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 Kekulé “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.
Use the data to show that benzene is more stable than the hypothetical cyclohexa-1,3,5-triene, and find the size of the stabilisation.
Step 1. Predict the triene’s value from cyclohexene:
3 × (−120) = −360 kJ mol−1 expected for three isolated C=C
Step 2. Compare with what benzene actually releases:
−208 kJ mol−1 measured
Step 3. The difference is the extra stability:
−360 − (−208) = −152, so benzene is 152 kJ mol−1 more stable
The sentence to write: benzene’s enthalpy of hydrogenation is 152 kJ mol−1 less exothermic than expected for three isolated double bonds, because the π electrons are delocalised: benzene starts lower in energy, so it is more stable than the hypothetical triene.
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 costs the structure mark.)
- Examiner reports note a low mean mark on the extended-response comparison of benzene with cyclohexa-1,3,5-triene. The valid comparison points are exactly four: equal C–C bond lengths, both planar, hydrogenation less exothermic than expected, and delocalised electrons.
- The stability mark comes from delocalisation, named as such. A generic “resonance makes it stable” explains nothing and scores nothing.
- Quote the numbers with their signs and the direction of the comparison: −208 kJ mol−1 is less exothermic than −360 kJ mol−1, so benzene starts lower in energy.
Exam questions: real AQA past-paper questions
The figure below shows enthalpy of hydrogenation data for cyclohexene and benzene. It also shows predicted data for the theoretical molecule cyclohexa-1,3,5-triene.
Compare benzene and the theoretical molecule cyclohexa-1,3,5-triene in terms of:
- stability
- shape
- carbon–carbon bond lengths.
For each of these properties, suggest reasons for any differences. Use data from the figure above in your answer.
Show answer
M1: benzene is more stable than cyclohexatriene. 1 mark
M2: the enthalpy of hydrogenation of benzene is (152 kJ mol−1) less / less exothermic. 1 mark
M3: due to the delocalisation of electrons in benzene. 1 mark
M4: both are planar / hexagonal. 1 mark
M5: benzene has equal C–C bond lengths (a regular hexagon) whereas cyclohexa-1,3,5-triene has bonds of different / varied length (the hexagon is distorted / irregular). 1 mark
M4 and M5 could be shown in a clear diagram.
Data about the hydrogenation of cyclohexene and of benzene are given.
The enthalpy of hydrogenation of cyclohexa-1,3-diene is not exactly double that of cyclohexene. Suggest a value for the enthalpy of hydrogenation of cyclohexa-1,3-diene and justify your value.
Show answer
Any value within the range −239 to −121 kJ mol−1 (i.e. slightly less exothermic than 2 × (−120), for example −230 kJ mol−1). 1 mark
The two double bonds are separated by one single bond / alternating. 1 mark
This allows some delocalisation / overlap of p orbitals. 1 mark
A value outside the range (including any positive value) scores zero for the whole part. The wording “close enough to allow delocalisation” would score the second and third marks together.
Kekulé suggested a structure for benzene with alternating double and single bonds. Benzene is now represented by a hexagon with a circle. Figure 1 shows the relative stability of the Kekulé structure compared with benzene.
Use Figure 1 and the data shown in the table below to calculate ΔH2.
| ΔH / kJ mol−1 | |
|---|---|
| Enthalpy of atomisation for carbon | +715 |
| Enthalpy of atomisation for hydrogen | +218 |
| Bond enthalpy (C–C) | +348 |
| Bond enthalpy (C=C) | +612 |
| Bond enthalpy (C–H) | +412 |
Show answer
M1: bonds in the Kekulé structure (broken to reach the atoms):
(3 × 612) + (3 × 348) + (6 × 412) = 5352 1 mark
M2: atomising the elements up to the same atoms:
(6 × 715) + (6 × 218) = 5598 1 mark
M3. Close the cycle:
ΔH2 = 5598 − 5352 − 83 = +163 kJ mol−1 1 mark
+163 kJ mol−1 is a bond-enthalpy estimate of the same stability gap the hydrogenation data gave (152 kJ mol−1): two independent routes to the delocalisation energy.
Explain, in terms of structure and bonding, why benzene is more thermodynamically stable than the Kekulé structure.
Show answer
The (π) electrons are delocalised. 1 mark
Source: AQA A-Level Chemistry past papers.
Naming aromatic compounds
Almost every aromatic compound is named as a derivative of benzene, so benzene is the root of the name. The systematic rules you met at AS still apply: number the ring to give the lowest locants and list substituents in alphabetical order (see Naming: the IUPAC rules). Two patterns cover nearly everything at A-level.
- The ring as the root. A single low-priority group (alkyl, halogeno or nitro) becomes a prefix on benzene: methylbenzene, chlorobenzene, nitrobenzene.
- The ring as a substituent: phenyl, C6H5–. When a C=C, a carbonyl, an amine or a longer carbon chain is the parent, the ring hangs off it as a phenyl group: phenylethene, phenylethanone, phenylamine, 2-phenylbutane.
- A few keep special names you simply learn: phenol (C6H5OH), benzaldehyde (C6H5CHO) and benzenecarboxylic acid / benzoic acid (C6H5COOH).
