OCR A Level Chemistry A: Aromatic compounds (H432) – Revision Notes

For full explanations and worked examples, use the aromatic compounds study guide. These notes condense section 6.1.1, Aromatic compounds, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover outcomes 6.1.1(a) to (l), including the PAG7 link on phenol acidity. Module 6 is examined in components 02 and 03, according to the specification.

Other units sit on the OCR chemistry hub, and the printable H432 checklist keeps a record of what you know. When you have finished, try the aromatic compounds practice questions or a 10-minute diagnostic.

Key definitions

• Arene / aromatic compound: contains a benzene ring.
• Delocalised electrons: shared by more than two atoms; in benzene, six π-electrons spread round the ring.
• π-system: the region formed by sideways overlap of p-orbitals, above and below the ring plane.
• Electrophile: an electron pair acceptor (NO₂⁺, Br⁺, R⁺, RCO⁺).
• Electrophilic substitution: an electrophile replaces a hydrogen on the ring; the ring’s delocalisation is kept.
• Halogen carrier: Fe, FeBr₃, FeCl₃, AlCl₃ or AlBr₃; it generates the stronger electrophile.
• Directing group: a substituent that decides which ring position the next group takes.

6.1.1(a): Kekulé versus delocalised

Feature	Kekulé model	Delocalised model
Bonds in ring	3 C=C and 3 C–C, alternating	6 identical bonds
π-electrons	Localised in three π-bonds	Delocalised over six carbons
Shape	Flat ring	Flat ring, 120° angles
Predicted reactions	Electrophilic addition, like alkenes	Substitution; resists addition

How the π-system forms: each carbon uses three electrons in σ-bonds and has one electron in a p-orbital. Adjacent p-orbitals overlap sideways all round the ring.

6.1.1(b): Three pieces of evidence

1. Bond lengths: all six equal, between C–C and C=C.
2. Hydrogenation: less exothermic than three times the value for cyclohexene, so benzene is more stable than Kekulé predicts.
3. Resistance to reaction: no electrophilic addition with bromine at room temperature.

Method for the hydrogenation comparison: multiply the cyclohexene value by three, then subtract it from the measured benzene value. The positive difference is the extra stability.

6.1.1(c): Naming in four rules

1. Prefix + benzene: bromobenzene, 1,4-dichlorobenzene.
2. Parent names: methylbenzene, phenol, phenylamine, benzoic acid. The parent group is on carbon 1.
3. Lowest locants; prefixes in alphabetical order.
4. Ring as a substituent: phenyl (e.g. phenylethanone).

Check: 2,4-dinitromethylbenzene has CH₃ on carbon 1 and NO₂ on carbons 2 and 4.

6.1.1(d): Reactions to learn

Reaction	Reagents	Conditions	Electrophile	Product with benzene
Nitration	Conc. HNO₃ + conc. H₂SO₄	About 50 °C	NO₂⁺	Nitrobenzene + H₂O
Bromination	Br₂	Halogen carrier, room temp	Br⁺	Bromobenzene + HBr
Chlorination	Cl₂	Halogen carrier, room temp	Cl⁺	Chlorobenzene + HCl
Alkylation	Haloalkane, e.g. CH₃Cl	AlCl₃	R⁺	Alkylbenzene + HCl
Acylation	Acyl chloride, e.g. C₂H₅COCl	AlCl₃	RCO⁺	Phenyl ketone + HCl

Friedel–Crafts reactions form a new C–C bond to the ring, so they are used to build carbon skeletons in synthesis (see 6.2.4).

6.1.1(e): Mechanism in steps

Form the electrophile.

• HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
• Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻ (or Br₂ + AlBr₃ → Br⁺ + AlBr₄⁻)

Three arrows on the ring.

1. Arrow from the π-system (circle) to the electrophile.
2. Intermediate: broken horseshoe over five carbons, + inside, H and E on the sixth carbon.
3. C–H electron pair returns to the ring (arrow); H⁺ is lost.

Regenerate the catalyst.

• H⁺ + HSO₄⁻ → H₂SO₄
• H⁺ + FeBr₄⁻ → FeBr₃ + HBr

6.1.1(f): Benzene versus alkene with bromine

	Alkene (e.g. cyclohexene)	Benzene
π-electrons	Localised in one C=C	Delocalised over the ring
Electron density	High between two carbons	Lower at any one bond
Effect on Br₂	Induces a dipole; Br₂ reacts	Cannot polarise Br₂ enough
Result	Bromine water decolourised	No reaction without a halogen carrier

6.1.1(g): Unfamiliar reactions

Spot the δ+ or positive species in the reagents. That is the electrophile. Use the same three arrows. The ring loses one H⁺. Read any extra information the question gives you, such as which atom carries the δ+ charge.

6.1.1(h): Phenol is a weak acid (PAG7)

• C₆H₅OH ⇌ C₆H₅O⁻ + H⁺ (partial dissociation)
• With NaOH(aq): C₆H₅OH + NaOH → C₆H₅ONa + H₂O
• With Na₂CO₃(aq): no reaction, no effervescence

Test	Ethanol	Phenol	Carboxylic acid
NaOH(aq)	No reaction	Reacts (dissolves)	Reacts
Na₂CO₃(aq)	No reaction	No reaction	Effervescence (CO₂)

6.1.1(i): Reactions of phenol

Reagent	Conditions	Observation	Product(s)
Bromine water	Room temperature, no carrier	Decolourised; white precipitate	2,4,6-tribromophenol + 3HBr
Dilute HNO₃	Room temperature, no H₂SO₄	Mixture of two isomers forms	2-nitrophenol and 4-nitrophenol + H₂O

Equations: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr; C₆H₅OH + HNO₃ → C₆H₄(OH)NO₂ + H₂O.

