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CIE 9701 Chemistry · A Level · Topic 30

Hydrocarbons

Clear, syllabus-mapped CIE 9701 Chemistry revision notes on hydrocarbons: explanations, worked examples and exam technique, then a free targeted practice drill.

CIE 9701 ChemistryA LevelFree revision notes
Contents: 8 sections

The single subtopic here, 30.1 Arenes, is printed under "A Level subject content" in the 9701 syllabus and all four of its objectives carry the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 14 of the same name, which covers alkanes and alkenes. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

Syllabus points

30.1 Arenes

Why arenes substitute rather than add

Benzene has a delocalised pi system of six electrons spread over all six carbons, and that delocalisation makes it about 152 kJ mol⁻¹ more stable than a molecule with three isolated double bonds would be, as calculated in topic 29.

An addition reaction across the ring would use up two of the pi electrons in forming sigma bonds, so the delocalised system would be destroyed and that stabilisation lost. A substitution reaction replaces a hydrogen atom and leaves the delocalised ring intact, so the stabilisation is retained.

That is why benzene, despite having a high electron density, does not behave like an alkene. It does not decolourise bromine water, and it requires a catalyst and forcing conditions where an alkene reacts on contact.

The ring is electron rich, so the species that attack it are electrophiles, electron-pair acceptors. Every reaction in this topic except the last two is electrophilic substitution.

The reactions of benzene

ReactionReagents and conditionsProduct
ChlorinationCl₂ with AlCl₃ catalystChlorobenzene
BrominationBr₂ with AlBr₃ catalystBromobenzene
NitrationConcentrated HNO₃ with concentrated H₂SO₄, 25 to 60 °CNitrobenzene
Friedel-Crafts alkylationCH₃Cl with AlCl₃, heatMethylbenzene
Friedel-Crafts acylationCH₃COCl with AlCl₃, heatPhenylethanone
HydrogenationH₂ with Pt or Ni catalyst, heatCyclohexane

Two conditions carry marks on their own. The nitration temperature must be between 25 °C and 60 °C: below that the reaction is too slow, and above about 60 °C a second nitro group is introduced and dinitrobenzene forms. And the aluminium halide catalyst is a halogen carrier, sometimes written as anhydrous, because water destroys it.

Hydrogenation is the one addition reaction, and it needs the harshest conditions of the lot, which is itself evidence of how stable the ring is.

The mechanism of electrophilic substitution

The mechanism has the same three stages every time, and only the electrophile changes.

Stage 1: generate the electrophile.

For nitration, sulfuric acid protonates nitric acid, which then loses water to give the nitronium ion:

HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻

For bromination, the aluminium bromide polarises the bromine molecule and generates Br⁺, or a strongly polarised species that behaves as one:

Br₂ + AlBr₃ → Br⁺ + AlBr₄⁻

Stage 2: the electrophile attacks the ring. Two of the delocalised pi electrons form a bond to the electrophile, giving a positively charged intermediate in which the delocalisation is broken over one carbon and the remaining four pi electrons are delocalised over five carbons. Draw this intermediate with a partial ring and a positive charge on it, and with the electrophile and the hydrogen both attached to the same carbon by full lines.

Stage 3: lose a proton and restore the ring. The carbon-to-hydrogen bond breaks heterolytically, and the pair of electrons returns to the ring, restoring the delocalised system. The H⁺ released regenerates the catalyst:

AlBr₄⁻ + H⁺ → AlBr₃ + HBr

The mechanism marks are almost always for the same four things: the correct electrophile, a curly arrow from the ring to the electrophile, the intermediate drawn with a partial ring and a positive charge, and a curly arrow from the C-H bond back into the ring.

Side-chain or ring? The conditions decide

Methylbenzene has two sites that halogen can attack, and which one reacts is controlled entirely by the conditions.

The reason is the mechanism each condition initiates. A halogen carrier generates an electrophile, which is attracted to the electron-rich ring. Ultraviolet light causes homolytic fission of the halogen molecule into free radicals, which attack the C-H bonds of the alkyl side-chain just as they would in an alkane.

Naming the condition and naming the mechanism together is what a full answer needs.

Oxidation of the side-chain

Hot alkaline potassium manganate(VII), followed by acidification with dilute acid, oxidises any alkyl side-chain on a benzene ring completely to a carboxylic acid group attached directly to the ring.

Methylbenzene gives benzoic acid. So does ethylbenzene, and so does propylbenzene: the whole side-chain is cut back to a single COOH group whatever its length, with the extra carbons lost as carbon dioxide. That result surprises people and is examined for exactly that reason.

The ring itself is untouched, which is another demonstration of its stability.

Directing effects

When an arene that already carries a substituent undergoes electrophilic substitution, the new group does not go in at random. The existing substituent directs it.

Existing substituentDirects toEffect on rate
-NH₂2, 4 and 6Activating, faster than benzene
-OH2, 4 and 6Activating
-R (alkyl)2, 4 and 6Activating
-NO₂3 and 5Deactivating, slower than benzene
-COOH3 and 5Deactivating
-COR3 and 5Deactivating

The pattern is easier to hold once you see the reason. Groups that release electron density into the ring, whether by a lone pair delocalising into the pi system, as with -NH₂ and -OH, or by an inductive effect, as with an alkyl group, make the ring more electron rich and therefore more attractive to an electrophile. They activate it, and they concentrate the extra density at positions 2, 4 and 6.

Groups that withdraw electron density from the ring, as -NO₂, -COOH and -COR do, make it less electron rich and therefore less reactive, and what density remains is relatively greater at positions 3 and 5.

So: electron-donating groups are 2,4-directing and activating; electron-withdrawing groups are 3-directing and deactivating. Learn one example of each and derive the rest.

This has practical consequences for synthesis. To make 3-nitrobenzoic acid, oxidise the side-chain of methylbenzene to COOH first and then nitrate, because COOH directs to position 3. To make 4-nitromethylbenzene, nitrate first, because the methyl group directs to position 4. The order of the steps decides the product, and questions on synthesis routes turn on exactly this.

Common mistakes

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