OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2024: Question 21

✅ Correct Answer

An electron pair acceptor.

💡 Key Knowledge

Make sure you memorise precise definitions. Saying "electron seeker" or "electron-deficient species" will lose the mark. Examiners strictly require electron pair acceptor (or gains an electron pair ).

✅ Correct Answer / Outline

• Generation of electrophile: AlCl₃ + CH₃CH₂Cl → CH₃CH₂⁺ + AlCl₄⁻
• First curly arrow: Starts from the delocalised benzene pi-bond ring and points directly to the CH₃CH₂⁺ carbon.
• Intermediate: Horseshoe-shaped intermediate with a positive charge inside the incomplete circle, and both H and CH₂CH₃ on the junction carbon.
• Second curly arrow: Starts from the C-H bond and goes back into the ring to restore the delocalised system.
• Regeneration of catalyst: H⁺ + AlCl₄⁻ → AlCl₃ + HCl

🧠 Exam Technique & Arrow Rules

• The first curly arrow must start from the circle (pi-bond) and point to the plus charge on the carbocation.
• The horseshoe in the intermediate must have the open end facing the carbon holding the alkyl group, with the positive charge clearly located inside the remaining horseshoe part of the ring.

❌ Common Errors

• Drawing the first curly arrow starting from a carbon atom instead of the ring's pi-electron cloud.
• Failing to show the regeneration step of the AlCl₃ catalyst alongside HCl production.

✅ Correct Answer

• For the -CN group (3-directing / meta-directing): The incoming Cl must substitute onto the carbon at position 3 relative to the CN group.
• For the -NHCOCH₃ group (2- and 4-directing / ortho- and para-directing): You must draw both the 2-substituted (ortho) product and the 4-substituted (para) product.

💡 Key Knowledge

Look carefully at the directing table provided in the prompt: -CN is 3-directing, while -NHCOCH₃ directs to the 2- and 4-positions.

❌ Common Errors

Forgetting to draw both isomers when a group directs to multiple positions (ortho and para). Only drawing one structure when two products are formed will lose credit.

✅ Correct Answer

Organic product structure: 2,4,6-trichlorophenylamine (amine group with Cl atoms at both ortho positions and the para position).

Balanced equation:
C₆H₅NH₂ + 3Cl₂ → C₆H₂Cl₃NH₂ + 3HCl

🧠 Exam Technique

Ensure your stoichiometric coefficients balance properly. You need 3Cl₂ to introduce three chlorine atoms, which also generates 3HCl as a byproduct.

✅ Correct Answer (Mark Scheme Points)

• Point 1: The lone pair of electrons on the nitrogen (in -NH₂ ) is partially delocalised / donated into the pi-system of the benzene ring.
• Point 2: This increases the electron density on the ring compared to benzene (or makes electron density higher).
• Point 3: Therefore, phenylamine is more susceptible to electrophilic attack / polarises the incoming Cl₂ molecule more effectively.

💡 Key Knowledge

Examiners look for comparative language when answering "why" questions comparing two species. You must explicitly contrast phenylamine with benzene (e.g., "higher electron density than benzene").

❌ Common Errors

Stating that the lone pair belongs to the whole -NH₂ group rather than specifically originating from the nitrogen atom. The mark scheme explicitly insists it must be clear the electrons come from N.