The nitration you may carry out in the lab is the preparation of methyl 3-nitrobenzoate from methyl benzoate. The nitro group enters at position 3 because the ester group already on the ring directs it there. The crude solid is then purified by recrystallisation and checked by measuring its melting point: sharp and at the book value if it is pure, lower and over a range if it is not. The technique is the same one set out for Required practical 10.
With more than one substituent, number the ring so the substituents get the lowest set of locants, then cite them alphabetically: 1,3-dimethylbenzene. The classic synoptic example is TNT, made by nitrating methylbenzene three times (2-methyl-1,3,5-trinitrobenzene), which is exactly why nitration matters for explosives.
methylbenzene
(toluene)
chlorobenzene
nitrobenzene
benzenecarboxylic acid
(benzoic acid)
phenylamine
(aniline)
phenol
phenylethanone
(acetophenone)
phenylethene
(styrene)
- Answer with the systematic name (methylbenzene, phenylamine, phenylethene), but recognise the traditional names, because AQA drops toluene, aniline and styrene into question stems.
- phenol, benzaldehyde and benzenecarboxylic acid (benzoic acid) are the accepted names in their own right. There is no “benzenol”.
- The Friedel–Crafts product C6H5COCH3 is phenylethanone (the phenyl is the substituent, ethanone the parent), a commonly mis-named structure.
Quick check: can you name these?
A quick self-test (our own, not an exam question). Name each, then reveal the answers.
- The product of nitrating benzene, C6H5NO2.
- The Friedel–Crafts acylation product C6H5COCH3.
- The systematic name for aniline.
- A benzene ring with a methyl group on C1 and a methyl group on C3.
Show answers
- nitrobenzene: the ring is the root, nitro is the prefix.
- phenylethanone: phenyl substituent on the ethanone parent.
- phenylamine: the amine is the parent, phenyl the substituent.
- 1,3-dimethylbenzene (traditional name m-xylene): lowest locants 1 and 3.
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.
Benzene is warmed with concentrated nitric acid and concentrated sulfuric acid. Write an equation for the formation of the electrophile, the overall equation, and name the organic product.
Step 1. Generate the electrophile (the H2SO4 protonates the HNO3):
HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−
Step 2. The overall equation swaps one ring H for NO2 (H2SO4 is a catalyst, so it does not appear):
C6H6 + HNO3 → C6H5NO2 + H2O
Step 3. Name the product: nitrobenzene.
- Examiner reports found it disappointing how few students could write the overall equation for a familiar electrophilic substitution and name the product. More scored on the catalyst and mechanism. Learn the two reagents, the electrophile-generation equation and the mechanism as three separate items, and practise the overall equation too.
- Concentrated matters: the reagents are concentrated HNO3 and concentrated H2SO4. The mark scheme docks a mark when “concentrated” is missing.
- The generation equation must balance with 2H2SO4 (the one-H2SO4 version HNO3 + H2SO4 → NO2+ + H2O + HSO4− is also accepted, but nothing unbalanced is).
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 under dry (anhydrous) conditions, because both the acyl chloride and the AlCl3 react with water:
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.
Benzene reacts with ethanoyl chloride in the presence of AlCl3. Write an equation for the formation of the electrophile, the overall equation, and name the organic product.
Step 1. Generate the electrophile (AlCl3 pulls the Cl off as AlCl4−):
CH3COCl + AlCl3 → CH3CO+ + AlCl4−
Step 2. The overall equation swaps one ring H for the acyl group:
C6H6 + CH3COCl → C6H5COCH3 + HCl
Step 3. Name the product: phenylethanone, an aromatic ketone.
Step 4. Account for the catalyst: the H+ released reacts with AlCl4− to regenerate AlCl3 and form the HCl in the overall equation.
Interactive: one mechanism, two electrophiles
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 now has four single bonds: it 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.
AQA’s mark schemes state the geometry of the intermediate explicitly. The four rules:
- The intermediate is a horseshoe, not a complete circle. A full circle wrongly implies all six carbons still share the delocalised electrons, but C1 now has four single bonds and has left the delocalised ring.
- The horseshoe is centred on C1 and must not extend beyond C2 and C6 (it can be drawn smaller, but never wrapping round past C1).
- The + charge sits in the centre of the horseshoe, well away from C1. (The mark scheme allows it on or inside a line joining C2 to C6; the centre is the unambiguous place.) A + written on the H loses the structure mark.
- 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 now has four single bonds). Same-looking rings, completely different meanings.
Exam questions: real AQA past-paper questions
CH3CH2COCl reacts with benzene in the presence of AlCl3 in an electrophilic substitution reaction. Give an equation for the reaction of CH3CH2COCl with AlCl3 to form the electrophile. Outline a mechanism for the reaction of this electrophile with benzene.
Show answer
M1. The equation:
CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4− 1 mark
The + may sit on the C or the O in the equation, but it must be on the C in the mechanism.
