AQA A-Level Chemistry: Aromatic chemistry (7405)

This guide teaches section 3.3.10 Aromatic chemistry of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards. It covers both sub-sections: 3.3.10.1 Bonding and 3.3.10.2 Electrophilic substitution. The whole section is A-level only, so it is not part of the AS course. The specification places all of section 3.3 on A-level Paper 2, and A-level Paper 3 can test any content.

For quick recall, use the aromatic chemistry revision notes. To test yourself, try the aromatic chemistry practice questions. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are. This topic builds on electrophilic addition in alkenes, enthalpy changes in energetics, and naming and mechanism conventions in introduction to organic chemistry.

What this topic covers

Spec section	What you must be able to do	Status
3.3.10.1	Describe the bonding in a benzene ring: planar structure, bond length between single and double	A-level only
3.3.10.1	Explain that delocalisation of p electrons makes benzene more stable than cyclohexa-1,3,5-triene	A-level only
3.3.10.1	Use enthalpies of hydrogenation to account for this extra stability	A-level only
3.3.10.1	Explain why benzene undergoes substitution rather than addition	A-level only
3.3.10.2	State that electrophilic attack on benzene gives substitution (monosubstitution only)	A-level only
3.3.10.2	State that nitration is an important step in synthesis, including explosives and amines	A-level only
3.3.10.2	State that Friedel–Crafts acylation is an important step in synthesis	A-level only
3.3.10.2	Outline the mechanism of nitration, including how the nitronium ion forms	A-level only
3.3.10.2	Outline the mechanism of acylation using AlCl₃ as a catalyst	A-level only

Section 3.3.10 lists no required practical of its own. As a skills opportunity, the specification suggests preparing methyl 3-nitrobenzoate by nitrating methyl benzoate, then purifying it by recrystallisation and measuring its melting point. That uses the same techniques as Required practical 10 (preparation of a pure organic solid and a test of its purity), which the specification lists under 3.3.9.2.

The bonding in benzene (3.3.10.1)

Benzene has the formula C₆H₆. Its six carbon atoms form a regular hexagon, and each carbon is bonded to one hydrogen atom.

Planar ring, equal bonds

Each carbon atom uses three of its outer electrons to form single (σ) bonds: two to neighbouring carbons and one to hydrogen. Three bonding pairs repel equally, so the bond angles are 120° and the whole molecule is planar.

That leaves each carbon with one electron in a p orbital, at right angles to the plane of the ring. The six p orbitals overlap sideways with each other all the way round. The six p electrons are not fixed between particular pairs of atoms. They are delocalised: spread over all six carbons, in a ring of electron density above and below the plane.

Because every carbon–carbon bond shares the same delocalised electrons, all six bonds are the same length. That length is intermediate between a C–C single bond and a C=C double bond. This is why benzene is drawn as a hexagon with a circle inside, not with three separate double bonds.

The theoretical model: cyclohexa-1,3,5-triene

A model with three localised C=C bonds alternating with three C–C bonds describes cyclohexa-1,3,5-triene, which would have two different bond lengths. It is a useful theoretical comparison, but not the real structure.

Thermochemical evidence: enthalpies of hydrogenation

Hydrogenation adds H₂ across C=C bonds. When cyclohexene (one C=C) is hydrogenated to cyclohexane, a fixed amount of energy is released per C=C bond. If benzene were cyclohexa-1,3,5-triene, you would expect three times that amount when it is hydrogenated to cyclohexane.

The measured value for benzene is much less exothermic than the prediction. Less energy is released because benzene starts lower in energy. The difference is the extra stability that comes from delocalisation.

Worked example 1. You are given these enthalpies of hydrogenation:

• cyclohexene + H₂ → cyclohexane, ΔH = −120 kJ mol⁻¹
• benzene + 3H₂ → cyclohexane, ΔH = −208 kJ mol⁻¹

Use the data to show that benzene is more stable than cyclohexa-1,3,5-triene.

