Hydrocarbons: Arenes

0. What Examiners Want

This guide covers Topic 30 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. It builds on Topic 29 (the shape and stability of benzene) and prepares you for Topic 31 (halogenoarenes), Topic 32 (phenol) and Topic 34 (phenylamine).

1. Arenes at a Glance: Why Substitution?

An arene is a hydrocarbon that contains a benzene ring. In Topic 29 you saw that every carbon in benzene is sp² hybridised and that the six p electrons form one delocalised π ring above and below the plane. That ring is the key to all the chemistry in this topic. The figure below is a map of the reactions you must know. Benzene sits in the middle; each arrow gives the reagents and conditions.

The reactions of benzene and methylbenzene in this topic. Four are electrophilic substitutions of the ring, one is addition of hydrogen, and one is oxidation of a side chain.

Why does benzene substitute instead of add? An alkene such as cyclohexene is attacked by electrophiles such as Br₂ and adds across the C=C bond. If benzene did the same, the product would lose the delocalised π system. From Topic 29, the enthalpy change of hydrogenation of benzene is −208 kJ mol⁻¹, but three isolated C=C bonds would give 3 × (−120) = −360 kJ mol⁻¹. The difference, 152 kJ mol⁻¹, is the delocalisation (aromatic stabilisation) energy. An addition reaction would give this up. In substitution, an H atom is replaced and the ring is restored, so the stabilisation is kept.

Substitution keeps the delocalised ring. Addition would destroy it and make the product about 150 kJ mol⁻¹ less stable than benzene.

Why does benzene react more slowly than an alkene? The π electrons in benzene are spread over six carbons, so the electron density at any one carbon is lower than in a localised C=C bond. Benzene cannot polarise a non-polar molecule such as Br₂ strongly enough, and it does not decolourise bromine water. It needs a stronger electrophile, which is why each reaction below begins by making one.

A simple test that distinguishes benzene, methylbenzene and cyclohexene.

The general mechanism: electrophilic substitution. All the ring reactions in sections 2 to 4 follow the same two steps. (Electrophile = a species that accepts a pair of electrons.)

1. Attack. A pair of π electrons from the ring forms a bond to the electrophile E⁺. This is the slow step. The ring now has a positive charge spread over five carbons (the intermediate), and the delocalisation is broken, with one carbon bonded to both E and H.
2. Loss of H⁺. The electrons of the C–H bond return to the ring, H⁺ leaves, and the delocalised ring is restored. The overall change is that H is replaced by E.

Worked example. Explain why benzene reacts with bromine by substitution, not addition. [3]

(1) Benzene has a delocalised π system that makes it unusually stable (aromatic stabilisation). (2) Addition would destroy the delocalisation, so the product would be much less stable. (3) In substitution the ring is restored after H⁺ is lost, so the stabilisation is kept.

2. Halogenation of the Ring

Benzene reacts with chlorine or bromine in the presence of a halogen carrier, which is anhydrous aluminium chloride (AlCl₃) or aluminium bromide (AlBr₃). Iron(III) halides, FeCl₃ and FeBr₃, work in the same way and are accepted in examination answers; iron filings added to bromine make FeBr₃ in the flask.

Making the electrophile. A bromine molecule is non-polar and is not a strong enough electrophile for benzene. The Lewis acid AlBr₃ has an empty orbital and accepts a lone pair from one bromine atom. This polarises the Br–Br bond so strongly that the reaction behaves as if Br⁺ were produced:

Br₂ + AlBr₃ → Br⁺ + AlBr₄⁻

Bromination of benzene. The first curly arrow starts at the π ring and ends at Br⁺. The second starts at the C–H bond and ends inside the ring.

Drawing the mechanism (what earns the marks).

1. A curly arrow from the π ring (the circle) to the electrophile.
2. The intermediate: a hexagon with a partial circle (horseshoe) covering five carbons, a + sign, and Br and H both on the same carbon.
3. A curly arrow from the C–H bond back into the ring, giving H⁺ and bromobenzene.
4. H⁺ reacts with AlBr₄⁻: H⁺ + AlBr₄⁻ → HBr + AlBr₃. The catalyst is regenerated.

Methylbenzene. Under the same conditions the CH₃ group is already on the ring, and the bromine enters at positions 2 and 4 (Section 7), giving a mixture of 2-bromomethylbenzene and 4-bromomethylbenzene. Methylbenzene reacts faster than benzene.

Worked example. Write an equation for the formation of the electrophile in the chlorination of benzene and state the role of AlCl₃. [2]

Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻ [1]. AlCl₃ is a halogen carrier (Lewis acid catalyst): it accepts a lone pair from chlorine to make a stronger electrophile and is regenerated [1].

3. Nitration

Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid at a temperature between 25 °C and 60 °C to form nitrobenzene, a pale yellow liquid.

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

HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺

Nitration of benzene. The electrophile is NO₂⁺. Sulfuric acid is regenerated, so it acts as a catalyst.