M2: arrow from inside the hexagon to the C (or to a + on the C). 1 mark
M3. Structure of the intermediate: horseshoe centred on C1, not extending beyond C2 and C6 (it can be smaller); the + not too close to C1 (on or below a line from C2 to C6). 1 mark
M4: arrow from the C–H bond into the hexagon. 1 mark
A + written on the H in the intermediate loses M3, not M4. Cl− or AlCl4− shown removing the H+ is ignored.
Benzene reacts with methanoyl chloride (HCOCl) in the presence of a catalyst. Give an equation for the overall reaction when benzene reacts with methanoyl chloride. Name the organic product.
Show answer
C6H6 + HCOCl → C6H5CHO + HCl 1 mark
The product is benzaldehyde. 1 mark
“Phenyl methanal” and “benzenecarbaldehyde” are allowed; the equation may also be given with structural formulae.
Identify the catalyst needed in this reaction. Give an equation to show how the catalyst is used to form the electrophile, [HCO]+.
Show answer
Catalyst: AlCl3. 1 mark
HCOCl + AlCl3 → [HCO]+ + [AlCl4]− 1 mark
Iron(III) chloride or bromide is also allowed as the catalyst; the + may sit on the C or the O in the equation.
Outline the mechanism for the reaction of benzene with the electrophile, [HCO]+.
Show answer
M1: arrow from inside the hexagon to the C (or to a + on the C). 1 mark
M2. Structure of the intermediate: horseshoe centred on C1, not beyond C2 and C6; the + not too close to C1. 1 mark
M3: arrow from the C–H bond into the hexagon. 1 mark
Figure 2 shows an intermediate formed in the first step of a reaction mechanism of methylbenzene.
Complete Figure 2 to show the reactant species and any curly arrows involved in the formation of the intermediate. Draw a curly arrow on the intermediate to show how the product is formed. Give the name of the reaction mechanism.
Show answer
M1: structures of the reactant species, including the + on the N of +NO2. 1 mark
M2: arrow from the ring (inside the hexagon) to the N (or to the + on the N). 1 mark
M3: arrow from the C–H bond into the hexagon. 1 mark
M4: electrophilic substitution. 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 HNO3/H2SO4 C6H5NO2 Sn/HCl C6H5NH2
- Drawing the intermediate as a full circle instead of a horseshoe, or letting the horseshoe wrap past C1, loses the structure mark.
- Writing the nitronium generation unbalanced. The two-acid version 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 flag multi-step synthesis as a repeated weakness: students recall single reactions but struggle to move through a route, naming mechanisms and conditions at each step. Practise the whole benzene → nitrobenzene → phenylamine chain, not the steps in isolation.
Exam questions: the route, as AQA asked it
Two steps in the synthesis of an aromatic amine are shown.
State the two reagents needed for Step 1. Give an equation to show the formation of the reactive intermediate from these two reagents.
Show answer
M1: concentrated nitric acid AND concentrated sulfuric acid. 1 mark
M2. The equation:
HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4− 1 mark
If “concentrated” is missing from both acids, only one mark. Also accepted: HNO3 + H2SO4 → H2NO3+ + HSO4− then H2NO3+ → NO2+ + H2O, or HNO3 + H2SO4 → H2O + NO2+ + HSO4−.
Outline a mechanism for Step 1.
Show answer
M1: the + must be on the N, and the arrow runs from inside the hexagon to the N (or to the + on the N). 1 mark
M2: the intermediate shows the horseshoe and the positive charge: horseshoe centred on C1, not extending beyond C2 and C6; + not too close to C1. 1 mark
M3: arrow from the C–H bond back into the hexagon. 1 mark
State the reagent(s) needed for Step 2.
Show answer
Sn / HCl (tin and hydrochloric acid). 1 mark
H2 with Pt or Ni, and HCl with Fe, are allowed. NaOH used after the Sn/HCl is ignored, but NaOH written in at the same time as the Sn/HCl is penalised.
State a possible use for the amine formed in Step 2.
Show answer
Manufacture of dyes (or cationic surfactants / fabric softener). 1 mark
Source: AQA A-Level Chemistry past papers.
Capstone quiz: six past-paper questions
Six real AQA multiple-choice questions on benzene and its chemistry, in the style that opens Paper 3. Pick one answer each. The reasoning appears once you commit.
Which substance does not have any bond angles of 120°?
Which substance has no delocalised electrons?
In which molecule are all the atoms in the same plane?
What is the relative molecular mass (Mr) of benzene-1,4-dicarboxylic acid?
The reaction between propanoyl chloride and benzene is an example of acylation. Which is a correct representation of part of the mechanism of this reaction?
Which type of reaction is not involved in this reaction sequence?
Source: AQA A-Level Chemistry past papers.
The nitrobenzene → phenylamine reduction you met above is the opening move of amines (3.3.11), where the base strength and reactions of aromatic amines get their own page.
- 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− (dry conditions); the acylium ion substitutes onto the ring to make an aromatic ketone (the exception: HCOCl gives the aldehyde, benzaldehyde). The H+ released regenerates the catalyst: H+ + AlCl4− → AlCl3 + HCl.
- The intermediate is a horseshoe over five carbons (electrophile on C1, which now has four single bonds), 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.