1. The triene has three C=C bonds, so its expected value is 3 × (−120) = −360 kJ mol⁻¹.
2. The measured value for benzene is −208 kJ mol⁻¹.
3. Difference: −208 − (−360) = +152 kJ mol⁻¹. Benzene releases 152 kJ mol⁻¹ less energy than predicted.
4. Both reactions give the same product, cyclohexane. So benzene must be 152 kJ mol⁻¹ lower in energy (more stable) than the triene would be.
5. The reason is delocalisation of the p electrons in benzene.

On an energy level diagram, cyclohexane is at the bottom, benzene 208 kJ mol⁻¹ above it and the triene 360 kJ mol⁻¹ above it. The 152 kJ mol⁻¹ gap is the extra stability.

Why substitution, not addition

Addition across a ring bond would break up the delocalised system. The product would lose the extra stability shown above, so addition is energetically unfavourable. In substitution, one hydrogen atom is replaced and the delocalised ring is kept intact. So benzene undergoes substitution reactions in preference to addition reactions.

In an alkene, the electron density of the C=C bond is concentrated between two carbons. In benzene it is spread over six. So benzene needs a strong electrophile, usually a positive ion made in the reaction mixture.

Electrophilic substitution (3.3.10.2)

Both reactions you need follow the same three-stage pattern. The specification limits this to monosubstitution: one hydrogen on the ring is replaced.

1. Make the electrophile (a positive ion) from the reagents.
2. Attack: a curly arrow goes from the delocalised ring to the electrophile. The ring’s delocalisation is partly broken and a positive intermediate forms.
3. Lose H⁺: a curly arrow goes from the C–H bond back into the ring. Delocalisation is restored and H⁺ is released.

Drawing the mechanism

• Arrow 1 starts on or inside the circle of the ring and points to the atom of the electrophile that forms the new bond (N in NO₂⁺, C in RCO⁺).
• The intermediate: the carbon under attack now carries both H and the new group. Draw a broken horseshoe (partial circle) over the other five carbons, with a + inside it. The gap in the horseshoe faces the carbon under attack.
• Arrow 2 starts on the C–H bond of that carbon and points into the ring.
• Product: a full circle again, with the new group in place of H.

Nitration

Reagents: concentrated nitric acid and concentrated sulfuric acid.

Making the electrophile. Sulfuric acid is the stronger acid. It protonates nitric acid, which then loses water to give the nitronium ion, NO₂⁺:

HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻

An equivalent form is HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.

Mechanism. Arrow from the ring to the N of NO₂⁺; horseshoe intermediate with + inside, the attacked carbon carrying H and NO₂; arrow from the C–H bond into the ring. The product is nitrobenzene, C₆H₅NO₂.

Overall equation: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O.

Regenerating the catalyst. The H⁺ lost from the ring combines with HSO₄⁻: H⁺ + HSO₄⁻ → H₂SO₄. Sulfuric acid is used up in step 1 and re-formed here, so it acts as a catalyst.

Why nitration matters. It is an important step in synthesis.

• Explosives: compounds with several nitro groups on an aromatic ring, such as TNT (2,4,6-trinitromethylbenzene), are explosives. You only need to show monosubstitution.
• Amines: a nitro compound can be reduced to an aromatic amine, such as phenylamine from nitrobenzene. Aromatic amines are used to make dyes. That reduction belongs to section 3.3.11.1 Amines.

Friedel–Crafts acylation

Reagents: an acyl chloride, RCOCl, with aluminium chloride, AlCl₃, as the catalyst. Conditions must be dry, because aluminium chloride reacts with water.

Making the electrophile. AlCl₃ accepts a chloride ion from the acyl chloride, forming an acylium ion, RCO⁺:

RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻

For ethanoyl chloride: CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻.

Mechanism. Arrow from the ring to the carbonyl carbon of RCO⁺ (the carbon with the + charge); horseshoe intermediate with + inside, the attacked carbon carrying H and COR; arrow from the C–H bond into the ring. The product is a phenyl ketone, C₆H₅COR.

