Why benzene reacts by substitution, not addition
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describe the mechanism of electrophilic substitution in arenes, with regards to the effect of delocalisation (aromatic stabilisation) of electrons in arenes to explain the predomination of substitution over addition (syllabus 30, learning outcome 2)
From bonding picture to chemical behaviour
The previous note gave you benzene's bonding: six sp² carbons, a flat hexagon, six π electrons delocalised around the ring — plus the rule that follows from it: benzene reacts by substitution, never addition. This section turns that rule into an argument you can reproduce under exam conditions:
- the deposit-and-refund picture of what the delocalised π system contributes while a reaction runs;
- the exact two-sided sentence mark schemes reward when they ask why addition does not occur;
- and the contrast with alkene chemistry that §06 later turns into a conditions fork.
Everything here is one short step beyond the bonding you already know — and these are the steps examiners pay for.
One idea drives the whole topic. The delocalised π system IS benzene's stability — think of it as a large energy deposit built into the molecule.
- Stage 1 of the mechanism spends it: the ring donates a π pair to the electrophile , leaving only four π electrons over five carbons (the arenium intermediate).
- Stage 3 refunds it: the C–H pair pushes back into the ring as leaves, restoring all six delocalised electrons. That refund only happens because a hydrogen LEFT — substitution, not addition.
- A true addition (E and H both kept) would leave a cyclohexadiene: two ordinary localised double bonds, deposit never refunded. Benzene declines any deal that costs it the delocalisation energy.
An alkene accepts the same electrophile happily because a localised π bond has almost nothing to lose — remember this contrast, it returns in §06.
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Say what addition would produce: BOTH atoms of the reagent kept, giving a product with only localised (ordinary) double bonds — a cyclohexadiene-type ring.
Naming the addition product's bonding sets up the comparison the marker is looking for.
- 2
State the key fact: the SUBSTITUTION product is stabilised by delocalisation of its π electrons — the aromatic system survives.
This clause appears in mark schemes nearly word-for-word (9701/42 M/J 2023 Q5(b)); it is the mark.
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Give the other half: the ADDITION product is not stabilised by delocalisation — addition removes the delocalised π system.
'Either/or' credit exists, but writing both halves guarantees the mark regardless of which wording the scheme leads with.
Worked demo — two exits from the intermediate
An electrophile has attacked a benzene ring, giving the arenium ion in which E and H are both bonded to one carbon. Two routes now lie open: loss of to give a substituted ring, or gain of an anion to give an addition product carrying both E and X on the ring. Predict which route dominates, and explain your answer.
Show full working
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Route 1 — lose : the C–H electron pair falls back into the ring, restoring the full six-electron delocalised π system in the substituted product.
Deprotonation refunds the deposit: aromaticity returns exactly as it was.
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Route 2 — add : the ring keeps both new groups but retains only two localised double bonds — the delocalised π system is destroyed for good.
Addition spends the deposit and never gets it back; what is left is an ordinary cyclohexadiene.
- 3
Prediction: substitution dominates. The substituted product retains the stabilisation that the addition product throws away, so losing is the thermodynamic route.
The comparison IS the explanation mark: retained versus destroyed delocalisation, stated in one sentence.
Substitution (loss of H⁺): the delocalised π system is restored; an addition product would keep only localised double bonds
Every 'why substitution, not addition' question is this comparison wearing different words.
Why no addition with an electrophile
When methylbenzene reacts with an electrophile, a substitution reaction occurs. No addition reaction takes place under these conditions. Explain why no addition reaction takes place.
Show full working
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Consider what addition would mean: the electrophile AND a hydrogen both stay on the ring, so the ring is left with only localised double bonds — the delocalised π system is gone.
Set up both products before comparing them; the marker wants the contrast, not a slogan.
- 2
The substitution product is stabilised by delocalisation of its π electrons — aromaticity is retained.
Mark scheme wording, first credited half.
- 3
Equivalently: the addition product is not stabilised by delocalisation — forming it removes the delocalised π system entirely.
