OCR A Level Chemistry A: Aromatic compounds (H432)

This study guide teaches 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. It opens Module 6, Organic chemistry and analysis, and all twelve lettered outcomes, 6.1.1(a) to (l), are taught below. According to the specification, Module 6 content is examined in component 02 (Synthesis and analytical techniques) and component 03 (Unified chemistry). PAG7 is linked to outcome 6.1.1(h).

Browse other H432 units from the OCR chemistry hub, mark progress against the H432 checklist, or find gaps with a free 10-minute diagnostic.

Outcomes in 6.1.1

Outcome	What you must do	Additional guidance
6.1.1(a)	Compare Kekulé and delocalised models (p-orbital overlap, π-system)	Either structure may be drawn; HSW1,7
6.1.1(b)	Bond lengths, hydrogenation enthalpy and resistance to reaction as evidence	HSW11
6.1.1(c)	Name substituted aromatics	Locants, e.g. 2,4-dinitromethylbenzene; HSW8
6.1.1(d)	Substitution by (i) conc. HNO₃/H₂SO₄, (ii) halogen + carrier, (iii) Friedel–Crafts	Halogen carriers: iron, iron halides, aluminium halides
6.1.1(e)	Nitration and halogenation mechanisms	Show NO₂⁺ formation; take X⁺ as the halogen electrophile
6.1.1(f)	Benzene’s resistance to bromination versus alkenes	HSW2,5
6.1.1(g)	Unfamiliar substitutions and mechanisms	Extra information may be provided
6.1.1(h)	Weak acidity of phenols: NaOH but not carbonates	PAG7
6.1.1(i)	Phenol with (i) bromine, (ii) dilute HNO₃	No conc. acids needed
6.1.1(j)	Why phenol reacts more easily than benzene	Susceptibility to attack only
6.1.1(k)	2,4-directing OH, NH₂; 3-directing NO₂	Other groups: data supplied
6.1.1(l)	Predict products; importance in synthesis	See 6.2.5

6.1.1(a): Two models of benzene

The Kekulé model (proposed by August Kekulé in 1865) puts three C=C double bonds and three C–C single bonds alternately round the ring. It predicts two bond lengths and alkene-like reactions.

The delocalised model describes the bonding like this:

• Each carbon forms three σ-bonds: two to neighbouring carbons and one to hydrogen. The bond angles are 120°, so the molecule is flat.
• Each carbon has one remaining electron in a p-orbital, sitting above and below the plane of the ring.
• The p-orbitals overlap sideways with both neighbours, forming a π-system above and below the plane.
• The six π-electrons are delocalised over all six carbons.

You may draw benzene as a Kekulé hexagon or a hexagon with a circle (the π-system). The change of model shows science revising a model as evidence builds (HSW1,7).

6.1.1(b): Evidence for delocalisation

Three kinds of evidence led chemists to accept the delocalised model (HSW11).

1. Bond lengths. X-ray diffraction shows six equal carbon–carbon bonds in benzene, each intermediate in length between C–C and C=C. Kekulé predicts three short bonds and three long ones.
2. Enthalpy change of hydrogenation. Benzene releases far less energy on hydrogenation than three isolated C=C bonds would. Benzene is more stable than the Kekulé model predicts.
3. Resistance to reaction. Benzene does not readily take part in electrophilic addition. It does not decolourise bromine water at room temperature, while alkenes do (see 6.1.1(f)).

Worked example 1: reading the evidence

Use these values. Bond lengths: C–C in ethane 153 pm; C=C in ethene 134 pm; every ring bond in benzene 140 pm. Enthalpy changes of hydrogenation: cyclohexene −119 kJ mol⁻¹; benzene −207 kJ mol⁻¹.

Step 1: bond lengths. One value, 140 pm, lies between 134 pm and 153 pm. All bonds are identical, so there are no separate single and double bonds.

Step 2: predicted value for Kekulé benzene. Three C=C bonds: 3 × (−119) = −357 kJ mol⁻¹.

Step 3: compare with experiment. −207 − (−357) = +150. Benzene releases 150 kJ mol⁻¹ less energy than predicted, so it is 150 kJ mol⁻¹ more stable than the Kekulé structure. The difference is the extra stability from delocalisation.

