Electrophilic aromatic substitution: mechanism, reagents and directing effects
Every EAS reaction is the same two steps: the ring's π electrons attack a strong electrophile to give a resonance-stabilised carbocation (the arenium ion), then a base removes the proton from that carbon to restore aromaticity. Only the electrophile changes between nitration, halogenation, sulfonation and the Friedel–Crafts reactions — and the substituent already on the ring decides where the new group lands.
Why the ring substitutes instead of adding
Benzene is an alkene that refuses to behave like one. An ordinary alkene reacts with Br₂ instantly, adding across the double bond. Benzene does not react with Br₂ at all without a catalyst, and when it finally does, it substitutes — swapping an H for a Br and keeping the ring intact.
The reason is aromatic stabilisation. Addition would permanently destroy the aromatic system; substitution borrows it for one step and gives it back. So the intermediate arenium ion — positively charged, with the charge spread over three ring carbons — is a very high-energy species, and the fastest way out is to lose a proton and get the aromaticity back.
That also explains the catalyst. Benzene's π system is far less nucleophilic than an alkene's, so an ordinary electrophile is not enough. Every EAS reaction begins by making an unusually strong electrophile.
The two-step mechanism, once
- Attack. Two π electrons from the ring form a bond to the electrophile. That carbon becomes sp³, and the remaining four π electrons are delocalised over the other five carbons — draw the three resonance structures, because mark schemes usually ask for them. This step is slow: it costs the aromaticity.
- Deprotonation. A weak base (usually the counterion generated alongside the electrophile) removes the proton from the sp³ carbon. The C–H electrons drop back into the ring and aromaticity returns. This step is fast and strongly downhill.
Why no nucleophile adds to the arenium ion. It could — that would be addition, exactly as with an alkene. But the product would be a non-aromatic cyclohexadiene, which is far higher in energy than the aromatic substitution product. The thermodynamics decide it.
The five reactions and their electrophiles
| Reaction | Reagents | The actual electrophile | Group installed |
|---|---|---|---|
| Nitration | HNO₃ / H₂SO₄ | NO₂⁺ (nitronium) | –NO₂ |
| Halogenation | Br₂ / FeBr₃ or Cl₂ / AlCl₃ | Br⁺-like, polarised by the Lewis acid | –Br, –Cl |
| Sulfonation | Fuming H₂SO₄ (SO₃) | SO₃ or HSO₃⁺ | –SO₃H (and it is reversible) |
| Friedel–Crafts alkylation | RCl / AlCl₃ | R⁺ (carbocation) | –R |
| Friedel–Crafts acylation | RCOCl / AlCl₃ | RCO⁺ (acylium) | –COR |
Sulfuric acid's job in nitration is to protonate nitric acid so that water leaves, generating NO₂⁺. The Lewis acid's job in halogenation and Friedel–Crafts is to pull electron density off the halogen and make the other end sharply electrophilic. Both are the same trick.
Directing effects: where does the new group go?
If the ring already carries a substituent, that substituent controls both how fast the reaction runs and which position reacts. The rule comes straight from the arenium ion: a substituent that can stabilise the positive charge directs to the positions where the charge appears.
- Activators are ortho/para directors. Groups that donate electrons — –NH₂, –OH, –OR, –NHCOR, and alkyl groups — put extra electron density at the ortho and para positions and stabilise the cation when the attack happens there. The strong donors (amino, hydroxy) are powerful enough that the reaction may need moderating.
- Deactivators are meta directors. Groups that withdraw electrons — –NO₂, –CN, –SO₃H, –COR, –COOH, –CF₃ — destabilise the cation everywhere, but destabilise it most at ortho and para, where the positive charge lands directly next to the withdrawing group. Meta becomes the least-bad option.
- The halogens are the exception. –F, –Cl, –Br and –I are deactivating (they withdraw inductively, so the reaction is slow) but ortho/para directing (their lone pairs donate by resonance, which stabilises the cation at ortho and para). Learn this pair as a special case; it is a favourite exam question.
Worked example: nitration of toluene
- Sulfuric acid protonates nitric acid; water leaves; NO₂⁺ forms.
- Toluene's π electrons attack the nitrogen of NO₂⁺. Methyl is a weak activator and an ortho/para director, so attack at those positions gives an arenium ion with the positive charge on the carbon bearing the methyl — a tertiary-like cation, which is the stabilised option.
- HSO₄⁻ removes the proton from the sp³ carbon and aromaticity is restored.
Products: 2-nitrotoluene and 4-nitrotoluene, with meta a minor product. Because there are two ortho positions and only one para, ortho takes the larger share of the mixture for a substituent as small as methyl — but the para:ortho ratio climbs steeply as the alkyl group gets bulkier, and a t-butyl group gives almost entirely para.
The Friedel–Crafts limitations you will be examined on
- Carbocation rearrangement. Alkylation with 1-chloropropane does not give propylbenzene; the primary cation rearranges by a hydride shift to the secondary one, so you get isopropylbenzene. Acylation avoids this entirely — the acylium ion is resonance-stabilised and does not rearrange.
- Polyalkylation. The alkyl group you install is an activator, so the product reacts faster than the starting material and you get multiple substitutions. Acylation self-limits, because the ketone installed is deactivating.
- Deactivated rings do not react. Neither Friedel–Crafts reaction works on a ring bearing a strong deactivator such as nitro, and amino groups poison the Lewis acid catalyst.
- The standard workaround. To get a straight-chain alkyl group, acylate first and then reduce the ketone (Clemmensen or Wolff–Kishner). Two steps, no rearrangement, no polysubstitution.
Planning a disubstituted ring: order matters
To make 3-nitroacetophenone you acylate first, then nitrate — the ketone is a meta director, so the nitro group lands where you want it. Reverse the order and the nitro group both directs the acyl group to meta and deactivates the ring so badly that Friedel–Crafts fails altogether. Whenever a question gives you a disubstituted target, work out the directing consequences of each order before committing.
How Organic Chemistry AI helps here
EAS problems fail in predictable places: the electrophile was not identified, the wrong directing effect was applied, or the two steps were sequenced in the wrong order. A solve names the reaction type, works through generating the electrophile, and states the directing rule that chose the position — so you can see which of those three steps you got wrong. The bundled study library includes an aromatic chapter, and the mechanism library holds named EAS reactions you can browse offline.
The mechanism library holds 106 named reactions, each tagged with its reaction type and searchable offline.
Frequently asked
Why are halogens deactivating but ortho/para directing?
Two opposing effects act on different things. Inductive withdrawal, which dominates the ground state, pulls electron density out of the ring and makes it react slowly. Resonance donation from the halogen's lone pairs only matters in the arenium ion, where it stabilises the positive charge specifically at the ortho and para positions. So the ring is less reactive, but reacts at those positions when it does.
Why does Friedel–Crafts acylation avoid rearrangement?
The acylium ion RCO⁺ is stabilised by resonance with the oxygen lone pair, which gives it a C≡O⁺ character. It is already low enough in energy that there is nothing to gain from a hydride or alkyl shift, so the carbon skeleton you draw is the one you get.
Does the order of two substitutions matter?
Almost always. The first group installed directs the second, and it also changes how reactive the ring is. Acylating then nitrating gives the meta product; nitrating then attempting Friedel–Crafts usually fails entirely, because a nitro group deactivates the ring too much for the reaction to run.
Work through an aromatic problem
Photograph the ring and the reagents, and get the electrophile, the intermediate and the directing call written out.
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