The mechanism has the same two steps as in Section 1: NO₂⁺ accepts a pair of π electrons to give the intermediate, then H⁺ is lost. The H⁺ combines with HSO₄⁻ to regenerate H₂SO₄.

Why keep the temperature below 60 °C? The nitro group deactivates the ring, but at higher temperatures a second NO₂ group can still enter, giving 1,3-dinitrobenzene (the second group goes to the 3-position, Section 7). Controlling the temperature keeps the product mainly mononitro.

Methylbenzene is nitrated more easily than benzene because the CH₃ group activates the ring. The product is a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene. With more forcing conditions the ring takes three nitro groups to give 2,4,6-trinitromethylbenzene (TNT).

Worked example. State the reagents and conditions for converting benzene into nitrobenzene, and write an equation for the formation of the electrophile. [3]

Concentrated HNO₃ and concentrated H₂SO₄ [1], at a temperature between 25 °C and 60 °C [1]. HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺ [1].

4. Friedel–Crafts Alkylation and Acylation

The Friedel–Crafts reactions form a new carbon–carbon bond to the ring. Both need anhydrous AlCl₃ as the catalyst and heat.

Friedel–Crafts alkylation. AlCl₃ helps the C–Cl bond to break, and is regenerated in the last step.

Friedel–Crafts acylation. The electrophile is the acylium ion CH₃CO⁺.

The two mechanisms have three stages: (1) AlCl₃ removes Cl⁻ from the halogenoalkane or acyl chloride to make the electrophile; (2) the electrophile attacks the ring, giving the intermediate, and H⁺ is lost; (3) H⁺ + AlCl₄⁻ → HCl + AlCl₃.

The same idea works for other halogenoalkanes and other halogen carriers. For example, BrCH₂CH₂CH₂CH₂Br with FeBr₃ gives the electrophile BrCH₂CH₂CH₂CH₂⁺ (and FeBr₄⁻), which can attack a ring in the same two-step way.

Alkylation can go on too long. An alkyl group activates the ring, so the product (methylbenzene) reacts faster than benzene and some dimethylbenzenes form. Using an excess of benzene keeps this small. Acylation does not have this problem: the C=O group deactivates the ring, so only one acyl group is added.

Methylbenzene. Alkylation or acylation of methylbenzene puts the new group at positions 2 and 4 (Section 7).

Worked example. Name the organic product when benzene reacts with CH₃COCl and AlCl₃ on heating, and write an equation for the formation of the electrophile. [2]

Phenylethanone [1]. CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻ [1].

5. Side-Chain Oxidation and Hydrogenation

Oxidation of the side chain. An alkyl side chain on the ring is oxidised to a carboxylic acid group by hot alkaline KMnO₄, followed by dilute acid. Whatever the length of the chain, the product is benzoic acid, because the whole chain except the carbon nearest the ring is lost (as carbon dioxide or carbonate). The ring is not affected.

Any alkyl side chain with a hydrogen on the carbon next to the ring gives benzoic acid. tert-Butylbenzene has no such hydrogen and is not oxidised.

For methylbenzene: C₆H₅CH₃ + 2KMnO₄ → C₆H₅COOK + 2MnO₂ + KOH + H₂O. The purple solution turns to a brown precipitate of MnO₂. Adding dilute acid then converts the benzoate into benzoic acid, a white solid: C₆H₅COO⁻ + H⁺ → C₆H₅COOH.

Worked example. 4.60 g of methylbenzene is oxidised to benzoic acid and 4.20 g of benzoic acid is obtained. Calculate the percentage yield. [3]

M_r(C₆H₅CH₃) = 92.13 and M_r(C₆H₅COOH) = 122.12. Amount of methylbenzene = 4.60 / 92.13 = 0.0499 mol [1]. The mole ratio is 1 : 1, so the theoretical mass of benzoic acid = 0.0499 × 122.12 = 6.10 g [1]. Percentage yield = 4.20 / 6.10 × 100 = 68.9 % [1].

Hydrogenation of the ring. Benzene adds hydrogen when heated with hydrogen gas and a nickel or platinum catalyst. All three π bonds are saturated and the product is cyclohexane. Methylbenzene gives methylcyclohexane.

Hydrogenation needs a catalyst and heat. It is an addition reaction, but the ring is lost because all three π bonds react.

C₆H₆ + 3H₂ → C₆H₁₂   (H₂, Ni or Pt, heat). One mole of benzene needs three moles of H₂.

6. Ring or Side Chain?

Methylbenzene has two places for a halogen to react: the ring hydrogens and the CH₃ hydrogens. The conditions decide which, because the two reactions use different mechanisms.

The same reagent, Br₂, gives different products depending on the conditions.

The side-chain reaction is the same free-radical substitution you met for alkanes (Topic 14), with the benzene ring as a large substituent:

• Initiation: Br₂ → 2Br• (UV light)
• Propagation: C₆H₅CH₃ + Br• → C₆H₅CH₂• + HBr, then C₆H₅CH₂• + Br₂ → C₆H₅CH₂Br + Br•
• Termination: two radicals combine, for example 2Br• → Br₂

If bromine is in excess, further H atoms on the side chain can also be replaced, giving C₆H₅CHBr₂ and C₆H₅CBr₃.

