OxfordAQA A-Level Chemistry: Aromatic chemistry (9620)

This study guide teaches 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. It covers 3.3.10.1 Bonding and 3.3.10.2 Electrophilic substitution. Because the specification lists it with the International A2 organic topics, everything here is International A-level only. It is examined in Unit 4 (Organic 2 and Physical 2), and Unit 5 (Practical and synoptic) covers all content.

Besides nitration and acylation, the 9620 section asks for sulfonation, alkylation, methylbenzene and chlorine radicals.

Pair this guide with the revision notes and the practice questions. Background you need: curly arrows and electrophiles from alkenes, radical steps from alkanes, nucleophilic substitution from halogenoalkanes and enthalpy diagrams from energetics. The organic synthesis guide uses these reactions in routes. For the wider course, use the hub page, the printable checklist and the free diagnostics.

Specification checklist

Ref	Learning outcome	Stage
3.3.10.1	Benzene bonding: planar ring, bond length between single and double	International A-level only
3.3.10.1	Delocalised p electrons make benzene more stable than cyclohexa-1,3,5-triene; show this with enthalpies of hydrogenation	International A-level only
3.3.10.1	Explain why substitution happens in preference to addition	International A-level only
3.3.10.2	Electrophilic attack gives substitution (monosubstitution only); importance of nitration, sulfonation and Friedel–Crafts acylation in synthesis	International A-level only
3.3.10.2	Outline nitration and sulfonation of benzene and methylbenzene, and name the products	International A-level only
3.3.10.2	Outline acylation and alkylation with an AlCl₃ catalyst, and name the products	International A-level only
3.3.10.2	Outline free-radical attack of chlorine on benzene (ring addition) and methylbenzene (side-chain substitution), and name the products	International A-level only
3.3.10.2	Explain the different reactivities of chlorine on the ring and on the side chain	International A-level only

No required practical is attached to 3.3.10.

Bonding in benzene (3.3.10.1)

Benzene is C₆H₆. Each carbon is bonded to two other carbons and one hydrogen by σ bonds. Three bonding pairs around each carbon spread out to 120°, so the ring is a flat hexagon: the molecule is planar.

Each carbon has one more outer electron, in a p orbital that sticks up and down at right angles to the ring. The six p orbitals overlap sideways around the whole ring. Their six electrons are delocalised over the whole ring, above and below the plane.

Every carbon–carbon bond shares this delocalised density equally, so all six bonds are identical. Their length is between that of a single and a double bond. Measured values are about 139 pm in benzene, against roughly 154 pm for a C–C single bond and 134 pm for a C=C double bond.

The structure with three fixed C=C bonds is a theoretical molecule called cyclohexa-1,3,5-triene. It would have two bond lengths. Benzene does not.

Hydrogenation evidence for extra stability

Comparing the energy released when benzene and a compound with one C=C bond are hydrogenated shows how much more stable benzene is than the triene.

Worked example 1. Use these enthalpies of hydrogenation:

• cyclohexene + H₂ → cyclohexane, ΔH = −118 kJ mol⁻¹
• benzene + 3H₂ → cyclohexane, ΔH = −206 kJ mol⁻¹

Use them to compare the stability of benzene with that of the triene.

1. Cyclohexene has one C=C, so −118 kJ mol⁻¹ is released per C=C hydrogenated.
2. The triene would have three C=C bonds: predicted ΔH = 3 × (−118) = −354 kJ mol⁻¹.
3. Benzene’s measured value is −206 kJ mol⁻¹, which is 354 − 206 = 148 kJ mol⁻¹ less exothermic.
4. Both reactions end at cyclohexane. If benzene releases less energy reaching the same product, it must start lower down. So benzene is 148 kJ mol⁻¹ more stable than the triene.
5. Reason: delocalisation of the p electrons around the ring.

On an energy diagram, put cyclohexane at the bottom. Benzene + 3H₂ sits 206 kJ mol⁻¹ above it and the triene + 3H₂ sits 354 kJ mol⁻¹ above it. The 148 kJ mol⁻¹ gap between those two upper levels is the extra stability.

