observations. [4]

7. Benzene reacts with bromine in the presence of iron(III) bromide.

(a) Show, with an equation, how the iron(III) bromide generates the electrophile. [1] (b) Describe the mechanism of the reaction of this electrophile with benzene. Include curly arrows and the structure of the intermediate. [3] (c) Write an equation to show that iron(III) bromide acts as a catalyst. [1]

8. A student nitrates 5.20 g of benzene.

(a) Give the reagents and the temperature used. [2] (b) Give the equation by which the two acids produce the electrophile. [1] (c) Describe the mechanism by which this electrophile substitutes into the ring. [3] (d) The student obtains 6.36 g of nitrobenzene. Calculate the percentage yield. [3] (e) State one large-scale use of nitrobenzene. [1]

9. Benzene is warmed with fuming sulfuric acid.

(a) Name the species in fuming sulfuric acid that reacts with benzene. [1] (b) Give a balanced equation for the sulfonation, naming the aromatic product. [1] (c) State one use of compounds made by this type of reaction. [1]

10. Two Friedel-Crafts reactions of benzene are carried out with aluminium chloride.

(a) Benzene reacts with 2-chlorobutane. Write an equation and name the organic product. [2] (b) Explain why the aluminium chloride and the apparatus must be dry. [1] (c) Benzene reacts with pentanoyl chloride, CH₃CH₂CH₂CH₂COCl. Write an equation for the formation of the electrophile, then describe the mechanism of its reaction with benzene. [4] (d) 2.73 g of benzene gives 4.10 g of 1-phenylpentan-1-one, C₆H₅COCH₂CH₂CH₂CH₃. Calculate the percentage yield. [3]

11. 1.41 g of phenol is dissolved in water and bromine water is added until no further change occurs.

(a) State two observations. [2] (b) Give a balanced equation for this bromination. [1] (c) Explain why phenol reacts with bromine water without a catalyst, but benzene does not. [3] (d) Calculate the maximum mass of organic product. [2] (e) The bromine water contains 0.150 mol dm⁻³ Br₂. Calculate the minimum volume needed. [2]

12. Three unlabelled bottles contain cyclohexene, methylbenzene and aqueous phenol. Using only bromine water, describe how you would identify each one. Give the expected observation for each and the type of reaction where one occurs. [5]

Answers

1. (a) Every ring carbon has a singly occupied p-orbital standing perpendicular to the ring [1]; these orbitals overlap sideways all round the ring, each with the orbitals on either side, so six π-electrons are delocalised over both faces of the ring [1]. (b) 120° [1] Examiner insight: Say “both neighbours” or “all six carbons”; overlap with one neighbour describes a localised C=C, not delocalisation.

2. (a) Three bonds of about 0.154 nm alternating with three of about 0.134 nm [1]. (b) All six bonds are equal in length [1]; the length lies between a single and a double bond, so the π-electrons are delocalised over the whole ring rather than in three localised double bonds [1]. Examiner insight: Use the figures: “between 0.134 and 0.154 nm” earns more than “all bonds are the same”.

3. (a) C₆H₆ + 3H₂ → C₆H₁₂ [1] (b) 3 × (−120) = −360 kJ mol⁻¹ [1] (c) Benzene releases 360 − 205 = 155 kJ mol⁻¹ less energy than predicted [1]; so benzene is more stable than the Kekulé structure, because its π-electrons are delocalised [1]. Examiner insight: State which substance is more stable and by how much; a correct difference with no conclusion does not answer “comment”.

4. Cyclohexene shows an alkene C=C stretching absorption at about 1650 cm⁻¹ [1]; benzene has no alkene C=C absorption but shows aromatic ring absorptions at about 1450–1600 cm⁻¹ [1]. Examiner insight: Both compounds show C–H stretches near 3000 cm⁻¹, so quoting that band does not distinguish them.

5. (a) C₇H₈ + 9O₂ → 7CO₂ + 4H₂O [1] (b) Methylbenzene: 84.0 ÷ 92.0 × 100 = 91.3% [1]; heptane: 84.0 ÷ 100.0 × 100 = 84.0% [1]. (c) Methylbenzene has a higher carbon to hydrogen ratio, so combustion is incomplete in air and unburnt carbon forms soot [1]. Examiner insight: Link the high carbon content to incomplete combustion and soot; “it is aromatic” alone explains nothing.

6. Cyclohexene’s π-electron pair is concentrated between just two carbon atoms, a region of high electron density [1]; this induces a dipole in Br₂ so electrophilic addition happens without a catalyst [1]. Benzene’s π-electrons are spread round six carbons, so no single region is electron-rich enough to polarise Br₂; a catalyst is needed to make Br⁺ [1]. Substitution then occurs rather than addition because it keeps the stable delocalised ring [1]. Examiner insight: Four marks need four separate ideas; cover both compounds and both observations (no reaction, then substitution).

