An electrophile pulls a pair of electrons out of the delocalised ring, a positive intermediate forms, and losing H+ gives the ring back its full delocalisation. Groups already on the ring speed this up or slow it down and decide where the next group goes.
In electrophilic substitution a strong electrophile is made first: for nitration, HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−. A curly arrow from the delocalised π system to the electrophile forms a new C–E bond, giving a positive intermediate (an arenium ion) in which the delocalisation is partly broken: four electrons over five carbons. A curly arrow from the C–H bond back into the ring then removes H+ and restores the aromatic ring. Benzene substitutes rather than adds because addition would destroy the stable delocalised system. Substituents already on the ring direct the next group: –OH, –NH2, –CH3 and –Cl send it to the 2- and 4-positions, while –NO2 and –COOH send it to the 3-position. Electron-donating groups activate the ring; electron-withdrawing groups (and halogens) deactivate it.
Look at the delocalised ring in the simulator: adding across one C=C, as ethene does with bromine, would destroy that delocalisation, so benzene substitutes instead and keeps the ring.
| Ethene | Benzene | |
|---|---|---|
| π electrons | localised between two carbons | delocalised over six carbons |
| electron density | high, polarises Br2 by itself | spread out, needs a halogen carrier |
| reaction with Br2 | addition: CH2BrCH2Br | substitution: C6H5Br + HBr |
| after the intermediate | Br− adds to the carbocation | H+ is lost, ring restored |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What is the electrophile in the nitration of benzene?
In the second step of electrophilic substitution, what is lost from the intermediate?
Why does benzene undergo substitution rather than addition?
Methylbenzene is nitrated. Where does the NO2 group mainly go?
Every electrophilic substitution has the same three steps. Fig. 1 shows nitration; swap NO2+ for Br+, CH3+ or CH3CO+ and the arrows do not change.
| Step | What happens | Curly arrow |
|---|---|---|
| 1. make the electrophile | HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4− | none needed on the ring |
| 2. attack | two π electrons form a C–N bond; positive intermediate | from the delocalised ring (circle) to the N of NO2+ |
| 3. lose H+ | C–H electrons return to the ring; aromaticity restored | from the C–H bond into the horseshoe |
A group already on the ring changes both the rate and the position of a second substitution. Number the ring from the carbon carrying the first group.
| Group on the ring | Effect on ring | Directs to | Example |
|---|---|---|---|
| –OH, –NH2 | strongly activating (lone pair donated into ring) | 2- and 4- | phenol + Br2(aq), no carrier: 2,4,6-tribromophenol |
| –CH3 (alkyl) | activating (electron-donating) | 2- and 4- | methylbenzene nitrates faster than benzene: 2- and 4-nitromethylbenzene |
| –Cl, –Br | deactivating (electron-withdrawing) | 2- and 4- | chlorobenzene nitrates more slowly: 1-chloro-2-nitrobenzene and 1-chloro-4-nitrobenzene |
| –NO2, –COOH | strongly deactivating (electron-withdrawing) | 3- | nitrobenzene + Br2/FeBr3: 1-bromo-3-nitrobenzene |
The first curly arrow starts from the delocalised ring (the circle), not from a C=C or a C–H bond, and ends at the electrophile's atom. The intermediate's horseshoe must not include the sp3 carbon, and the + goes inside it. The second arrow starts from the C–H bond, and the product is H+, not H2 or H−. In the nitronium equation it is 2H2SO4. Halogens direct to 2 and 4 even though they deactivate the ring; –NO2 and –COOH direct to 3.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
Which equation shows the formation of the electrophile in nitration?
Nitrobenzene is brominated with Br2 and FeBr3. What is the main product, and how does the rate compare with benzene?
Which statement about the intermediate in electrophilic substitution of benzene is correct?
Which compound reacts fastest with bromine in electrophilic substitution?
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