Worked example. Methylbenzene is mixed with bromine. Predict the main product (a) with AlBr₃ in the dark, and (b) in sunlight with no catalyst. [2]

(a) Ring substitution: 2-bromomethylbenzene and 4-bromomethylbenzene [1]. (b) Side-chain substitution: (bromomethyl)benzene, C₆H₅CH₂Br [1].

7. Directing Effects

When a benzene ring already has one substituent, the second group does not go in at random. The first group directs the new group to particular positions. We number the carbon holding the first group as C1.

The six substituents you must know, and the positions to which they direct.

One example of each type. The groups are not equally strong, so real reactions give mixtures, but the pattern shown is the one to quote.

Why do they direct differently? The –NH₂ and –OH groups have a lone pair on the atom bonded to the ring, which can be donated into the π system. An alkyl group pushes electrons into the ring through the sigma bond. In each case the electron density is increased, especially at positions 2, 4 and 6, so electrophiles attack there and the ring is more reactive than benzene. The –NO₂, –COOH and –COR groups have a C=O or N=O bond or a positively polarised atom next to the ring. They withdraw electrons, mostly from positions 2, 4 and 6, so the electron-poor ring reacts more slowly, and the electrophile attacks at positions 3 and 5, where the ring is least depleted of electrons.

Reactivity order. The reactivity of a ring towards electrophiles follows the activating or deactivating effect of the substituent: C₆H₅OH > C₆H₅CH₃ > C₆H₆ > C₆H₅COOH. This is why phenol is nitrated with dilute nitric acid at room temperature (Topic 32), while benzene needs concentrated acids and a higher temperature.

Worked example. Predict the structure of the main product when each compound is nitrated with concentrated HNO₃ and H₂SO₄ at 50 °C: (a) phenylethanone, (b) ethylbenzene. [2]

(a) –COCH₃ is a 3-directing group, so the product is 3-nitrophenylethanone (NO₂ on carbon 3) [1]. (b) The ethyl group is 2,4-directing, so the product is a mixture of 2-nitroethylbenzene and 4-nitroethylbenzene [1].

8. Planning a Synthesis

Directing effects control the order in which reactions must be carried out. If you want a particular isomer, ask: which group should go on first so that it directs the second group to the right place?

The same two reactions in a different order give different isomers.

The method is the same every time:

1. Draw the target and the starting material.
2. Decide which group is added by substitution and which is made by a side-chain reaction (for example, the COOH group by oxidation of CH₃).
3. Choose the order so that the first group directs the second to the position needed. For a 1,3-relationship that means a 3-directing group goes on first; for a 1,2 or 1,4 relationship, a 2,4-directing group goes on first.
4. Write the reagents and conditions at each step, and name the intermediates.

A typical three-step route from benzene is alkylation, then nitration, then reduction: benzene + RCl/AlCl₃ gives an alkylbenzene; conc. HNO₃/conc. H₂SO₄ puts NO₂ at the 4-position (and the 2-position); Sn with concentrated HCl, then NaOH(aq), reduces NO₂ to NH₂ (Topic 34). The alkyl group must go first, because it directs the nitro group to positions 2 and 4; the nitro group is then reduced after it has done its job. If you reduced first, the amino group would be attacked (oxidised or protonated) by the nitrating mixture, so nitration must come before the reduction.

Worked example. Suggest a two-step synthesis of 3-nitrobenzoic acid from methylbenzene, giving reagents and conditions. [4]

Step 1: oxidise the side chain with hot alkaline KMnO₄, then add dilute H₂SO₄, to give benzoic acid [2]. Step 2: nitrate with concentrated HNO₃ and concentrated H₂SO₄ [1]; the COOH group is 3-directing, so the NO₂ goes to position 3 [1]. (If nitration were done first, the CH₃ group would direct to the 2- and 4-positions, so the wrong isomers would form.)

9. Quick Sheet and Checklist

Checklist. Before the exam you should be able to:

• [ ] explain why benzene undergoes substitution rather than addition, using delocalisation energy
• [ ] give the reagents and conditions for halogenation, nitration, Friedel–Crafts alkylation and acylation, side-chain oxidation and hydrogenation
• [ ] write the equation for forming each electrophile (Br⁺, NO₂⁺, R⁺, RCO⁺)
• [ ] draw the mechanism of electrophilic substitution with curly arrows, the intermediate and the regeneration of H⁺ or the catalyst
• [ ] state what is formed from methylbenzene in each reaction, including the 2- and 4-isomers
• [ ] state which side chains are oxidised to benzoic acid and which are not
• [ ] predict ring or side-chain halogenation from the conditions
• [ ] state the directing effects of –NH₂, –OH, –R, –NO₂, –COOH and –COR, and predict the product of a second substitution
• [ ] plan a two- or three-step synthesis, giving the reagents, the intermediates and the reason for the order