Why substitution, not addition

If an electrophile added across the ring, the product would no longer have a complete delocalised ring, so it would lose that extra stability. Swapping one H for another group leaves the ring of delocalised electrons whole. So benzene’s reactions with electrophiles are substitutions. The spread-out electron density also means benzene needs a stronger electrophile than an alkene does, usually a positive ion or a highly polarised molecule.

Electrophilic substitution (3.3.10.2)

The specification limits this to monosubstitution: one ring hydrogen is replaced. Every mechanism has the same shape.

1. Form the electrophile.
2. Curly arrow from the delocalised ring (start it on the circle) to the electrophile’s attacking atom.
3. Intermediate: the attacked carbon carries both H and the new group. Draw a horseshoe over the other five carbons with a + inside it, opening towards the attacked carbon.
4. Curly arrow from that C–H bond back into the ring. H⁺ leaves and the full circle returns.

Nitration

Reagents: concentrated nitric acid and concentrated sulfuric acid. Sulfuric acid protonates nitric acid, which then loses water:

HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻

The electrophile formed is NO₂⁺, the nitronium ion.

The ring attacks the nitrogen of NO₂⁺. After the intermediate loses H⁺, benzene gives nitrobenzene, C₆H₅NO₂. The H⁺ reacts with HSO₄⁻ to re-form H₂SO₄, so sulfuric acid is a catalyst.

Methylbenzene. The methyl group pushes electron density into the ring, so methylbenzene is attacked faster than benzene and is nitrated at a lower temperature. Attack happens mainly at positions 2 and 4, giving a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene (also named 1-methyl-2-nitrobenzene and 1-methyl-4-nitrobenzene). The mechanism is unchanged.

Why nitration matters. Nitration is an important step in synthesis. Rings carrying several nitro groups are used as explosives. Reducing a nitro group gives an aromatic amine (section 3.3.11).

Sulfonation

Reagents: fuming sulfuric acid (concentrated sulfuric acid with sulfur trioxide dissolved in it), warmed; or concentrated sulfuric acid heated under reflux for a longer time. The electrophile is sulfur trioxide, SO₃. Its sulfur carries a large δ+ charge because three oxygen atoms pull electron density away from it. In concentrated acid it forms by:

2H₂SO₄ → SO₃ + H₃O⁺ + HSO₄⁻

The ring attacks the sulfur atom. In the intermediate, the attacked carbon carries H and an SO₃⁻ group, with the + in the horseshoe. Loss of H⁺ restores the ring, and the oxygen picks up a proton to give benzenesulfonic acid, C₆H₅SO₃H:

C₆H₆ + SO₃ → C₆H₅SO₃H

You may also see the electrophile written as HSO₃⁺. The pattern of curly arrows is the same.

Methylbenzene gives mainly 4-methylbenzenesulfonic acid, with some 2-methylbenzenesulfonic acid.

Why sulfonation matters. Sulfonation is an important step in making surfactants and sulfonamides. Many detergents are sodium salts of alkylbenzenesulfonic acids: a long alkyl chain is joined to the ring, the ring is sulfonated, and the acid is neutralised with sodium hydroxide. Sulfonamides include the sulfa drugs, an early group of synthetic antibacterial medicines.

Friedel–Crafts acylation and alkylation

Both reactions use aluminium chloride, AlCl₃, as a catalyst under dry conditions. AlCl₃ takes a chloride ion from the organic chloride, leaving a positive carbon electrophile.

	Acylation	Alkylation
Reagent	acyl chloride, RCOCl	haloalkane, RCl
Electrophile	acylium ion, RCO⁺	carbocation, R⁺
Forming it	RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻	RCl + AlCl₃ → R⁺ + AlCl₄⁻
Product from benzene	phenyl ketone, C₆H₅COR	alkylbenzene, C₆H₅R
Catalyst re-formed	H⁺ + AlCl₄⁻ → AlCl₃ + HCl	H⁺ + AlCl₄⁻ → AlCl₃ + HCl

Both reactions join a new carbon chain to the ring, which is why acylation is an important step in synthesis. An alkyl group makes the ring more reactive, so the alkylbenzene can be attacked again. A large excess of benzene is used so that the main product is monosubstituted.

Worked example 2. 11.7 g of benzene is heated with 2-chloropropane, CH₃CHClCH₃, and AlCl₃. The organic product weighs 13.5 g. Ar: H = 1.0, C = 12.0, Cl = 35.5.

