Explain electrophile generation and trace the electron pairs through aromatic substitution, including how the catalyst is regenerated.
Generate the electrophile before attacking the ring
For benzene nitration, concentrated nitric acid reacts with concentrated sulfuric acid to generate NO₂⁺, the nitronium ion. Sulfuric acid acts as the acid catalyst in the overall process. Controlled warming is used for mononitration; do not treat arbitrary high temperature as interchangeable with specified conditions.
One balanced representation is HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. The positive charge is essential: neutral NO₂ is not the nitronium electrophile. The overall substitution replaces one ring H with NO₂, forming nitrobenzene and water from benzene and nitric acid.
Follow the electron pair, then restore delocalisation
A full-headed curly arrow begins at the benzene π system and ends at the nitrogen of NO₂⁺. A new C–N σ bond forms. The attacked ring carbon temporarily has both H and NO₂ attached, so the intermediate is not fully aromatic and bears an overall positive charge delocalised over the remaining ring positions.
Next, a base such as HSO₄⁻ removes H⁺ from the substituted carbon. Draw a curly arrow from that C–H bond back into the ring to restore the π system. This is not an arrow from H into the ring: it is the bond’s electron pair that returns. Sulfuric acid is regenerated.
The complete story explains substitution: the electrophile has replaced H, the ring’s delocalisation returns and a catalyst is restored. An intermediate with an intact full aromatic circle and an extra bond at a carbon would misrepresent both the temporary loss of aromaticity and valency.
Diagram placeholder
Electrophilic substitution of benzene by NO₂⁺
Labels to include:
- Benzene π system
- NO₂⁺ with positive charge on the electrophile
- Full-headed arrow from π system to N
- Non-aromatic positively charged intermediate with H and NO₂ on one carbon
- Arrow from that C–H bond back into the ring
- Nitrobenzene, regenerated H₂SO₄ and restored aromatic circle
The first arrow uses ring electrons to make the C–N bond. The attacked carbon retains H in the positively charged intermediate. Removal of that H as H⁺ returns the C–H electron pair to the ring and restores aromaticity. The diagram must show the interrupted π system in the intermediate, not a complete benzene circle.
The halogen carrier strengthens the electrophile
Benzene reacts with Br₂ in the presence of a suitable halogen carrier such as FeBr₃, or iron that generates a carrier in situ. A useful formal representation is Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻; the course permits Br⁺ as the electrophile for the mechanism.
The aromatic π system attacks Br⁺, giving the analogous positive intermediate. Loss of H⁺ restores delocalisation; FeBr₄⁻ supplies the base role and regenerates FeBr₃ with HBr formed. Overall C₆H₆ + Br₂ → C₆H₅Br + HBr. For chlorine use an appropriate chlorine carrier and Cl⁺ model.
Do not import the alkene addition product with two bromines on adjacent carbons. The arene pathway replaces one H and retains the aromatic system. In unfamiliar questions, identify the supplied electrophile and use the same attack–intermediate–restoration logic.
Form a carbon–carbon bond to the ring
Alkylation uses a haloalkane and a halogen carrier such as anhydrous AlCl₃. For example benzene plus CH₃Cl gives methylbenzene and HCl. Acylation uses an acyl chloride: CH₃COCl and benzene give C₆H₅COCH₃ and HCl in the presence of an appropriate halogen carrier.
In alkylation the carbon from R–Cl attaches to the ring. In acylation the carbonyl carbon of RCOCl attaches, so the C=O remains in the ketone product. Count the incoming carbons before drawing the product: ethanoyl chloride adds a two-carbon acyl group, not just a methyl group.
The specification requires these transformations and their C–C bond-forming importance. Do not imply that every alkylation produces only one pure product under all circumstances. Selectivity and further substitution can matter in synthetic planning.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.
Q1. Balance the equation generating NO₂⁺ using one HNO₃ and one H₂SO₄.Show answer
HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. Atoms balance and the two ionic charges sum to zero.
Q2. Where does the first curly arrow begin in benzene nitration?Show answer
At the ring π electron system, ending at N of NO₂⁺. It shows movement of an electron pair from benzene to the electrophile, not movement of the electrophile towards the ring.
Q3. Why does the intermediate lose H⁺ rather than retain both H and the electrophile?Show answer
Removing H⁺ returns the C–H bond pair into the ring and restores delocalisation. The net outcome replaces H and preserves the aromatic system.
Q4. Give the organic product from benzene and propanoyl chloride with a suitable halogen carrier.Show answer
C₆H₅COCH₂CH₃, an aromatic ketone. The incoming acyl group retains its carbonyl and three carbons; HCl is also formed in the overall substitution.
Q5. An unfamiliar reagent supplies E⁺. What must a defensible substitution mechanism show?Show answer
An electron-pair arrow from the ring to E, a valid positively charged intermediate with H and E at the attacked carbon, and return of the C–H pair to restore aromaticity as H⁺ is lost. Use the given base/catalyst information rather than inventing reagents.
Sources
Sources and examiner guidance (reviewed 6 October 2026)
- OCR A H432 specification — version 3.1 — 6.1.1, printed pp. 54–55; outcomes and additional guidance, with relevant Module 1 practical skills.
- Chemrevise — OCR A 6.1.1 — Pages 1–6; secondary coverage cross-check. Lesson explanations, data and questions are original Finesse material.
- OCR H432/02 mark scheme — June 2025 — Q20(b–c); printed pp. 24–27. Question-specific evidence, not universal marking rules.
- OCR H432/02 examiner report — June 2025 — Q20(b–c); printed pp. 38–43. Read with the corresponding question context.
- OCR H432/02 question paper — June 2025 — Q20(b–c); context for the assessment references, not reproduced questions.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.