Regenerating the catalyst. H⁺ + AlCl₄⁻ → AlCl₃ + HCl. Aluminium chloride is re-formed, so it is a catalyst. Hydrogen chloride is the by-product.

Overall equation (ethanoyl chloride): C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl. The product is phenylethanone.

Why acylation matters. It forms a new carbon–carbon bond to the ring, so it is an important step in synthesis.

Worked example 2. 3.90 g of benzene is reacted with excess propanoyl chloride, CH₃CH₂COCl, in the presence of AlCl₃. The mass of organic product is 4.69 g. Use Ar: H = 1.0, C = 12.0, O = 16.0.

1. Product: C₆H₅COCH₂CH₃, 1-phenylpropan-1-one, C₉H₁₀O.
2. Mr(C₆H₆) = 78.0; n = 3.90 / 78.0 = 0.0500 mol.
3. Mole ratio benzene : product = 1 : 1, so the theoretical amount of product is 0.0500 mol.
4. Mr(C₉H₁₀O) = 134.0; theoretical mass = 0.0500 × 134.0 = 6.70 g.
5. Percentage yield = 4.69 / 6.70 × 100 = 70.0%.
6. Electrophile: CH₃CH₂CO⁺, formed by CH₃CH₂COCl + AlCl₃ → CH₃CH₂CO⁺ + AlCl₄⁻.

Comparing the two reactions

	Nitration	Friedel–Crafts acylation
Reagents	conc. HNO₃ + conc. H₂SO₄	RCOCl + AlCl₃ (dry)
Electrophile	NO₂⁺ (nitronium ion)	RCO⁺ (acylium ion)
Catalyst	H₂SO₄	AlCl₃
Catalyst regenerated by	H⁺ + HSO₄⁻ → H₂SO₄	H⁺ + AlCl₄⁻ → AlCl₃ + HCl
Product from benzene	nitrobenzene	phenyl ketone, C₆H₅COR
Use in synthesis	explosives; route to aromatic amines	adds a carbon chain to the ring

Practical opportunity: methyl 3-nitrobenzoate

The specification suggests nitrating methyl benzoate as a skills activity. In outline:

1. Dissolve methyl benzoate in concentrated sulfuric acid and cool the mixture in ice.
2. Add a cold mixture of concentrated nitric and sulfuric acids slowly, a few drops at a time, keeping the reaction mixture cold. Slow addition and cooling help to limit side reactions such as further substitution.
3. Pour the mixture onto crushed ice. The crude solid separates out.
4. Collect the solid by filtration under reduced pressure and wash it with cold water.
5. Recrystallise: dissolve in the minimum volume of hot solvent, filter hot if needed, cool, then filter off the crystals and dry them.
6. Measure the melting point. A pure solid melts sharply, close to the data-book value; an impure one melts lower and over a range.

Concentrated acids are corrosive: wear eye protection and gloves.

Common errors

• Drawing benzene with three fixed double bonds when the question asks about delocalisation or bond length.
• Working out 3 × (−120) but never saying which compound is more stable.
• Saying benzene “has no double bonds so it can’t do addition”. The correct reason is that addition would destroy the delocalised, stabilising ring.
• Starting arrow 1 from a single carbon atom instead of from the delocalised ring.
• Drawing a full circle in the intermediate, or putting the horseshoe gap on the wrong side.
• Leaving the + off the intermediate, or putting it on the electrophile’s atom.
• Pointing arrow 1 at an oxygen of NO₂⁺ instead of the nitrogen.
• Writing AlCl₃ + Cl⁻ in the equation for the electrophile but never showing the catalyst being re-formed.
• Calling the acylation product an aldehyde. Benzene + RCOCl gives a ketone.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.10 Aromatic chemistry (A-level only), covering 3.3.10.1 Bonding and 3.3.10.2 Electrophilic substitution.