Benzene’s six p electrons are spread into a single delocalised ring, and that delocalisation makes the molecule far more stable than you would predict from three separate double bonds. That stability sets everything else: benzene guards its ring, reacting by substitution (which keeps the ring intact) rather than the addition you saw with alkenes.
Bonding & stability of benzene
Benzene, C6H6, is a planar, regular hexagon. Each carbon is bonded to two neighbouring carbons and one hydrogen, with bond angles of 120°. Each carbon uses three of its electrons in these σ bonds and contributes its fourth electron, in a p orbital, to a shared system. Those six p orbitals overlap sideways, above and below the plane of the ring, to form a ring of delocalised π electrons.
- All six C–C bonds are the same length (about 0.139 nm) — between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm). The Kekéule “alternating single and double bonds” model predicts two different lengths, which is not what is observed.
- Benzene resists addition reactions that would break up the ring, and does not decolourise bromine water — unlike a real alkene.
The thermochemical evidence — enthalpy of hydrogenation
The clinching evidence is energetic. Hydrogenating one C=C in cyclohexene releases about −120 kJ mol−1. If benzene were really cyclohexa-1,3,5-triene — three isolated double bonds — you would expect three times this, about −360 kJ mol−1. The value actually measured for benzene is only about −208 kJ mol−1.
Benzene releases less energy on hydrogenation, so it must have started lower in energy — it is roughly 152 kJ mol−1 more stable than the hypothetical triene. That extra stability is the delocalisation (resonance) energy.
Represent benzene as a hexagon with a complete circle inside it — the circle stands for the six delocalised π electrons shared over all six carbons. (Keep this picture in mind: the reaction intermediate later is drawn very differently, and mixing the two up is a classic lost mark.)
🧪 Exam-style questions
Compare benzene and the theoretical molecule cyclohexa-1,3,5-triene in terms of stability, shape and carbon–carbon bond lengths, suggesting reasons for any differences. Use enthalpy-of-hydrogenation data in your answer.
Show answer
Benzene is more stable than cyclohexa-1,3,5-triene. 1 mark
Its enthalpy of hydrogenation is about 152 kJ mol−1 less exothermic than the value predicted for the triene (−208 vs −360). 1 mark
This is due to the delocalisation of the π electrons in benzene. 1 mark
Both are planar / hexagonal. 1 mark
Benzene has equal C–C bond lengths (a regular hexagon), whereas the triene would have alternating short (C=C) and long (C–C) bonds / a distorted ring. 1 mark
Source: AQA A-Level Chemistry past papers.
Electrophilic substitution
Because the delocalised ring is so stable, benzene will not give it up. An addition reaction would use up the delocalised electrons and destroy the stabilisation; a substitution replaces just one hydrogen and keeps the ring intact. So the ring, rich in π electrons, attracts electrophiles and reacts by electrophilic substitution.
Addition would break the delocalised system and lose the ~152 kJ mol−1 of stability. Substitution regenerates the aromatic ring, so it is strongly favoured. Every reaction below follows the same shape: make an electrophile, attack the ring, then lose H+ to restore aromaticity.
Nitration
Benzene is nitrated by a mixture of concentrated nitric acid and concentrated sulfuric acid (warmed to about 50 °C). The sulfuric acid is a catalyst that generates the electrophile, the nitronium ion NO2+:
HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−
The nitronium ion then substitutes onto the ring. Afterwards the H+ released reacts with HSO4− to regenerate the H2SO4 catalyst. Nitration matters because it is the first step in making explosives (e.g. TNT) and, by later reduction, aromatic amines and dyes.
Friedel–Crafts acylation
Acylation attaches an acyl group (RCO–) to the ring, making an aromatic ketone — a valuable synthetic step. The electrophile is an acylium ion, generated from an acyl chloride using an aluminium chloride (AlCl3) catalyst:
RCOCl + AlCl3 → RCO+ + AlCl4−
The acylium ion substitutes onto the ring by exactly the same mechanism as nitration. At the end, the H+ released reacts with AlCl4− to regenerate the AlCl3 catalyst and give off HCl. For example, ethanoyl chloride gives phenyl methyl ketone (acetophenone): C6H6 + CH3COCl → C6H5COCH3 + HCl.
