Aromatic Substitution Mechanisms
Aromatic rings react by mechanisms that exist nowhere else, because the ring insists on getting its aromaticity back. This part covers all four routes — the familiar electrophilic one, the addition–elimination that needs electron-withdrawing groups, the elimination–addition that goes through a triple bond inside a ring, and the radical chain that needs none of the above. It then does what most textbooks skip: shows how directing effects are measured rather than merely asserted. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.
The 9 sections in Part 8
- 1The two-step mechanism, the arenium ion, and the evidence that step 1 is rate-determining Free below
- 2Making the electrophile: nitration, halogenation, sulfonation and Friedel–Crafts
- 3Activating and deactivating groups — and why the halogens break the pattern
- 4Ortho:para ratios, competing substituents and the order of a synthesis
- 5Nucleophilic aromatic substitution by addition–elimination: the Meisenheimer complex
- 6Benzyne and the SRN1 radical chain — substitution without activation
- 7Partial rate factors — measuring a directing effect instead of asserting it
- 8The Hammett equation, σ+ and σ−
- 9What a change in ρ tells you — and where this goes next
The two-step mechanism, the arenium ion, and the evidence that step 1 is rate-determining
Free extractSection H.1 of Part 8, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.
Two steps, one intermediate, and one experiment that settles which step is slow. Get this section right and three-quarters of the topic follows.
Electrophilic aromatic substitution is two steps, never one. Not ‘usually two’ — two, with an intermediate that has been isolated, crystallised and structurally characterised.
Mechanism — electrophilic aromatic substitution (SEAr), general
- Step 1 — addition of the electrophile (slow, rate-determining). Two π electrons of the ring reach out to E+. One ring carbon becomes sp3, carrying both the incoming electrophile and its original hydrogen. The aromatic sextet is destroyed and a delocalised cation — the arenium ion — is formed. This step is strongly endothermic, because it is here that the 150 kJ mol−1 of aromatic stabilisation is surrendered.
- Step 2 — loss of the proton (fast). A base — the conjugate base of the acid used to make the electrophile, or the solvent, or the counterion AlCl4−/HSO4− — removes the hydrogen from the sp3 carbon. The C–H bonding pair drops into the ring, the sextet reforms, and the whole 150 kJ mol−1 comes back. Strongly exothermic and effectively irreversible.
- Net result: Ar–H + E+ → Ar–E + H+. The electrophile has taken the hydrogen’s place; the ring is aromatic again.
Four names, one species. Wheland intermediate honours G. W. Wheland, who first used it to rationalise directing effects. σ complex distinguishes it from the weakly bound π complex that precedes it. Benzenonium or benzenium ion is the systematic name of the parent C6H7+. Examiners use all four interchangeably.
Where the positive charge actually sits — and where it does not
This is the single most misdrawn structure in the whole of organic chemistry, so slow down here. The sp3 carbon — the one bearing the new electrophile — has four σ bonds and no p orbital. It cannot carry positive charge. It is not part of the delocalised system at all; it is an insulator sitting in the ring, and its whole function is to break the conjugation into a five-carbon chain.
The charge is spread over the other five carbons, and even then not evenly. A pentadienyl cation puts its charge on alternate atoms — positions 1, 3 and 5 of the five-carbon chain. Numbering the ring from the sp3 carbon, those are the two ortho carbons and the para carbon. The two meta carbons carry essentially none.
The same molecule, drawn by RDKit from an explicit Lewis structure, makes the point again — note which carbon carries the plus:
The energy profile: two humps, and the first one is taller
The isotope-effect test, in one page
Suppose you did not know which step was slow. There is a clean experiment. Deuterium is twice as heavy as protium, so a C–D bond vibrates more slowly than a C–H bond — about 2200 cm−1 against 3000 cm−1 — and therefore sits lower in its potential well by half that difference in vibrational wavenumber, which is the difference in zero-point energy. If the bond is being broken in the rate-determining step, the deuterated compound starts from further down and has a taller hill to climb, so it reacts more slowly.
The experiment was done by Lars Melander in the early 1950s. He nitrated and brominated benzene, toluene and naphthalene that had been labelled with deuterium and with tritium, and measured how much slower the labelled positions reacted. The answer was: they did not react more slowly at all. Within experimental error, kH/kD = 1.
