OxfordAQA A-Level Chemistry: Aromatic chemistry (9620) – Revision Notes

This page summarises section 3.3.10 (Aromatic chemistry) from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020: subsections 3.3.10.1 (Bonding) and 3.3.10.2 (Electrophilic substitution). Everything here is International A2 content and therefore International A-level only. Unit 4 (Organic 2 and Physical 2) examines it, and Unit 5 (Practical and synoptic) may also draw on it.

The study guide has the full teaching and worked examples, and the practice questions test it. Radical steps are revised in the alkanes revision notes and curly-arrow rules in the alkenes revision notes. To plan the rest of your revision, see the course hub, the printable checklist and the free diagnostics.

Key terms

Term	Meaning
Arene	Compound containing a benzene ring
Delocalised p electrons	The six electrons from the carbons’ p orbitals, spread over the whole ring above and below the plane
Cyclohexa-1,3,5-triene	Hypothetical C₆H₆ with alternating single and double bonds; not real
Electrophile	Electron pair acceptor
Electrophilic substitution	An electrophile replaces one ring H; the delocalised ring is kept
Monosubstitution	One ring hydrogen replaced: the limit set by the specification
Side chain	Carbon atoms attached to the ring but not part of it, such as the CH₃ in methylbenzene
Free radical	Species with an unpaired electron, such as Cl•

3.3.10.1 Bonding

• Each C forms three σ bonds (two C, one H) at 120°, so the molecule is planar.
• Each C has one p electron. The six p orbitals overlap sideways, so the electrons are delocalised over all six carbons.
• All six C–C bonds are equal, with length between single and double (about 139 pm, against roughly 154 pm and 134 pm).
• Result: benzene is lower in energy (more stable) than the triene.

Method: hydrogenation evidence

1. Take ΔH for hydrogenating one C=C (from cyclohexene).
2. Multiply by 3 to predict ΔH for the triene.
3. Compare with benzene’s measured ΔH, which is less exothermic.
4. Same product (cyclohexane) but less energy released, so benzene starts further down.
5. Size of the gap = stabilisation from delocalisation.

Worked reminder: the study guide’s data give 3 × (−118) = −354 kJ mol⁻¹ predicted, −206 kJ mol⁻¹ measured, so benzene is 148 kJ mol⁻¹ more stable. Always use the values a question gives you.

Substitution, not addition. An addition product has no complete delocalised ring, so the stabilisation is gone. Substitution leaves the ring whole, so it is preferred.

3.3.10.2 Electrophilic substitution

Method: drawing any electrophilic substitution

1. Equation to form the electrophile.
2. Arrow from the ring’s circle to the attacking atom (N of NO₂⁺, S of SO₃, C of RCO⁺ or R⁺).
3. Intermediate: attacked C carries H and the new group; horseshoe over five carbons, + inside, opening towards the attacked C.
4. Arrow from the C–H bond into the ring; H⁺ released.
5. Show the catalyst re-formed where there is one.

Reaction summary

Reaction	Reagents and conditions	Electrophile	Product from benzene
Nitration	conc. HNO₃ + conc. H₂SO₄	NO₂⁺	nitrobenzene, C₆H₅NO₂
Sulfonation	fuming H₂SO₄, warm (or conc. H₂SO₄, reflux)	SO₃	benzenesulfonic acid, C₆H₅SO₃H
Acylation	RCOCl + AlCl₃, dry	RCO⁺	phenyl ketone, C₆H₅COR
Alkylation	RCl + AlCl₃, dry	R⁺	alkylbenzene, C₆H₅R

Equations to learn

Purpose	Equation
Form NO₂⁺	HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻
Re-form H₂SO₄	H⁺ + HSO₄⁻ → H₂SO₄
Form SO₃ in conc. acid	2H₂SO₄ → SO₃ + H₃O⁺ + HSO₄⁻
Overall sulfonation	C₆H₆ + SO₃ → C₆H₅SO₃H
Form RCO⁺	RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻
Form R⁺	RCl + AlCl₃ → R⁺ + AlCl₄⁻
Re-form AlCl₃	H⁺ + AlCl₄⁻ → AlCl₃ + HCl

Methylbenzene

• The CH₃ group pushes electron density into the ring, so methylbenzene reacts faster than benzene (nitration is done at a lower temperature).
• Substitution happens mainly at positions 2 and 4.
• Nitration: 2-nitromethylbenzene and 4-nitromethylbenzene.
• Sulfonation: mainly 4-methylbenzenesulfonic acid, with some of the 2-isomer.

Why these reactions matter in synthesis

Reaction	Importance named in the specification
Nitration	manufacture of explosives; formation of amines (reduce the nitro group, 3.3.11)
Sulfonation	manufacture of surfactants (alkylbenzenesulfonate detergents) and sulfonamides (including antibacterial drugs)
Friedel–Crafts acylation	joins a carbon chain to the ring

Free-radical attack by chlorine

Conditions: Cl₂ with UV light, no catalyst. Initiation: Cl₂ → 2Cl•.