The horseshoe, drawn right
The step that costs the most marks is the intermediate. When the electrophile bonds to a carbon of the ring — call it C1 — that carbon becomes sp3: it now holds the electrophile and a hydrogen, and it drops out of the delocalised system. That leaves a positive charge spread over the other five carbons, which we draw as a horseshoe — a broken circle, not a full one.
- The intermediate is a horseshoe, not a complete circle. A full circle is the commonest error — it wrongly implies all six carbons still share the delocalised electrons, but C1 is now sp3 and out of the ring.
- The horseshoe is centred on C1 and must not extend beyond C2 and C6 (it can be drawn smaller, but never wrapping back round past C1).
- The + charge sits inside the horseshoe, not too close to C1 — place it on, or just below, a line joining C2 to C6.
- Restore aromaticity with a curly arrow from the C1–H bond back into the ring; the H+ then leaves.
Benzene itself → hexagon with a complete circle (six delocalised electrons over six carbons). The reaction intermediate → a horseshoe over five carbons plus a “+” (one carbon is now sp3). Same-looking rings, completely different meanings.
🧪 Exam-style questions
Which species is the electrophile in the nitration of benzene?
A mixture of concentrated nitric acid and concentrated sulfuric acid reacts with benzene. Name the mechanism, give an equation for the formation of the electrophile, and describe the two steps of the reaction of that electrophile with benzene.
Show answer
Mechanism: electrophilic substitution. 1 mark
Electrophile: HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−. 1 mark
Step 1: a pair of electrons from the ring attacks the N of NO2+, forming the positively charged horseshoe intermediate (electrophile on C1). Step 2: a curly arrow from the C1–H bond restores the ring, releasing H+. 1 mark
CH3CH2COCl reacts with benzene in the presence of AlCl3. Give an equation for the reaction of CH3CH2COCl with AlCl3 to form the electrophile, and outline the mechanism of the reaction of this electrophile with benzene.
Show answer
CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4−. 1 mark
A pair of electrons from the ring attacks the C of the acylium ion CH3CH2CO+ (arrow from inside the hexagon to the carbon). 1 mark
This gives the horseshoe intermediate: + charge centred on C1, not extending beyond C2/C6. 1 mark
A curly arrow from the C1–H bond back into the ring restores aromaticity and releases H+ (which reacts with AlCl4− to regenerate AlCl3 and form HCl). 1 mark
Source: AQA A-Level Chemistry past papers.
Benzene in synthesis
Nitration is the gateway to the whole aromatic-nitrogen family. Nitrate benzene to nitrobenzene, then reduce it (tin and concentrated hydrochloric acid, then NaOH) to phenylamine — the aromatic amine used to manufacture dyes. That reduction is the first thing you meet in the next topic.
C6H6 → nitration C6H5NO2 → reduction C6H5NH2
- Drawing the intermediate as a full circle instead of a horseshoe, or letting the horseshoe wrap past C1.
- Writing the nitronium generation unbalanced — it needs two H2SO4: HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−.
- Forgetting the second step: the arrow from the C–H bond that restores the aromatic ring.
- Calling the reaction “addition” — aromatic rings react by substitution to preserve the delocalised system.
Examiner reports return to the intermediate again and again: the horseshoe must be centred on C1 and stop at C2 and C6, with the positive charge placed away from C1. Candidates who draw a neat, correctly bounded horseshoe pick up the structure mark that many others lose.
- Benzene: planar regular hexagon, all C–C equal (0.139 nm, between single and double), six delocalised p electrons. Draw it as a hexagon with a complete circle.
- Stability: enthalpy of hydrogenation (−208) is ~152 kJ mol−1 less exothermic than the −360 predicted for cyclohexa-1,3,5-triene — the delocalisation energy. Benzene therefore reacts by substitution, not addition.
- Nitration: electrophile NO2+, generated by HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−. Uses: explosives, and making amines/dyes.
- Friedel–Crafts acylation: RCOCl + AlCl3 → RCO+ + AlCl4−; the acylium ion substitutes onto the ring to make an aromatic ketone.
- The intermediate is a horseshoe over five carbons (electrophile on C1, which becomes sp3) — not a full circle. Centre it on C1, do not extend beyond C2/C6, and restore aromaticity with an arrow from a C–H bond back into the ring.