Edexcel A-Level Chemistry Paper 2, June 2025: Question 8

✅ Correct Answer

C₁₀H₁₂O

❌ Common Errors

• Miscounting aromatic ring hydrogens: A 1,4-disubstituted ring has only 4 hydrogens (not 6).
• Writing numbers as superscripts ( C¹⁰H¹²O ), which loses the mark automatically.
• Forgetting the methyl group at the end of the propenyl chain.

✅ Correct Structure

Draw the (Z)-isomer (cis-isomer) around the aliphatic C=C double bond:

🧠 Exam Technique

The question provides the (E)-isomer (trans) where the high-priority aromatic group and the methyl group are on opposite sides of the double bond. Simply invert the geometry at the double bond so the bulky groups are cis to one another.

✅ Correct Answer

In anethole, each carbon of the C=C double bond is attached to two different groups / atoms, whereas in estragole, one carbon of the C=C double bond is attached to two identical atoms (two hydrogen atoms).

❌ Common Errors

• Vague phrasing like "estragole has identical groups on the same side" (must refer specifically to the same carbon atom).
• Writing only "there is restricted rotation around the C=C double bond". Both molecules possess restricted rotation; this does not explain the difference.

✅ Correct Answer: A

A: acidified potassium manganate(VII) (KMnO₄ / H₂SO₄)

💡 Distractor Analysis

• B & C (Aqueous/Ethanolic NaOH): Hydroxide ions are nucleophiles and are repelled by the electron-dense π-bond of the alkene.
• D (Steam + acid catalyst): Hydration of an alkene forms a mono-alcohol, not a vicinal diol.

✅ Model Answer (3 Marks Breakdown)

• σ-bonds (Mark 1): Formed by the head-on (direct) overlap of orbitals (sp² hybridised orbitals) between adjacent carbon atoms.
• π-bonds (Mark 2): Formed by the sideways overlap of adjacent p-orbitals (which are perpendicular to the plane of the ring).
• Delocalisation (Mark 3): The sideways overlap occurs in both directions across all 6 carbons, creating a delocalised π-electron ring system above and below the plane of the carbon ring.

🧠 Diagram Annotation Checklist

✅ Model Answer (4 Marks Breakdown)

Thermodynamic Measurement (2 Marks):

• The enthalpy of hydrogenation of benzene is -208 kJ mol⁻¹ , whereas the theoretical Kekulé cyclic triene (based on 3 × cyclohexene at -120 kJ mol⁻¹ ) would be -360 kJ mol⁻¹ . [1 mark]
• Benzene's enthalpy of hydrogenation is less exothermic than expected (by ~152 kJ mol⁻¹), proving the delocalised model is more energetically stable. [1 mark]

Physical Measurement (2 Marks):

• X-ray diffraction reveals that all carbon–carbon bond lengths in benzene are identical / equal (at 0.139 nm). [1 mark]
• The Kekulé structure would have alternating short C=C double bonds (0.134 nm) and longer C–C single bonds (0.154 nm), which is not observed. [1 mark]

❌ Common Misconceptions

• "Benzene has a smaller enthalpy change": Saying "smaller" or "lower" is ambiguous because -360 is mathematically smaller than -208. State clearly: "less exothermic" or "lower magnitude".
• Confusing the models: Forgetting to compare the actual benzene measurement explicitly with what the Kekulé model would predict.
• Vague physical measurement: Stating merely "spectroscopy" without specifying X-ray diffraction or exact equivalence of bond lengths.

✅ Model Answer (3 Marks Breakdown)

• Solubility deduction 1 (1 mark): Anethole is soluble in ethanol (as liquor is transparent/clear).
• Solubility deduction 2 (1 mark): Anethole is insoluble / poorly soluble in water (as adding water causes oily droplets and a cloudy emulsion).
• Structural justification (1 mark - any one):
  • Anethole has a large non-polar hydrocarbon region (benzene ring and alkyl/alkenyl chain) which readily interacts with the non-polar ethyl ( –CH₂CH₃ ) group of ethanol via London dispersion forces.
  • Anethole cannot form significant hydrogen bonds with water molecules due to its predominantly non-polar structure.

🧠 Exam Tip

Read the prompt carefully: "detailed descriptions of intermolecular forces are not required". You do not need a full dipole-dipole/H-bonding essay. Focus directly on the hydrophobic/non-polar hydrocarbon skeleton matching ethanol's non-polar part.

✅ Correct Answer: B (6 peaks)

Anisaldehyde has exactly 6 unique carbon environments.

📐 Step-by-Step Peak Count

1. C1 (ring): Carbon bonded to –OCH₃ . (Environment 1)
2. C2 & C6 (ring): Two symmetric carbons adjacent to C1. (Environment 2)
3. C3 & C5 (ring): Two symmetric carbons adjacent to C4. (Environment 3)
4. C4 (ring): Carbon bonded to –CHO . (Environment 4)
5. Methoxy Carbon: The –O C H₃ carbon. (Environment 5)
6. Aldehyde Carbon: The – C HO carbonyl carbon. (Environment 6)

Total unique peaks = 4 (ring) + 1 (methoxy) + 1 (aldehyde) = 6.