Bromination, chlorination, Friedel–Crafts: same answer.
Conclusion: the C–H bond is intact at the rate-determining transition state ⇒ proton loss is not the slow step ⇒ step 1 is rate-determining.
A student proposes that electrophilic aromatic substitution is a single concerted step in which E+ comes in as H+ leaves, like an SN2 at carbon. Design the experiment that kills this proposal, and state what result you expect. Easy
⚠ Common mistakes & exam traps
- Never put the positive charge on the carbon bearing the electrophile. That carbon is sp3, has four σ bonds and no p orbital, and is the one carbon in the arenium ion that cannot be positive. Charge lives on the two ortho carbons and the para carbon relative to it.
- Do not expect a primary isotope effect in ordinary SEAr. For nitration, bromination, chlorination and Friedel–Crafts, kH/kD = 1.0. If a question reports a KIE of 1 and asks what it tells you, the answer is ‘C–H cleavage is after the rate-determining step’, not ‘the reaction is not SEAr’.
- The arenium ion is an intermediate, not a transition state. It sits in a well, it can in favourable cases be isolated, and it must be drawn with full bonds and a full charge — never with dashed partial bonds and a double dagger.
- Do not draw four π electrons over six carbons. The delocalisation in the arenium ion runs over five carbons only. Drawing a circle inside the whole ring with a plus in the middle says the intermediate is aromatic, which is exactly the opposite of what makes step 1 slow.
- ‘Slow step’ is not ‘the step with the higher-energy product’. It is the step with the higher transition state. In SEAr the two happen to coincide, but the reasoning must be about TS-1.
Direct observation of the intermediate; the π complex that precedes it; the steady-state condition that decides whether an isotope effect appears; and the reactions where one does.
The arenium ion has been seen
Wheland proposed the intermediate in 1942 as a bookkeeping device. It is now a compound. The essential experimental move is to remove every base from the medium, so that step 2 has nothing to react with. In superacid media — HF/SbF5, FSO3H/SbF5 — the conjugate base is so feeble that the cation simply sits there.
- Benzene + HF/BF3 at low temperature gives a stable solution of the benzenium ion C6H7+. Its 1H NMR spectrum shows the two protons on the sp3 carbon far upfield of the ring protons, exactly as an sp3 CH2 in a non-aromatic ring should be, and the 13C spectrum shows the characteristic huge downfield shifts at the ortho and para carbons that carry the charge.
- Heptamethylbenzenium salts — from hexamethylbenzene plus a methylating agent — are crystalline. Their X-ray structures show the tetrahedral sp3 carbon and the alternating short/long bonds of a pentadienyl cation around the rest of the ring.
π complex, encounter complex, σ complex
Between free reactants and the arenium ion there is at least one shallower minimum, and distinguishing them matters for the very fast reactions in H.7.
| Species | Bonding | Geometry | Lifetime / role |
|---|---|---|---|
| Encounter pair | none — simply two molecules in the same solvent cage | unchanged | Sets the ceiling on rate. When every encounter leads to reaction the process is encounter-controlled and no substituent can make it faster |
| π complex | electrophile accepts electron density from the whole π cloud | electrophile sits above the ring plane; all six carbons still sp2 and equivalent | Weak (a few kJ mol−1), no positional selectivity, aromaticity preserved. Benzene·HCl and benzene·Br2 complexes are of this kind |
| σ complex = arenium ion | a full C–E σ bond | electrophile in the ring plane at one carbon, which is now sp3 | The real intermediate. All the positional selectivity is created on the way to this |
When does an isotope effect appear? The steady-state criterion
Write the mechanism with the second step made explicit as a base-mediated deprotonation, and apply the steady-state approximation to the arenium ion:
Two limits, and they are the whole story:
| Regime | Rate law collapses to | Isotope effect? | Base catalysis? |
|---|---|---|---|
| k2[B] >> k−1 (the arenium ion loses H+ faster than it can spit the electrophile back out) | rate = k1[ArH][E+] | None — kH/kD = 1 | None — rate independent of [B] |
| k−1 >> k2[B] (step 1 is a fast pre-equilibrium; deprotonation is the bottleneck) | rate = (k1/k−1)k2[ArH][E+][B] | Yes — a full primary effect, 2–7 | Yes — rate first order in base |
So the question ‘why is there no isotope effect?’ becomes the sharper question ‘why is k−1 so small?’ — and the answer is that for a good electrophile the C–E bond just formed is strong and the reverse step would have to expel a high-energy cation. Reversal of nitration would mean expelling NO2+ from a neutral compound; it does not happen. The exceptions are precisely the cases where the C–E bond is weak, or the electrophile is stable enough to leave again, and something slows the deprotonation down.
