Hydrocarbons

📚The chemistry

The benzene ring is electron-rich (its delocalised π system), so it attracts electrophiles, like an alkene. Unlike an alkene, it does not add them: it substitutes, because keeping the delocalised ring is worth about 150 kJ mol⁻¹ of stability (topic 29). Benzene is also less reactive than an alkene: its π electrons are spread over six carbons rather than concentrated between two, so it needs a stronger electrophile, which is what the catalysts below provide.

The reactions of benzene and methylbenzene

Halogenation

Cl₂ or Br₂ with a halogen carrier catalyst, AlCl₃ or AlBr₃ (or FeBr₃), at room temperature, in the absence of UV light:

Nitration

A mixture of concentrated HNO₃ and concentrated H₂SO₄, at 25–60 °C. Above that, further nitration gives dinitrobenzene.

Friedel–Crafts alkylation and acylation

Both form a new C–C bond to the ring, using AlCl₃ and heat:

Oxidation of the side chain

Heat with hot alkaline KMnO₄, then acidify with dilute acid. The whole alkyl side chain, however long, is oxidised to a COOH group on the ring, giving benzoic acid. The ring itself is not attacked.

Hydrogenation

H₂ with a Pt or Ni catalyst and heat adds across the ring, destroying the delocalisation and forming a cyclohexane ring. It needs harsher conditions than alkene hydrogenation.

The mechanism: electrophilic substitution

Every substitution follows the same three stages: generate the electrophile, attack the ring, lose H⁺ to restore it.

Nitration

1. Electrophile: the nitronium ion, NO₂⁺, formed by the sulfuric acid: \[ \mathrm{HNO_3 + 2H_2SO_4 \rightarrow NO_2^{+} + H_3O^{+} + 2HSO_4^{-}} \]
2. Attack: curly arrow from the delocalised π system (from the circle) to the N of NO₂⁺. This forms an intermediate in which one carbon is sp³ (carrying both H and NO₂) and the positive charge is spread over the other five carbons — draw it as a horseshoe (a partial ring, opening towards the sp³ carbon) with a + inside. The delocalisation is temporarily broken.
3. Loss of H⁺: curly arrow from the C–H bond back into the ring. The H⁺ is removed (by HSO₄⁻, regenerating H₂SO₄, which is therefore a catalyst), and the full delocalised ring is restored: nitrobenzene.

Bromination

1. Electrophile: the halogen carrier polarises Br₂ so strongly that it gives Br⁺: \[ \mathrm{Br_2 + AlBr_3 \rightarrow Br^{+} + AlBr_4^{-}} \]
2. Attack: curly arrow from the π system to Br⁺, giving the horseshoe intermediate.
3. Loss of H⁺: arrow from C–H into the ring; the H⁺ combines with AlBr₄⁻ to form HBr and regenerate AlBr₃: \[ \mathrm{H^{+} + AlBr_4^{-} \rightarrow HBr + AlBr_3} \]

Friedel–Crafts reactions work the same way: AlCl₃ generates CH₃⁺ from CH₃Cl, or CH₃CO⁺ from CH₃COCl.

Side chain or ring?

Methylbenzene has two places a halogen can go, and the conditions decide which:

• UV light (or boiling, no catalyst): free-radical substitution in the side chain, exactly as for an alkane (topic 14). C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl, giving (chloromethyl)benzene; further substitution gives C₆H₅CHCl₂ and C₆H₅CCl₃.
• AlCl₃ catalyst, room temperature, no UV: electrophilic substitution in the ring, giving a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.

Directing effects

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

The pattern to remember: groups that donate electron density into the ring (a lone pair on N or O, or an alkyl group’s inductive effect) direct to 2 and 4 and make the ring more reactive. Groups that withdraw electron density (containing a C=O or N=O attached to the ring) direct to 3 and make it less reactive. So nitrating methylbenzene gives 2- and 4-nitromethylbenzene, easily; nitrating benzoic acid gives 3-nitrobenzoic acid, more slowly.