6.1.1(j): Why phenol reacts faster

Oxygen lone pair (in a p-orbital) → partly delocalised into the ring → ring electron density increases → ring more susceptible to electrophilic attack → Br₂ polarised without a carrier.

Do not explain it through the stability of the intermediate.

6.1.1(k) and (l): Directing effects

Existing group	Effect on ring	New group goes to
–OH	Electron-donating	2 and 4 (and 6)
–NH₂	Electron-donating	2 and 4 (and 6)
–NO₂	Electron-withdrawing	3 (and 5)

Positions 2 and 6 are equivalent, as are 3 and 5. Any other group’s effect will be supplied in the question.

Synthesis rule: put the directing group on first if you want its pattern. Nitrate before you halogenate to get a 3-substituted nitro compound.

Small worked reminders

• Nitrating nitrobenzene gives 1,3-dinitrobenzene.
• Bromine water with phenylamine gives 2,4,6-tribromophenylamine and 3HBr.
• 3-nitrophenol cannot be made by nitrating phenol, because OH sends NO₂ to 2 and 4.
• Mole ratio phenol : Br₂ is 1 : 3; phenol : NaOH is 1 : 1.
• Benzene with chloromethane and AlCl₃ gives methylbenzene, C₆H₅CH₃, and HCl: a new C–C bond to the ring.
• Iodination with ICl: chlorine is more electronegative, so iodine is δ+ and becomes the electrophile; the product is iodobenzene.
• Phenol and dilute HNO₃ give two isomers, so the products must be separated before either is used in a later step.

For the contrast with electrophilic addition to C=C, revisit alkenes for OCR.

Must-know distinctions

• Addition vs substitution: alkenes add Br₂; benzene substitutes Br.
• Concentrated vs dilute HNO₃: benzene needs concentrated HNO₃ and H₂SO₄; phenol needs dilute HNO₃ only.
• Phenol vs carboxylic acid: both react with NaOH; only the acid reacts with carbonate.
• Alkylation vs acylation: haloalkane gives an alkylbenzene; acyl chloride gives a phenyl ketone.
• 2,4- vs 3-directing: OH and NH₂ donate electron density; NO₂ withdraws it.

Quick self-test

1. State what overlaps to form the π-system in benzene.
2. Give the bond angle in benzene.
3. Name the compound with CH₃ on carbon 1 and Cl on carbon 3 of the ring.
4. Write the equation for forming NO₂⁺.
5. Name a halogen carrier for chlorination.
6. Give the organic product of benzene with ethanoyl chloride and AlCl₃.
7. Why does cyclohexene decolourise bromine water without a catalyst?
8. What do you see when phenol is added to sodium carbonate solution?
9. Calculate the volume of 1.00 mol dm⁻³ NaOH that reacts with 2.35 g of phenol (Mr 94.0).
10. Predict the product when 4-methylphenol reacts with excess bromine water. How many moles of Br₂ react per mole?
11. Which position does bromine take when nitrobenzene is brominated with FeBr₃?
12. Why does phenol react with bromine without a halogen carrier?

Answers

1. p-orbitals on adjacent carbons overlap sideways (one electron from each carbon).
2. 120°.
3. 3-chloromethylbenzene.
4. HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.
5. AlCl₃ (or FeCl₃, or iron).
6. Phenylethanone, C₆H₅COCH₃ (plus HCl).
7. Its localised π-bond has high electron density, which polarises Br₂.
8. No effervescence; no visible reaction.
9. n = 2.35 ÷ 94.0 = 0.0250 mol; volume = 0.0250 dm³ = 25.0 cm³.
10. 2,6-dibromo-4-methylphenol; position 4 is already taken, so 2 mol Br₂ react.
11. Position 3: 1-bromo-3-nitrobenzene.
12. An oxygen p-orbital lone pair is partly donated into the ring, raising its electron density so it polarises Br₂.

Where marks are usually lost

• Describing p-orbitals overlapping “end-on” or forming only three π-bonds.
• Quoting hydrogenation evidence without saying benzene is more stable than predicted.
• Naming with locants that are not the lowest set, or ignoring alphabetical order.
• Missing “concentrated” for nitration of benzene, or using it for phenol.
• Starting the first curly arrow on a single carbon instead of the π-system.
• Drawing the intermediate’s horseshoe over all six carbons or omitting its + charge.
• Forgetting to regenerate H₂SO₄ or FeBr₃.
• Writing “benzene has no double bonds so it can’t react” instead of comparing electron density.
• Predicting 2,4,6-substitution for phenol when position 4 is already occupied.
• Treating Na₂CO₃ as a test that phenol passes.

Official syllabus

OCR A Level Chemistry A (H432), specification Version 3.1 (May 2026), for first assessment in 2017, published by OCR: section 6.1.1 Aromatic compounds, learning outcomes (a) to (l).