Aromatic hydrocarbons

Aromatic hydrocarbons (arenes) are hydrocarbons whose molecules contain a benzene ring. Although three double bonds are drawn in benzene’s formula, it does not decolourise bromine water or KMnO₄ solution as alkenes do. The reason is a single π-electron cloud formed by the six p-electrons of the ring, which gives the molecule special stability (aromaticity). So arenes typically undergo substitution rather than addition, and the ring is kept. Arenes and their derivatives (phenol, phenylamine, nitro compounds) underpin the manufacture of medicines, dyes, explosives, plastics and pesticides. This topic completes the hydrocarbons, after alkanes, alkenes and alkynes.

§1. The structure of benzene

The C–C bond length in benzene lies between a single (0.154 nm) and a double (0.134 nm) bond. Each carbon uses three sp² hybrid orbitals to form σ-bonds to two neighbouring carbons and one hydrogen, while its unhybridised p-orbital stands perpendicular to the ring. The six p-orbitals overlap sideways to form a shared aromatic sextet (6 π-electrons) above and below the ring. The Kekulé formula (three alternating double bonds) is only a convention; a circle inside the ring is often drawn instead.

Homologues of benzene: CₙH₂ₙ₋₆ (n ≥ 6) — methylbenzene (toluene) C₆H₅CH₃, ethylbenzene C₆H₅C₂H₅, dimethylbenzenes (xylenes) C₆H₄(CH₃)₂. Two substituents can be arranged in three ways: ortho (1,2), meta (1,3) and para (1,4). The C₆H₅– group is called phenyl.

Physical properties: benzene is a colourless liquid with a characteristic smell, less dense than water (boils at 80 °C, freezes at 5.5 °C), insoluble in water and a good organic solvent. Its vapour is toxic and carcinogenic.

Worked example

A homologue of benzene has a molar mass of 106 g/mol. Find its formula and count its isomers.

1. 14n − 6 = 106 → n = 8: C₈H₁₀.
2. Isomers: ethylbenzene, 1,2-, 1,3- and 1,4-dimethylbenzene.
3. In total, 4 aromatic isomers.

§2. Electrophilic substitution

The aromatic ring is rich in electrons, so an electrophile (Br⁺, NO₂⁺, R⁺) attacks it; hydrogen leaves as a proton and the aromatic system is restored.

The catalyst’s job is to create a strong electrophile: for example, FeBr₃ polarises the bromine molecule.

§3. Directing effects of substituents

If the ring already carries a substituent, a second one does not enter just anywhere but at particular positions:

• Group I (ortho/para directors): –CH₃, –OH, –NH₂, –OR, halogens. Most of them release electron density into the ring and speed up the reaction (halogens slow it slightly but still direct to ortho and para).
• Group II (meta directors): –NO₂, –COOH, –CHO, –SO₃H, –CN. They withdraw electron density from the ring and slow the reaction.

That is why methylbenzene is nitrated more easily than benzene, and three nitro groups enter positions 2, 4 and 6:

Nitrobenzene is brominated with more difficulty and gives mainly 1-bromo-3-nitrobenzene (meta). The order of introducing substituents therefore matters when planning a synthesis: to make the meta product, nitrate first and then brominate; the reverse order gives ortho and para isomers.

§4. Addition and oxidation

The aromatic system is hard to destroy, so addition happens only under harsh conditions:

• hydrogenation: C₆H₆ + 3H₂ → C₆H₁₂ (Ni, heat, pressure) — cyclohexane;
• chlorination in light: C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (hexachlorocyclohexane, once used as an insecticide).

Benzene is not oxidised by KMnO₄. In its homologues the side chain is oxidised and the carbon attached to the ring becomes a carboxyl group:

Whatever the length of the side chain, one –COOH forms: ethylbenzene also gives benzoic acid (the remaining carbon becomes CO₂). This reaction distinguishes methylbenzene from benzene.

• Combustion: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O — the high carbon content gives a smoky flame.

Worked example

What mass of benzoic acid forms when 9.2 g of methylbenzene is completely oxidised?

1. n(C₆H₅CH₃) = 9.2/92 = 0.1 mol.
2. Each methylbenzene molecule gives one benzoic acid molecule: n(C₆H₅COOH) = 0.1 mol.
3. m = 0.1·122 = 12.2 g.

§5. Preparation, and phenol

Phenol C₆H₅OH has the hydroxyl group attached directly to the ring. The ring draws in the oxygen lone pair and the O–H bond is polarised, so phenol, unlike alcohols, is a weak acid (carbolic acid):

Phenol is weaker even than carbonic acid: CO₂ displaces phenol from a phenoxide solution. Phenylamine also gives a white precipitate (2,4,6-tribromophenylamine) with bromine water, because –NH₂, like –OH, activates the ring.

Industrially, phenol is made mainly by the cumene process: benzene is alkylated with propene to cumene, cumene is oxidised by air, and the hydroperoxide splits into phenol and propanone. Phenol is used to make phenol–formaldehyde resins, caprolactam (for nylon-6), medicines (aspirin) and dyes; its dilute solution was once a disinfectant. Phenol is toxic and burns the skin. Benzene is also converted into styrene (for polystyrene), nitrobenzene and phenylamine, cumene and many other products.

§6. An official test item

Find the sum of the oxidation states of all carbon atoms in 2,4,6-tribromophenylamine.

Worked example

1. The carbon bonded to nitrogen: +1 from the C–N bond (C–C bonds are not counted).
2. Three C–Br carbons: +1 each → +3 in total.
3. Two C–H carbons: −1 each → −2 in total.
4. Sum: +1 + 3 − 2 = +2.

Check yourself

1. What mass of nitrobenzene is obtained by nitrating 15.6 g of benzene at 80 % yield?

2. What forms when methylbenzene is oxidised by acidified KMnO₄?

3. How do you tell phenol from benzene?

4. Which isomer forms mainly when nitrobenzene is brominated?

5. Does benzene decolourise bromine water?

6. What mass of NaOH neutralises 9.4 g of phenol?

7. What volume of H₂ (STP) is needed to hydrogenate 7.8 g of benzene completely?

8. Into which positions does the nitro group mainly go when methylbenzene is nitrated?

Common mistakes

• Writing that benzene decolourises bromine water and KMnO₄.
• Thinking phenol, like alcohols, does not react with alkalis.
• Mixing up directing effects: –NO₂, –COOH are meta directors; –CH₃, –OH are ortho/para directors.
• Writing phenylethanoic acid as the oxidation product of ethylbenzene: benzoic acid forms.
• Thinking bromine water is used to brominate benzene: liquid bromine with FeBr₃ is needed.