The three genuine exceptions
1. Sulfonation. Sulfonation of benzene shows a small but real primary isotope effect, of order kH/kD ≈ 1.5–2. Two things conspire. The arenium ion formed from SO3 is a zwitterion, ArH(SO3−)+, in which the negative sulfonate end is a perfectly good leaving group — so k−1 is large. And SO3 leaves as a stable neutral molecule rather than as a high-energy cation, which is exactly what nitration cannot do. The result is k−1 a non-negligible fraction of k2[B] — about a seventh of it, as the worked example below shows — which is exactly the intermediate regime: a partial isotope effect, well below the maximum of 7. The very same large k−1 is why sulfonation is the one classical electrophilic aromatic substitution that is reversible overall — the strategic consequence of which is H.4 in the next module.
2. Diazo coupling. An arenediazonium ion ArN2+ is a feeble electrophile — it will only attack rings as electron-rich as phenoxide or a naphthylamine. Feeble electrophile means weak new C–N bond means large k−1. Zollinger’s classic study coupled diazotised sulfanilic acid with 2-naphthol-6,8-disulfonate, where coupling occurs at C-1, flanked by the bulky 8-sulfonate group. The steric crowding slows the removal of the C-1 hydrogen, cutting k2. Both effects push the system into the pre-equilibrium regime, and the observed isotope effect climbs to kH/kD ≈ 6.5 — close to the theoretical maximum — accompanied, as the table predicts, by clean general base catalysis by pyridine. Unhindered couplings (for example with 1-naphthol) show no isotope effect at all: the exception is structural, not a property of diazo coupling as a class.
3. Sterically obstructed halogenation and some iodinations. Brominate 1,3,5-tri-tert-butylbenzene and an isotope effect appears, because the flanking tert-butyl groups make the hydrogen on the sp3 carbon very hard for a base to reach. Iodination is the other recurring case: I+-type electrophiles are weak and the C–I bond is weak, so k−1 is large and iodinations of activated substrates such as phenol and aniline commonly show kH/kD in the range 3–4 together with base catalysis.
| Reaction | Electrophile | kH/kD | Interpretation |
|---|---|---|---|
| Nitration | NO2+ | ≈ 1.0 | k2[B] >> k−1; step 1 fully rate-determining |
| Bromination (ordinary substrates) | Br2·FeBr3 | ≈ 1.0 | as above |
| Friedel–Crafts alkylation / acylation | R+ / RCO+ | ≈ 1.0 | as above |
| Sulfonation | SO3 | ≈ 1.5–2 | intermediate regime; k−1 a non-negligible fraction of k2[B] — and hence overall reversibility |
| Iodination (activated arenes) | I+ equivalent | ≈ 3–4 | weak electrophile, weak C–I bond, large k−1 |
| Diazo coupling, hindered | ArN2+ | up to ≈ 6.5 | pre-equilibrium regime; base catalysis observed |
| Bromination of 1,3,5-tri-t-Bu-benzene | Br2 | large (> 3) | deprotonation sterically obstructed, k2 small |
The sulfonation of benzene shows kH/kD = 1.7, whereas its nitration shows 1.0. A student concludes that sulfonation must go by a different mechanism. Assess this, and derive the correct conclusion quantitatively. Hard
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Sources for the arenium-ion mechanism and the isotope-effect evidence:
- Clayden, Greeves & Warren, Organic Chemistry (Oxford University Press) — the chapter on electrophilic aromatic substitution gives the arenium-ion picture and the aromatic-stabilisation argument.
Read the rest of Part 8
The remaining 8 sections of this part — Making the electrophile: nitration, halogenation, sulfonation and Friedel–Crafts, Activating and deactivating groups — and why the halogens break the pattern, Ortho:para ratios, competing… — and all nine parts of Reaction Mechanisms are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.
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