The scheme allows EITHER statement; quoting both makes the mark unloseable.
substitution product keeps (is stabilised by) the delocalised π system; addition product would lose that delocalisation — so addition does not occur
Treating addition to benzene as merely SLOW — possible in principle with a suitable catalyst
Addition does not compete at all: it would destroy the delocalised π system, while substitution restores it.
The distinction is thermodynamic (which product keeps the stabilisation), never kinetic (how fast).
Writing 'addition does not happen because benzene is unreactive'
Name the thermodynamics: the substitution product retains the delocalised π stabilisation; addition would destroy it.
'Unreactive' is a conclusion, not an explanation — mark schemes want the delocalisation argument in either direction.
Your turn
Two applications of the same argument on fresh ground — no past-paper crutch this time.
- 13 marks
Benzene does not decolourise bromine water at room temperature. In the presence of , however, benzene reacts readily with bromine to give bromobenzene and HBr.
Explain both observations.Stuck? Show hint
One explanation covers both halves: how reactive Br⁺ is compared with unaided Br₂, and which product type the ring will accept.
Show solution
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Bromine water offers no strong electrophile: unpolarised cannot draw a pair out of the stable delocalised ring, so nothing visible happens.
Contrast with an alkene, whose localised π pair polarises instantly — AS knowledge doing work here.
- 2
With the situation changes: the carrier generates genuine (), an electrophile strong enough to attack the ring.
§02 develops this generation equation as a full-mark mechanism answer.
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Even then the attack ends in SUBSTITUTION: losing restores the delocalised π system, whereas adding both bromine atoms would destroy it.
'Retains/is stabilised by the delocalised system' is the sentence mark schemes print nearly verbatim.
Answerno strong enough electrophile in Br₂(aq) · AlBr₃ generates Br⁺ · reaction proceeds by substitution so the delocalised π system survives
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- 23 marks
Phenol reacts instantly with bromine water — far faster than benzene — yet the organic products are still SUBSTITUTION compounds (2-, 4- and 6-bromophenol), never addition products.
Explain both facts.Stuck? Show hint
Speed comes from electron density; substitution-versus-addition comes from somewhere else entirely.
Show solution
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Faster: the oxygen lone pair delocalises INTO the ring, raising its electron density, so the ring attacks electrophiles far more readily — no carrier is needed.
Activation explains RATE only; it says nothing about which product type forms.
- 2
Still substitution: once the electrophile has attacked, the same deposit-and-refund rule applies — loss of restores the delocalised π system, addition would annihilate it.
The thermodynamic preference for retaining aromaticity is unchanged by activation.
- 3
So rate and product type are independent levers: activation speeds the reaction up; aromaticity decides the reaction remains substitution.
This separation is exactly what examiners probe when 'faster BUT still substitution' appears in one question.
AnswerO lone-pair donation raises ring electron density → faster; substitution persists because only it restores the delocalised π system
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The rest of this note
Can you do all of these?
I can explain why benzene substitutes rather than adds using retention versus destruction of the delocalised π system
I can write the electrophile-generation equation for halogenation, nitration (both accepted forms) and both Friedel–Crafts reactions
I can write each catalyst-regeneration equation: H⁺ + FeBr₄⁻ and HSO₄⁻ + H⁺
I can draw the complete EAS mechanism with the first arrow from inside the hexagon and the second from the C–H bond into the ring
I can state conditions for each named reaction, including conc/conc at 25–60 °C for mono-nitration and dry solvent for Friedel–Crafts
I can rank phenol > benzene > benzoic acid for electrophilic substitution and justify it with lone-pair donation versus withdrawal
I can predict FC products, including rearrangement of primary carbocations and the polyalkylation-versus-polyacylation contrast
I can balance side-chain oxidations with [O], including carbons lost as CO₂, and state H₂/Pt-Ni heat conditions for ring hydrogenation
I can predict ring versus side-chain halogenation from the conditions, and give the AgNO₃ observations for a haloarene versus a benzylic halide
I can classify every director as 2,4- or 3-directing with its electronic reason, and name the minor isomer when asked
I can plan a multi-step benzene synthesis in the correct ORDER of steps and justify the order through directing effects