6.1.1(c): Naming aromatic compounds

• Substituents become prefixes on “benzene”: chlorobenzene, nitrobenzene, 1,3-dibromobenzene.
• Some groups make a parent name: methylbenzene (C₆H₅CH₃), phenol (C₆H₅OH), phenylamine (C₆H₅NH₂), benzoic acid (C₆H₅COOH). The carbon bearing that group is carbon 1.
• Number round the ring to give the lowest locants. List prefixes in alphabetical order, ignoring di- and tri-.
• When the ring is a substituent on a chain, it is “phenyl”: phenylethene, phenylethanone.

Worked example 2: naming

Description	Name
Methylbenzene with NO₂ on carbons 2 and 4	2,4-dinitromethylbenzene (the specification’s own example)
Benzene ring with Cl and Br on adjacent carbons	1-bromo-2-chlorobenzene (b before c gets locant 1)
Phenol with Br on both carbons adjacent to the C–OH	2,6-dibromophenol
Phenylamine with NO₂ opposite the NH₂	4-nitrophenylamine

6.1.1(d): Electrophilic substitution reactions

Substituting a hydrogen keeps the delocalised ring intact.

Reaction	Reagents and conditions	Equation (with benzene)
(i) Nitration	Conc. HNO₃ and conc. H₂SO₄; warm to about 50 °C	C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
(ii) Halogenation	Br₂ or Cl₂ with a halogen carrier (Fe, FeBr₃, AlCl₃, etc.); room temperature	C₆H₆ + Br₂ → C₆H₅Br + HBr
(iii) Friedel–Crafts alkylation	Haloalkane with a halogen carrier such as AlCl₃	C₆H₆ + CH₃CH₂Cl → C₆H₅CH₂CH₃ + HCl
(iii) Friedel–Crafts acylation	Acyl chloride with a halogen carrier such as AlCl₃	C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl

Keeping nitration warm, not hot, limits further nitration. Iron works as a carrier because it forms the iron(III) halide with the halogen.

Friedel–Crafts reactions matter in synthesis because they make a new C–C bond to the ring. That is one of few ways to attach a carbon chain directly to an arene (see 6.2.4).

Worked example 3: a Friedel–Crafts yield

6.24 g of benzene reacts with excess 3-methylbutanoyl chloride, (CH₃)₂CHCH₂COCl, and AlCl₃. 9.72 g of 3-methyl-1-phenylbutan-1-one, C₁₁H₁₄O, is collected. Ar: C 12.0, O 16.0, H 1.0.

1. Equation: C₆H₆ + (CH₃)₂CHCH₂COCl → C₆H₅COCH₂CH(CH₃)₂ + HCl
2. n(benzene) = 6.24 ÷ 78.0 = 0.0800 mol
3. Mr(C₁₁H₁₄O) = 162.0, so theoretical mass = 0.0800 × 162.0 = 12.96 g
4. Percentage yield = 9.72 ÷ 12.96 × 100 = 75.0%

6.1.1(e): Mechanisms for nitration and halogenation

Making the electrophiles

• Nitronium ion: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. Sulfuric acid is the stronger acid and acts as a catalyst. It is regenerated at the end: H⁺ + HSO₄⁻ → H₂SO₄.
• Bromine cation: Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻. The carrier is regenerated: H⁺ + FeBr₄⁻ → FeBr₃ + HBr. For halogenation you may take the electrophile as X⁺.

Worked example 4: nitration mechanism in steps

1. A curly arrow goes from the delocalised π-system to the nitrogen of NO₂⁺.
2. This forms an intermediate. Draw it as a hexagon with a broken horseshoe covering five carbons, a + charge inside, and H and NO₂ both attached to the sixth carbon.
3. The C–H bond breaks: its electron pair moves back into the ring (curly arrow), and the delocalised system is complete again.
4. Products: C₆H₅NO₂ and H⁺, which then reacts with HSO₄⁻.

Bromination follows the same three arrows with Br⁺ in place of NO₂⁺.

6.1.1(f): Why benzene resists bromination

An alkene’s π-bond is localised between two carbons. That gives a region of high electron density, which induces a dipole in an approaching Br₂ molecule. Br₂ is polarised enough to react, so cyclohexene decolourises bromine at room temperature.

Benzene’s π-electrons are delocalised over six carbons. The electron density between any two carbons is lower. It cannot polarise Br₂ enough, so no reaction occurs unless a halogen carrier provides the stronger electrophile Br⁺.