Compound	What happens	Product(s)
Benzene	ring addition, one Cl at a time	1,2,3,4,5,6-hexachlorocyclohexane, C₆H₆Cl₆
Methylbenzene	side-chain substitution; ring untouched	(chloromethyl)benzene, then (dichloromethyl)benzene and (trichloromethyl)benzene

Propagation for methylbenzene:

C₆H₅CH₃ + Cl• → C₆H₅CH₂• + HCl
C₆H₅CH₂• + Cl₂ → C₆H₅CH₂Cl + Cl•

Termination: any two radicals join, for example 2C₆H₅CH₂• → C₆H₅CH₂CH₂C₆H₅.

Ring chlorine versus side-chain chlorine

	Chlorobenzene, C₆H₅Cl	(Chloromethyl)benzene, C₆H₅CH₂Cl
Where Cl is	bonded to a ring carbon	bonded to an sp³ side-chain carbon
C–Cl bond	Cl lone pair overlaps with the ring: partial double-bond character, shorter and stronger	ordinary polar C–Cl
Nucleophiles	repelled by the delocalised electrons	attack the δ+ carbon
Reactivity	no nucleophilic substitution under normal conditions	substituted readily, like a halogenoalkane

Must-know distinctions

• Catalyst vs light: AlCl₃ or conc. acid gives electrophilic substitution on the ring; UV light gives radical reactions.
• Benzene vs methylbenzene in UV light: benzene adds chlorine to the ring; methylbenzene substitutes on the side chain.
• Acylation vs alkylation: RCO⁺ gives a ketone; R⁺ gives an alkylbenzene. Both give HCl.
• NO₂⁺ vs SO₃: the nitronium ion is charged; sulfur trioxide is neutral, with a strongly δ+ sulfur.
• Predicted vs measured: the triene value is calculated; benzene’s value is measured.

Quick self-test

Use Ar: H = 1.0, C = 12.0, O = 16.0, S = 32.1, Cl = 35.5.

1. How many electrons are delocalised in a benzene molecule?
2. Put these in order of increasing length: C–C single bond, C=C double bond, carbon–carbon bond in benzene.
3. Benzene’s enthalpy of hydrogenation is −206 kJ mol⁻¹ and it is 148 kJ mol⁻¹ more stable than the triene. Deduce ΔH for hydrogenating one C=C.
4. Name the electrophile in sulfonation and the atom the ring attacks.
5. Name the two main products when methylbenzene is nitrated.
6. Give the conditions and the product formula when chlorine adds to benzene.
7. Calculate the percentage atom economy of C₆H₆ + SO₃ → C₆H₅SO₃H and of C₆H₆ + H₂SO₄ → C₆H₅SO₃H + H₂O.
8. Write an equation for the electrophile formed from chloroethane and AlCl₃.
9. Calculate the percentage atom economy of C₆H₆ + CH₃CH₂Cl → C₆H₅CH₂CH₃ + HCl.
10. Which of chlorobenzene and (chloromethyl)benzene reacts with warm aqueous hydroxide ions? Give one reason.
11. Calculate Mr of 4-methylbenzenesulfonic acid, CH₃C₆H₄SO₃H.
12. Using −206 kJ mol⁻¹, calculate the heat released when 3.12 g of benzene is fully hydrogenated.

Answers

1. Six (one p electron from each carbon).
2. C=C double < benzene bond < C–C single.
3. (−206 − 148) / 3 = −118 kJ mol⁻¹.
4. Sulfur trioxide, SO₃; the ring attacks the sulfur atom.
5. 2-nitromethylbenzene and 4-nitromethylbenzene.
6. UV light, no catalyst; C₆H₆Cl₆ (1,2,3,4,5,6-hexachlorocyclohexane).
7. With SO₃ there is one product: 100%. With H₂SO₄: Mr C₆H₅SO₃H = 158.1; 158.1 / (78.0 + 98.1) × 100 = 89.8%.
8. CH₃CH₂Cl + AlCl₃ → CH₃CH₂⁺ + AlCl₄⁻.
9. Mr C₈H₁₀ = 106.0; 106.0 / (78.0 + 64.5) × 100 = 74.4%.
10. (Chloromethyl)benzene: its Cl is on a side-chain carbon with an ordinary C–Cl bond; in chlorobenzene the C–Cl bond is strengthened by overlap with the ring.
11. C₇H₈SO₃: 84.0 + 8.0 + 32.1 + 48.0 = 172.1.
12. n = 3.12 / 78.0 = 0.0400 mol; 0.0400 × 206 = 8.24 kJ.

Where marks are usually lost

• Saying the bonds are “in between” but never naming bond length.
• Giving the stability figure without saying benzene is the more stable one, or without naming delocalisation.
• Starting the first arrow on a ring carbon or C–H bond instead of the delocalised ring.
• Horseshoe drawn as a full circle, missing the +, or opening away from the attacked carbon.
• Pointing the arrow at an oxygen of NO₂⁺ or SO₃.
• Naming 3-nitromethylbenzene as a main product of methylbenzene nitration.
• Writing chlorobenzene as the product of benzene with chlorine in UV light.
• Showing the ring being chlorinated when methylbenzene reacts in UV light.
• Leaving out HCl in Friedel–Crafts equations, or calling it the catalyst.
• Explaining the low reactivity of chlorobenzene by bond polarity alone, with no mention of the lone pair overlapping with the ring.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.10 Aromatic chemistry (International A2): subsections 3.3.10.1 (Bonding) and 3.3.10.2 (Electrophilic substitution).