1. Electrophile: CH₃CHClCH₃ + AlCl₃ → CH₃C⁺HCH₃ + AlCl₄⁻. The positive carbon is the middle one.
2. The ring attacks that carbon. The product is (1-methylethyl)benzene, C₆H₅CH(CH₃)₂, formula C₉H₁₂.
3. Equation: C₆H₆ + CH₃CHClCH₃ → C₆H₅CH(CH₃)₂ + HCl.
4. n(benzene) = 11.7 / 78.0 = 0.150 mol. Mr(C₉H₁₂) = 120.0, so theoretical mass = 0.150 × 120.0 = 18.0 g.
5. Percentage yield = 13.5 / 18.0 × 100 = 75.0%.
6. Minimum mass of 2-chloropropane (Mr 78.5): 0.150 × 78.5 = 11.8 g.

Free-radical attack by chlorine (3.3.10.2)

Without a catalyst but with ultraviolet light, chlorine reacts by a radical mechanism, not an electrophilic one. Initiation is the same as for alkanes: Cl₂ → 2Cl• (UV light).

Benzene: ring addition. Chlorine radicals add to the ring one at a time. Once the delocalised ring has been broken, further addition follows until all six carbons carry a chlorine. The product is 1,2,3,4,5,6-hexachlorocyclohexane:

C₆H₆ + 3Cl₂ → C₆H₆Cl₆        (UV light)

Methylbenzene: side-chain substitution. The radical takes a hydrogen from the methyl group, not from the ring, so the ring stays intact:

Propagation   C₆H₅CH₃ + Cl• → C₆H₅CH₂• + HCl
              C₆H₅CH₂• + Cl₂ → C₆H₅CH₂Cl + Cl•
Termination   C₆H₅CH₂• + Cl• → C₆H₅CH₂Cl
              2C₆H₅CH₂• → C₆H₅CH₂CH₂C₆H₅
              2Cl• → Cl₂

The first product is (chloromethyl)benzene. With more chlorine, the other two side-chain hydrogens are replaced in turn, giving (dichloromethyl)benzene and (trichloromethyl)benzene.

Worked example 3. Methylbenzene reacts with excess chlorine in UV light. One product has Mr = 195.5. Identify it. Ar: H = 1.0, C = 12.0, Cl = 35.5.

1. Each side-chain substitution swaps H (1.0) for Cl (35.5), adding 34.5 to Mr.
2. Methylbenzene, C₇H₈, has Mr 92.0. One, two and three substitutions give 126.5, 161.0 and 195.5.
3. So the product is C₇H₅Cl₃: (trichloromethyl)benzene, C₆H₅CCl₃, with all three methyl hydrogens replaced and the ring untouched.

Chlorine on the ring versus on the side chain

Compare (chloromethyl)benzene, C₆H₅CH₂Cl, with chlorobenzene, C₆H₅Cl, where the chlorine is bonded directly to a ring carbon (chlorobenzene is made by a ring substitution that 3.3.10 does not ask you to describe).

• Side chain. The Cl is on an sp³ carbon outside the ring. The C–Cl bond is polar and has normal strength, so nucleophiles such as OH⁻ substitute it, as in any halogenoalkane.
• Ring. One lone pair on Cl overlaps with the ring’s delocalised electrons. The C–Cl bond gains partial π character; it becomes shorter and harder to break. The cloud of delocalised electrons also repels an approaching nucleophile. So chlorobenzene does not undergo nucleophilic substitution under normal conditions.

Chlorine in the side chain is therefore much more reactive than chlorine on the ring.

Common errors

• Quoting 148 kJ mol⁻¹ without saying which compound is more stable.
• Starting the first curly arrow on a carbon atom instead of the delocalised ring.
• Giving 3-nitromethylbenzene as the main product from methylbenzene.
• Using UV light when ring substitution is wanted, or AlCl₃ when side-chain chlorination is wanted.
• Saying benzene + Cl₂ in UV light gives chlorobenzene. It gives the addition product C₆H₆Cl₆.
• Calling SO₃ a positive ion. It is neutral; its sulfur atom is strongly δ+.

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).