OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 17

✅ Correct Answer

• Product with B: 1,3,5-trimethylcyclohexene reacted with Cl₂ via electrophilic addition across the C=C double bond, yielding a 1,2-dichlorocyclohexane derivative with two chlorine atoms attached to the adjacent carbons of the former double bond.
• Product with C: 1,3,5-trimethylbenzene (mesitylene) reacted with Cl₂ in the presence of a halogen carrier via electrophilic aromatic substitution, yielding 2-chloro-1,3,5-trimethylbenzene (one chlorine atom substituted onto the benzene ring).

🧠 Exam Technique

• Ensure skeletal or structural formulae clearly show attachment points.
• Double-check that Compound B undergoes addition (loss of double bond, gain of two Cl atoms) while Compound C undergoes substitution (retention of the stable delocalized ring, replacement of one H with Cl).

✅ Correct Answer (Mark Scheme Points)

• In Compound B, the ᴨ-electrons are localized in a C=C double bond.
• In Compound C, the ᴨ-electrons are delocalized across the benzene ring.
• Compound B has a higher electron density at the double bond compared to the delocalized ring of C.
• Therefore, Compound B is more susceptible to electrophilic attack / polarises the Cl₂ molecule more readily than C.

❌ Common Errors

• Vague references to "charge density" or "electronegativity" without linking back to electron localization or delocalization.
• Stating that C is stable without explaining why delocalization lowers reactivity toward electrophilic addition.

💡 Key Knowledge (The 4 Stages)

• 1. Generation of electrophile: AlCl₃ + Cl₂ ➔ AlCl₄⁻ + Cl⁺
• 2. Attack: Curly arrow starting from the delocalized ring (or close to it) pointing directly to the Cl⁺ ion.
• 3. Intermediate: Horseshoe-shaped partial ring inside the hexagon with a positive charge (⁺) , and both H and Cl bonded to the same carbon.
• 4. Regeneration of catalyst: Curly arrow from the C–H bond back into the ring, plus equation: H⁺ + AlCl₄⁻ ➔ AlCl₃ + HCl .

🧠 Exam Technique & Cautions

• Crucial Arrow Rule: The first curly arrow must start from the circle/pi-bond and must not cross the entire ring haphazardly; it must point clearly to the incoming electrophile.
• The horseshoe intermediate must show the positive charge clearly residing inside the incomplete ring, with the gap centered around the carbon bonding the chlorine.

✅ Correct Answer

3C₃H₆O ➔ C₉H₁₂ + 3H₂O

• Compound C has 9 carbon atoms (mesitylene, C₉H₁₂ ). Propanone is C₃H₆O .
• Trimerisation combines 3 molecules of propanone, eliminating 3 molecules of water (3H₂O) .

❌ Common Errors

• Using structural or displayed formulae when the question specifically asks for molecular formulae.
• Failing to balance water molecules or miscounting hydrogen atoms in mesitylene.

✅ Correct Answer

• Compound C: 3 peaks
• Compound D: 8 peaks

💡 Symmetry Analysis

• Compound C (Mesitylene): Has high molecular symmetry due to its 1,3,5-trisubstituted methyl groups. Environments: methyl carbons (1 peak), aromatic carbons with attached methyls (1 peak), aromatic CH carbons (1 peak) = 3 peaks total.
• Compound D: The attachment of the amide group disrupts the 3-fold symmetry, making almost all ring carbon environments chemically unique, resulting in 8 distinct signals.

💡 Reagents, Catalysts, and Intermediates

• Step 1 (Nitration of Compound C):
  • Reagent: Concentrated HNO₃
  • Catalyst: Concentrated H₂SO₄
  • Intermediate product: 2-nitro-1,3,5-trimethylbenzene (benzene ring with -NO₂ group).
• Step 2 (Reduction of Nitro group):
  • Conditions given: Sn + HCl followed by neutralisation.
  • Intermediate product: 2-amino-1,3,5-trimethylbenzene (benzene ring with -NH₂ group).
• Step 3 (Acylation to form Compound D):
  • Reagent: CH₃COCl (ethanoyl chloride) or (CH₃CO)₂O (ethanoic anhydride).
  • Final product: Compound D (amide linked as shown in prompt).

🧠 Exam Technique & Guidance

• Ensure structural/skeletal drawings clearly show functional group transformations ( -NO₂ ➔ -NH₂ ➔ -NHCOCH₃ ).
• Note that HNO₃ and H₂SO₄ labels around the arrow are interchangeable for the nitration step.