Generate the nitronium ion and follow electron pairs through the arenium intermediate and catalyst regeneration.
The electrophile is generated by the acid mixture
Benzene reacts with concentrated nitric acid in the presence of concentrated sulfuric acid, using controlled warming around 50–60 °C for the standard monosubstitution example. Sulfuric acid protonates nitric acid; loss of water forms the nitronium ion, NO₂⁺. Sulfuric acid acts as the acid catalyst.
The nitronium ion is an electron-pair acceptor. Show its positive charge on nitrogen in the mechanism and connect the incoming ring bond to N, not O. The reaction is an electrophilic substitution, not addition of nitric acid across one isolated alkene bond.
The ring donates an electron pair
Draw a curly arrow from the π system inside the benzene ring towards nitrogen of NO₂⁺. The attacked carbon now carries both H and NO₂ and is tetrahedral. The intermediate has a positive charge delocalised over the remaining ring positions; it is not fully aromatic.
If using the horseshoe representation, leave the attacked carbon outside the horseshoe and do not put an intact circle in the intermediate. The intermediate’s positive charge belongs to the ring system, not to the H that is still covalently attached. A complete arrow describes a pair of electrons, so it starts at the ring electron density rather than at an arbitrary carbon nucleus.
Loss of a proton restores delocalisation
A base such as HSO₄⁻ accepts the proton from the attacked carbon. Draw the C–H bond electron pair moving back into the ring. The π system is restored and nitrobenzene forms. Combining H⁺ with HSO₄⁻ regenerates H₂SO₄, explaining the catalyst’s role rather than just naming it.
June 2023 Paper 2 Q03.2 used methylbenzene and specifically tested the arrow to nitronium nitrogen and the C–H bond arrow back into the ring. This same electron accounting applies to the benzene example taught here.
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Benzene nitration mechanism to add
Labels to include:
- Benzene circle and NO₂⁺ with positive charge on N
- Arrow from inside ring to nitronium N
- Intermediate: attacked C bonded to H and NO₂
- Partial horseshoe over the other five ring carbons and overall + charge
- Arrow from C–H bond into ring
- Nitrobenzene with complete aromatic circle
- HSO₄⁻ accepts H⁺ and H₂SO₄ is regenerated
The first intermediate must not retain an intact aromatic circle. Both H and the newly bonded NO₂ are shown on the same attacked carbon until deprotonation.
Conditions and context matter
Control temperature and reagent addition because nitration is exothermic; the standard conditions favour the intended monosubstituted product. More forcing conditions can lead to further reaction, so do not extend a monosubstitution equation to every possible mixture. Benzene and the concentrated-acid mixture require controlled laboratory handling.
Nitroarenes are useful intermediates: reducing the nitro group can make aromatic amines used in dye manufacture. Nitration also occurs in industrial manufacture of some energetic materials, but the learning task here is the specified monosubstitution mechanism.
Constructed yield example: 0.0500 mol benzene gives at most 0.0500 mol nitrobenzene. With Mᵣ = 123.0, the theoretical mass is 6.15 g. If 4.92 g is isolated, yield = 4.92/6.15 × 100 = 80.0%. Catalyst mass is not part of that 1:1 reactant-to-product ratio.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Identify the nitrating electrophile and its attacking atom.Show answer
NO₂⁺, the nitronium ion. Benzene forms its new bond to nitrogen.
Q2. Write the equation generating NO₂⁺ from nitric and sulfuric acids.Show answer
HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. Atoms and overall charge balance.
Q3. Why is a full circle incorrect in the nitration intermediate?Show answer
The attacked carbon has become tetrahedral, interrupting full cyclic p-orbital overlap. Complete aromatic delocalisation is restored only after proton loss.
Q4. Where does the final ring arrow start?Show answer
At the C–H bond on the carbon carrying the new substituent. Its electron pair returns to the ring as H⁺ is removed.
Q5. What would 0.0300 mol benzene yield theoretically as nitrobenzene, using Mᵣ = 123.0?Show answer
The 1:1 ratio gives 0.0300 mol product. Theoretical mass = 0.0300 × 123.0 = 3.69 g.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- AQA 7405 organic chemistry specification — 3.3.10 coverage and required skills.
- Chemrevise: Aromatic chemistry — Coverage checklist; explanations, data exercises and quick checks on this page are original Finesse material.
- AQA June 2023 Paper 2 mark scheme — Q03.2 p14 and report Q03.2 p3: electrophilic substitution, arrow to nitronium nitrogen and C–H bond return into the ring.
- AQA June 2023 Paper 2 examiner report — Read alongside the question-specific marking guidance; not a universal wording checklist.
- AQA June 2022 Paper 2 mark scheme — Q08.1–08.3 pp28–29 and report Q08 p6: supplied acylating reagent, AlCl₃, electrophile and ring intermediate.
- AQA June 2022 Paper 2 examiner report — Read with the matching question context described in the mark-scheme source.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.