AQA A-Level Chemistry Paper 2, 2024: Question 10
12 marks · Medium difficulty · State/Explain/Describe
Benzene Stability & Reactivity: Enthalpy Evidence, Electrophilic Nitration, and Reduction to Aromatic Amines
Practise this questionQuestion
Question text
10 Figure 5 shows enthalpy of hydrogenation data for cyclohexene and benzene.
It also shows predicted data for the theoretical molecule cyclohexa-1,3,5-triene.
Figure 5
10.1 Compare benzene and the theoretical molecule cyclohexa-1,3,5-triene in terms of:
• stability
• shape
• carbon–carbon bond lengths.
For each of these properties, suggest reasons for any differences.
Use data from Figure 5 in your answer.
[5 marks]
Two steps in the synthesis of an aromatic amine are shown.
10.2 State the two reagents needed for Step 1.
Give an equation to show the formation of the reactive intermediate from these two
*25* reagents.
[2 marks]
Reagents
Equation
10.3 Outline a mechanism for Step 1.
[3 marks]
10.4 State the reagent(s) needed for Step 2.
[1 mark]
10.5 State a possible use for the amine formed in Step 2.
[1 mark]
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidelines Mark
M1 Benzene is more stable than cyclohexatriene
M2 The enthalpy of hydrogenation of benzene is (152 kJ mol–1) less / less exothermic
M3 Due to the delocalisation of electrons in benzene
10.1 M4 Both are planar / hexagonal (2 x AO1
M4 and M5 could 3 x AO3)
M5 Benzene has equal C-C bond lengths or regular hexagon whereas Cyclohexa-1,3,5-triene has be shown in a
bonds of different/varied length or the hexagon is distorted/irregular clear diagram
– A-LEVEL CHEMISTRY – – JUNE 2024
M1 Concentrated nitric acid AND concentrated sulfuric acid / conc.
HNO3 AND conc. H2SO4
M2
HNO + 2 H SO ⟶ NO + + H O+ + 2 HSO –
32 4 2 3 4
10.2 OR (1 x AO1
1 x AO2)
HNO + H SO ⟶ H NO + + HSO – then H NO + ⟶ NO + + H O
32 4 2 3 4 2 3 2 2
OR
HNO + H SO ⟶ H O + NO + + HSO –
32 4 2 2 4
M1 Positive must be on N and arrow from inside
hexagon to N or + on N
M2 Structure showing horseshoe and positive
charge.
10.3 • Horseshoe centred on C1 but must not
(3 x AO1)
extend beyond C2 and C6
• + in intermediate not too close to C1 (allow
on or “above” a line from C2 to C6)
M3 Arrow from C-H bond back into hexagon
– A-LEVEL CHEMISTRY – –
Sn/HCl Allow H2 with Pt/Ni
Allow HCl with Fe 1
10.4
Ignore references to NaOH used after Sn/HCl BUT penalise if NaOH (1 x AO1)
used at the same time as Sn/HCl
Manufacture of dyes/(cationic) surfactants/fabric softener Allow to make hair/fabric conditioner
10.5
(1 x AO1)
How to answer it
Benzene vs Cyclohexa-1,3,5-triene and Nitration → Aromatic Amine
What this question tests
Explaining benzene’s stability, shape and bond lengths using enthalpy of hydrogenation data, and applying the electrophilic substitution mechanism for nitration followed by reduction to an aromatic amine (including reagents and the nitronium-ion formation).
Compare benzene with “cyclohexa-1,3,5-triene”
✅ Correct points
- Stability: Benzene is more stable. Predicted triene would be −360 kJ mol⁻¹ if three isolated C=C bonds; actual benzene is −208 kJ mol⁻¹ → extra stability (~152 kJ mol⁻¹) due to π-electron delocalisation/aromaticity.
- Shape: Both are planar hexagons, but benzene is a regular hexagon; the triene would be distorted because alternating single/double bonds give different lengths.
- C–C bond lengths: Benzene has all equal C–C bonds (intermediate between single and double). The triene would show two lengths: short C=C and longer C–C.
💡 Key knowledge
“Resonance energy” is the extra stability of benzene versus a hypothetical triene with localised double bonds.
❌ Common errors
- Quoting the numbers but not stating that the less exothermic hydrogenation indicates greater stability.
- Saying benzene has alternating single and double bonds (it doesn’t; bonds are equivalent).
Reagents and formation of the electrophile
✅ Correct answers
- Reagents: conc. HNO₃ and conc. H₂SO₄.
- Electrophile formation: HNO₃ + 2 H₂SO₄ → NO₂⁺ + H₃O⁺ + 2 HSO₄⁻ (nitronium ion).
🧠 Exam technique
Any equivalent route to NO₂⁺ scores (e.g. via H₂NO₃⁺ → NO₂⁺ + H₂O). Make the overall charge and species correct.
Electrophilic substitution (SEAr) on the benzene ring
💡 Outline the mechanism
- Attack: π electrons of the ring attack NO₂⁺ to form a σ-complex/arenium ion (ring temporarily loses aromaticity). Show a horseshoe partial ring with a + inside (not at the carbon of attack).
- Deprotonation: HSO₄⁻ removes H⁺ from the carbon bearing NO₂; the C–H bond electrons reform the aromatic π system.
- Products: nitro-substituted ring + H₂SO₄.
❌ Common errors
- Placing the positive charge on the nitro group in the σ-complex, or outside the ring.
- Omitting the regeneration of H₂SO₄ or failing to show loss of H⁺.
Convert the nitro compound to an aromatic amine
✅ Reagents
- Sn/HCl, heat under reflux (then add NaOH to free the amine),
- or Fe/HCl,
- or catalytic hydrogenation H₂/Ni.
Use of the aromatic amine
✅ Correct answer
Manufacture of dyes (e.g. azo dye synthesis). Also acceptable: cationic surfactants/fabric softeners.
❌ Common errors
Stating that the amine itself is a dye rather than a feedstock used to make dyes.
Final takeaways
💡 Key knowledge
- Benzene’s lower (less exothermic) hydrogenation enthalpy evidences delocalisation energy.
- Nitration needs NO₂⁺ generated by conc. HNO₃/H₂SO₄; mechanism is SEAr.
- Nitro → amine via Sn/HCl (or Fe/HCl or H₂/Ni).
🧠 Exam technique
- Quote the ~152 kJ mol⁻¹ difference when discussing stability.
- In the σ-complex, draw the horseshoe, place + inside the ring (not on a carbon), and show loss of H⁺.
Topics
Organic Chemistry · 3.3.10 Aromatic Chemistry · 3.3.14 Organic Synthesis · 3.3.11 Amines
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.