Read the core explanations from Nitration of Benzene before testing yourself from memory.
1. Why Nitration of Benzene Matters
Benzene is famously unreactive towards the electrophilic addition reactions that alkenes undergo. Yet with the right reagent, you can attach a nitro group ($-\text{NO}_2$) to the ring — and that single transformation unlocks an enormous range of chemistry, from dyes and pharmaceuticals to explosives and amines.
Understanding nitration means understanding two things at once: how a powerful electrophile is generated, and why benzene's unique electronic structure forces the reaction to go via substitution rather than addition. Both ideas are examinable at A-level, and both reward careful mechanistic thinking.
2. Generating the Nitronium Ion
The nitrating mixture is made by combining concentrated nitric acid ($\text{HNO}_3$) with concentrated sulphuric acid ($\text{H}_2\text{SO}_4$). Sulphuric acid is a stronger acid than nitric acid, so it donates a proton to nitric acid — acting as the acid in a Brønsted–Lowry sense.
The protonated nitric acid then loses a water molecule, producing the nitronium ion, $\text{NO}_2^+$. This species carries a full positive charge and is a potent electrophile — exactly what is needed to attack the electron-rich benzene ring.
Notice that sulphuric acid is regenerated at the end of the overall reaction (it is consumed in generating $\text{NO}_2^+$ but the $\text{HSO}_4^-$ picks up the proton lost later in the mechanism). This is why sulphuric acid acts as a catalyst in the overall process.
3. The Electrophilic Substitution Mechanism
With the nitronium ion generated, the mechanism itself has two key stages. First, the electron-rich $\pi$ system of benzene donates two electrons to $\text{NO}_2^+$, forming a new C–N bond and creating a positively charged, non-aromatic intermediate. Second, that intermediate rapidly loses a proton to restore the aromatic ring.
The intermediate formed in the first stage is called the Wheland intermediate (sometimes called the arenium ion or sigma complex). At this point, the ring has lost its full delocalisation — four of the six $\pi$ electrons are now localised in two double bonds, and the positive charge is delocalised over three carbons of the ring (the arenium ion). The system is under energetic strain to recover aromaticity.
The overall equation for the nitration of benzene is clean and worth memorising for the exam:
3a. Why Substitution, Not Addition?
Alkenes undergo electrophilic addition because there is no special stability to lose — adding across the double bond converts $\text{sp}^2$ carbons to $\text{sp}^3$ carbons, and the product is more stable than the starting material. Benzene is fundamentally different.
The benzene ring possesses a delocalised $\pi$ system of six electrons spread over all six carbons. This delocalisation confers considerable thermodynamic stability — around 150 kJ mol$^{-1}$ more stable than a hypothetical cyclohexatriene. If addition occurred, that stabilisation would be permanently destroyed. Substitution, by contrast, restores the aromatic system in the second step, so the energetic cost is recovered.
The driving force for the second step (proton loss) is therefore the recovery of aromaticity. The Wheland intermediate is unstable precisely because it has lost full delocalisation, and losing $\text{H}^+$ is the fastest route back to stability.
4. Conditions for Nitration
The reaction is carried out at approximately 50 °C. This temperature is a deliberate compromise: warm enough to give a reasonable rate of reaction, but cool enough to prevent dinitration (the introduction of a second $-\text{NO}_2$ group). Above about 55 °C, the reaction becomes harder to control and dinitrobenzene forms in significant quantities.
- Reagents: concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ (the nitrating mixture)
- Temperature: approximately 50 °C (maintained carefully — use a water bath)
- Product: nitrobenzene ($\text{C}_6\text{H}_5\text{NO}_2$) — a pale yellow liquid
- Sulphuric acid is regenerated and acts as a catalyst overall
5. The Step-by-Step Mechanism Walkthrough
Mechanism questions in AQA A-level papers award marks for correct curly arrows, correct intermediates, and correct products. Here is a structured walkthrough of exactly what to draw and why each arrow goes where it does.
Worked example: Drawing the Nitration Mechanism for an AQA Exam
Outline the mechanism for the nitration of benzene using a mixture of concentrated nitric acid and concentrated sulphuric acid. Include the generation of the electrophile and all curly arrows. [5 marks]
- $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}$ — [B1] State that concentrated $\text{H}_2\text{SO}_4$ protonates $\text{HNO}_3$, which then loses water to give the nitronium ion $\text{NO}_2^+$. This is the electrophile. Always name it — examiners look for '$\text{NO}_2^+$' or 'nitronium ion'.