6.1.1(g): Unfamiliar electrophilic substitutions

Method:

1. Find the electrophile: the positive or δ+ species.
2. Write how it forms, if the question asks.
3. Apply the same three curly arrows as nitration.
4. Replace one ring H with the electrophile; the H leaves as H⁺.

Worked example 5: iodination with ICl

Iodine monochloride, ICl, substitutes benzene. Chlorine is more electronegative than iodine, so the bond is Iδ+–Clδ−. The ring attacks iodine: an arrow from the π-system to I, and an arrow from the I–Cl bond to Cl, releasing Cl⁻. The intermediate loses H⁺, which joins Cl⁻.

Overall: C₆H₆ + ICl → C₆H₅I + HCl. The product is iodobenzene.

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

Phenol, C₆H₅OH, partly dissociates in water: C₆H₅OH ⇌ C₆H₅O⁻ + H⁺. It is a weak acid (see 5.1.3).

• With sodium hydroxide it is neutralised: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. The product is sodium phenoxide, and phenol dissolves.
• With sodium carbonate there is no reaction and no effervescence. Phenol is not acidic enough to release CO₂ from a carbonate.

In PAG7 this tells a phenol from a carboxylic acid (which fizzes) and an alcohol (neither reacts).

6.1.1(i): Electrophilic substitution of phenol

(i) With bromine. Bromine water is decolourised at room temperature and a white precipitate of 2,4,6-tribromophenol forms. No halogen carrier is needed.

C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr

(ii) With dilute nitric acid. Phenol gives two isomers, 2-nitrophenol and 4-nitrophenol, at room temperature. No concentrated HNO₃ or concentrated H₂SO₄ is needed.

C₆H₅OH + HNO₃ → C₆H₄(OH)NO₂ + H₂O

Worked example 6: bromine needed

0.235 g of phenol reacts completely with 0.0500 mol dm⁻³ bromine water. Find the volume used and the mass of precipitate. Relative atomic masses: Br 79.9; O 16.0; C 12.0; H 1.0.

1. n(phenol) = 0.235 ÷ 94.0 = 0.00250 mol
2. n(Br₂) = 3 × 0.00250 = 0.00750 mol
3. Volume = 0.00750 ÷ 0.0500 = 0.150 dm³ = 150 cm³
4. Mr(C₆H₃Br₃O) = 330.7; mass = 0.00250 × 330.7 = 0.827 g

6.1.1(j): Why phenol reacts more easily

One lone pair on the oxygen sits in a p-orbital. It overlaps with the ring and is partly donated into the π-system. The electron density of the ring increases. The ring is now more susceptible to attack by electrophiles: it polarises Br₂ without a carrier and reacts with dilute nitric acid.

Keep the explanation to electron density and the ring’s susceptibility to attack. The specification does not ask you to discuss the stability of the intermediate.

6.1.1(k) and (l): Directing groups

A group already on the ring decides where the next substituent goes.

Group	Type	Directs to
–OH, –NH₂	Electron-donating	2- and 4- positions
–NO₂	Electron-withdrawing	3- position

For any other group, the exam supplies the directing information.

Worked example 7: predicting and planning

Predict: phenylamine with bromine water gives 2,4,6-tribromophenylamine (positions 2, 4 and 6 are all 2- or 4- relative to NH₂). Nitrobenzene with Br₂/FeBr₃ gives 1-bromo-3-nitrobenzene.

Plan: to make 1-bromo-3-nitrobenzene from benzene, nitrate first, then brominate. The NO₂ group sends bromine to position 3. If you brominated first, you would be told that Br directs to 2- and 4-, giving the wrong isomers.

Order matters in synthesis (6.2.5): the right sequence gives the target isomer as the main product.

Common errors

• Writing addition products, such as C₆H₆Br₂, for benzene.
• Forgetting the equation for forming NO₂⁺, or the regeneration of H₂SO₄.
• Saying phenol reacts with sodium carbonate.
• Using concentrated acids for phenol nitration.
• Explaining phenol’s reactivity by intermediate stability instead of electron density.
• Placing NO₂ products at 2- and 4- positions.

Keep going

• Condensed recall: aromatic compounds revision notes
• Test yourself: aromatic compounds practice questions
• Compare electrophilic addition: OCR alkenes study guide
• Find weak spots: Marlbridge diagnostics

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).