7. (a) Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻ [1] (b) Curly arrow from the delocalised ring to Br⁺ [1]; intermediate with H and Br on the same carbon, broken horseshoe over the other five carbons and a + charge [1]; the C–H bond pair returns to the ring (arrow) and H⁺ leaves, giving bromobenzene [1]. (c) H⁺ + FeBr₄⁻ → FeBr₃ + HBr [1] Examiner insight: Draw the horseshoe over five carbons only, with the + inside it, and keep the attacked carbon’s hydrogen.

8. (a) Concentrated nitric acid and concentrated sulfuric acid [1]; at or below about 55 °C [1]. (b) HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ [1] (c) Curly arrow from the ring to the N of NO₂⁺ [1]; intermediate with H and NO₂ on one carbon, horseshoe over five carbons, + charge [1]; curly arrow from C–H into the ring, H⁺ lost, giving nitrobenzene [1]. (d) n(benzene) = 5.20 ÷ 78.0 = 0.0667 mol [1]; Mr(C₆H₅NO₂) = 123.0, theoretical mass = 0.0667 × 123.0 = 8.20 g [1]; yield = 6.36 ÷ 8.20 × 100 = 77.6% [1]. (e) It is reduced to phenylamine, used to make dyes [1]. Examiner insight: “Concentrated” belongs with both acids; leaving it off either one makes the reagent description incomplete.

9. (a) Sulfur trioxide, SO₃ [1] (b) C₆H₆ + SO₃ → C₆H₅SO₃H; benzenesulfonic acid [1] (c) Making detergents (or dyes) [1]. Examiner insight: Write “fuming” sulfuric acid in any description of conditions; it is the reagent the specification names.

10. (a) C₆H₆ + CH₃CHClCH₂CH₃ → C₆H₅CH(CH₃)CH₂CH₃ + HCl [1]; 2-phenylbutane [1]. (b) Water reacts with aluminium chloride, so the catalyst would be destroyed [1]. (c) CH₃CH₂CH₂CH₂COCl + AlCl₃ → CH₃CH₂CH₂CH₂CO⁺ + AlCl₄⁻ [1]; curly arrow from the ring to the carbon of the acylium ion [1]; intermediate with H and COCH₂CH₂CH₂CH₃ on one carbon, horseshoe over five carbons, + charge [1]; curly arrow from C–H into the ring, H⁺ lost to give the ketone, and H⁺ + AlCl₄⁻ → AlCl₃ + HCl [1]. (d) n(benzene) = 2.73 ÷ 78.0 = 0.0350 mol [1]; Mr(C₁₁H₁₄O) = 162.0, theoretical mass = 0.0350 × 162.0 = 5.67 g [1]; yield = 4.10 ÷ 5.67 × 100 = 72.3% [1]. Examiner insight: Put the positive charge on the carbonyl carbon of the acylium ion, and point the first arrow at that carbon.

11. (a) The orange bromine water is decolourised [1]; a white precipitate forms [1]. (b) C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr [1] (c) A lone pair on the oxygen overlaps with the π-system of the ring [1]; it is partly delocalised into the ring, increasing the electron density [1]; so the ring can polarise Br₂ without a catalyst, which benzene’s ring cannot do [1]. (d) n(phenol) = 1.41 ÷ 94.0 = 0.0150 mol [1]; Mr(C₆H₃Br₃O) = 330.7, mass = 0.0150 × 330.7 = 4.96 g [1]. (e) n(Br₂) = 3 × 0.0150 = 0.0450 mol [1]; volume = 0.0450 ÷ 0.150 = 0.300 dm³ = 300 cm³ [1]. Examiner insight: Use the 1 : 3 ratio from your equation in both (d) and (e); treating the reaction as 1 : 1 gives a volume three times too small.

12. Add bromine water to a sample from each bottle [1]. Aqueous phenol: orange colour disappears and a white precipitate forms, by electrophilic substitution [1]. Cyclohexene: orange colour disappears with no precipitate [1], by electrophilic addition [1]. Methylbenzene: no reaction, so the orange colour remains (the bromine may pass into the organic layer) [1]. Examiner insight: Phenol and cyclohexene both decolourise bromine water, so the precipitate is the observation that separates them.

Where marks are usually lost

• Calling benzene’s ring bonds “single and double” or saying it “has no π-bonds”.
• Quoting hydrogenation or bond-length data without stating what they show.
• Using the C–H stretch to distinguish benzene from an alkene.
• Explaining benzene’s resistance to bromine by “stability” alone, without electron density.
• Forgetting “concentrated” for nitration acids, “fuming” for sulfonation or “anhydrous” for Friedel-Crafts.
• Charge on the wrong atom in NO₂⁺ or in the acylium ion.
• First curly arrow drawn from a carbon atom, or intermediate with a horseshoe over six carbons.
• No equation for catalyst regeneration when the question asks why something is a catalyst.
• Using a 1 : 1 ratio for phenol and bromine.
• Missing the white precipitate when describing phenol with bromine water.

Next steps

• Arenes revision notes
• Arenes study guide
• Alkenes practice questions for electrophilic addition
• Edexcel International A Level Chemistry hub
• Printable Edexcel Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 5, Topic 18: Organic Chemistry – Arenes, outcomes 18.1 to 18.6.