- Arrow from the benzene $\pi$ system (drawn as a circle or as alternating double bonds) pointing towards the nitrogen atom of $\text{NO}_2^+$. — [M1] The arrow must start from the electron-rich benzene ring and end at the electrophilic N. A common error is drawing the arrow from a specific C–C bond rather than the $\pi$ system — either is acceptable, but the arrow must clearly show electrons moving from ring to electrophile.
- A six-membered ring with the $\text{NO}_2$ group attached to one carbon, a hydrogen still on that carbon, a positive charge shown on the ring (delocalised over the remaining carbons), and two double bonds remaining in the ring. — [M1] The intermediate must show the ring is no longer fully aromatic — it has lost one $\pi$ electron pair to form the new C–N bond. The positive charge is delocalised over the remaining carbons. Examiners accept the charge shown on the carbon adjacent to the substituted carbon.
- Arrow from the C–H bond (on the carbon bearing $\text{NO}_2$) pointing towards H, and then H is accepted by $\text{HSO}_4^-$ (or shown leaving as $\text{H}^+$). — [M1] This step restores aromaticity. The arrow starts from the C–H bond and points to the H atom. Some mark schemes also credit an arrow from H to the base ($\text{HSO}_4^-$). Either way, the key idea is that proton loss regenerates the delocalised ring.
- Product: nitrobenzene $\text{C}_6\text{H}_5\text{NO}_2$. Sulphuric acid is regenerated: $\text{H}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4$. — [B1] Stating that $\text{H}_2\text{SO}_4$ is regenerated confirms your understanding that it acts as a catalyst. This is often worth a mark in longer questions asking you to 'explain the role of sulphuric acid'.
Answer: The nitronium ion ($\text{NO}_2^+$) attacks the benzene ring; the Wheland intermediate forms; loss of $\text{H}^+$ (accepted by $\text{HSO}_4^-$) restores aromaticity to give nitrobenzene. $\text{H}_2\text{SO}_4$ is regenerated.
6. Importance of Nitration in Synthesis
Nitration is not an end in itself — it is a gateway reaction. The $-\text{NO}_2$ group introduced onto the ring can be reduced to a primary amine ($-\text{NH}_2$), opening up a vast range of further chemistry. This reduction is typically carried out using tin and concentrated hydrochloric acid, or by catalytic hydrogenation.
Aromatic amines such as phenylamine (aniline, $\text{C}_6\text{H}_5\text{NH}_2$) are the starting materials for azo dyes and many pharmaceutical compounds. The sequence — nitration of benzene, then reduction of the nitro group — is one of the most important two-step routes in industrial organic chemistry.
At the other end of the application spectrum, polynitration of aromatic compounds produces explosives. Trinitrotoluene (TNT, 2,4,6-trinitrotoluene) is made by the controlled nitration of toluene (methylbenzene) three times in succession. The high nitrogen content and the energy released when the $\text{N}–\text{O}$ bonds break rapidly make these compounds powerful explosives. This is a direct industrial consequence of the electrophilic substitution chemistry you have just studied.
| Application | Starting material | Key transformation | Product / use |
|---|
| Amine synthesis | Nitrobenzene ($\text{C}_6\text{H}_5\text{NO}_2$) | Reduction of $-\text{NO}_2$ to $-\text{NH}_2$ | Phenylamine — dyes, pharmaceuticals |
| Explosives | Toluene (methylbenzene) | Repeated nitration (×3) | TNT — high explosive |
| Dye intermediates | Various arenes | Nitration then diazotisation | Azo dyes for textiles and food |
7. Exam Technique and Common Mistakes
Mechanism questions on nitration follow a predictable pattern in AQA papers. The marks are distributed across: naming or generating the electrophile, drawing curly arrows correctly, showing the Wheland intermediate, and showing proton loss restoring aromaticity. Missing any one of these stages costs marks even if the rest of the answer is perfect.
Before the next quiz block, connect the energy profile to the mechanism: the ring temporarily loses aromatic stability when the sigma complex forms, which is why the activation energy is relatively high.
The essential AQA points are generation of the nitronium ion electrophile, attack on the benzene ring to form a sigma complex, and loss of a proton to restore aromaticity.