From C7 Crude Oil, Fuels & Feedstock you already know crude oil is a mixture of hydrocarbons separated by fractional distillation, that cracking makes smaller molecules and alkenes, and that alkanes burn as fuels. A-Level keeps all of that and adds the detail: the pollutants of the internal combustion engine and how catalytic converters remove them, the conditions and economics of thermal vs catalytic cracking, and the free-radical substitution mechanism — with the UV condition the exam insists on.
Fractional distillation
Alkanes are saturated hydrocarbons — only single C–C and C–H bonds, general formula CnH2n+2. Crude oil (petroleum) is a mixture of them, and the first job of a refinery is to separate it by fractional distillation.
The mixture is vaporised and fed into a fractionating column that is hot at the bottom and cool at the top. As the vapours rise and cool, each group of hydrocarbons — a fraction — condenses and is drawn off at the height where the temperature matches its boiling-point range. Because boiling point rises with chain length (bigger molecules have stronger van der Waals forces), the largest molecules condense low down and the smallest gases leave the top. Nothing chemical happens: it is a physical separation of a mixture.
- A “fraction” is a group of hydrocarbons with similar boiling points — not a single pure compound.
- It is a physical separation — the hydrocarbons are not chemically combined, and no bonds are broken. Boiling point (not reactivity) does the sorting.
- Boiling point rises with chain length because bigger molecules have stronger van der Waals forces — and a straight chain boils higher than its branched isomer.
GCSE groundwork: C7 — fractional distillation.
🧪 Exam-style questions
Crude oil is separated into fractions by fractional distillation. State what is meant by a fraction.
Show answer
A group of hydrocarbons (compounds) with similar boiling points. 1 mark
Which of these alkanes has the highest boiling point? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Cracking
Fractional distillation has an economics problem: crude oil yields more long-chain fractions than the world wants, and not enough of the short-chain fuels it demands. Cracking fixes this by breaking C–C bonds in long alkanes to make shorter, more useful molecules — a smaller alkane plus one or more alkenes. Two versions are examined:
| Thermal cracking | Catalytic cracking | |
|---|---|---|
| Conditions | high temperature, high pressure | high temperature, slight pressure, zeolite catalyst |
| Main products | a high percentage of alkenes | motor fuels and aromatic hydrocarbons |
| Used for | alkenes for the polymer industry | branched/cyclic fuels; cheaper (catalyst lowers the temperature needed) |
The economic reason is supply and demand: cracking converts low-value surplus fractions into high-demand fuels and into the alkenes that make plastics. Whatever the products, the reaction must balance for atoms — every carbon and hydrogen in the long alkane reappears in the products.
- Balance the atoms. Examiner reports flag invented product formulae — count every C and H; a saturated alkane needs CnH2n+2, each alkene is CnH2n.
- Thermal = high pressure, alkenes; catalytic = zeolite, slight pressure, aromatics/fuels. Don’t swap the conditions.
- Cracking is a chemical reaction (C–C bonds broken) — unlike the physical separation of fractional distillation.
🧪 Exam-style questions
Which statement is correct about thermal cracking? Tick (✓) one box.
Hexadecane (C16H34) decomposes to form an alkane with eight carbon atoms and two different unsaturated compounds. Which equation could represent this reaction? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Combustion & pollutants
Alkanes are burned as fuels. With plenty of oxygen, combustion is complete; with too little, it is incomplete — and the difference is a matter of life and death, because incomplete combustion makes toxic carbon monoxide.
Complete (plenty of O2) → carbon dioxide and water:
CH4 + 2O2 → CO2 + 2H2O
Incomplete (limited O2) → carbon monoxide and/or carbon (soot), plus water:
2CH4 + 3O2 → 2CO + 4H2O (or, with even less O2, C is formed)
An internal combustion engine burns fuel fast and hot, so it releases a family of pollutants:
| Pollutant | Where it comes from | Problem |
|---|---|---|
| CO | incomplete combustion | toxic (binds to haemoglobin) |
| C / particulates | incomplete combustion | soot; respiratory problems |
| Unburned hydrocarbons | fuel that escapes unburnt | smog |
| NOx | N2 + O2 from the air react at the engine’s high temperature | acid rain, smog |
| SO2 | sulfur impurities in the fuel | acid rain |
Catalytic converters clean up three of these — CO, NOx and unburned hydrocarbons — on a platinum/rhodium honeycomb. In one reaction the CO is oxidised while the nitrogen oxide is reduced:
2CO + 2NO → 2CO2 + N2
The unburned hydrocarbons are oxidised too, to CO2 and water. So all three pollutants leave as harmless CO2, N2 and H2O — the H2O coming from the hydrogen in those hydrocarbons.
SO2 is a different problem: it comes from sulfur in the fuel and is removed not by a catalytic converter but from power-station flue gases, by reaction with calcium oxide or calcium carbonate, which are basic and neutralise the acidic gas (flue-gas desulfurisation):
CaO + SO2 → CaSO3
Complete combustion
CH4 + 2O2 → CO2 + 2H2O — every carbon leaves as CO2.
Incomplete combustion
2CH4 + 3O2 → 2CO + 4H2O — or, with even less O2, C. Water always forms.
Catalytic converter
CO and NOx: 2CO + 2NO → 2CO2 + N2. The hydrocarbons (HC) oxidise to CO2 + H2O — that’s where the H2O comes from.
Flue-gas desulfurisation
CaO + SO2 → CaSO3 — a base neutralising the acidic gas.
- Incomplete combustion makes CO and/or C — not CO2. Water is still a product.
- NOx forms in the engine from the nitrogen and oxygen in the air — not from the fuel — because it is hot enough for N2 and O2 to react.
- Catalytic converters do not remove SO2. SO2 comes from sulfur in the fuel and is removed from flue gases by CaO or CaCO3.
Combustion is oxidation in action — Oxidation & Reduction (3.1.7) puts numbers on it. The SO2 clean-up is the basic-oxide chemistry of Group 2 — uses.
🧪 Exam-style questions
Which statement is not correct about the pollutant sulfur dioxide? Tick (✓) one box.
Explain how oxides of nitrogen are formed in a car engine.
Show answer
Nitrogen and oxygen from the air react together. 1 mark
This happens at the high temperature inside the engine (which provides the activation energy). 1 mark
The nitrogen comes from the air, not the fuel — a common slip.
Source: AQA A-Level Chemistry past papers.
Free-radical substitution
Alkanes are unreactive, but in ultraviolet light they undergo substitution with halogens. Methane and chlorine give chloromethane and hydrogen chloride, by a free-radical mechanism — and the mechanism has three kinds of step. Radicals (species with an unpaired electron) are shown with a dot; no curly arrows are used.
Initiation — UV light breaks the Cl–Cl bond evenly (homolytic fission), making two radicals:
Cl2 UV light 2Cl•
Propagation — a radical reacts and makes a new radical, so the chain continues:
Cl• + CH4 → •CH3 + HCl
•CH3 + Cl2 → CH3Cl + Cl•
Termination — two radicals combine, ending a chain:
Cl• + Cl• → Cl2
•CH3 + Cl• → CH3Cl
•CH3 + •CH3 → C2H6
Interactive — name the step
Now build the whole mechanism for CH4 + Cl2 → CH3Cl + HCl — tap the four steps in order. Two of the tiles don’t belong.
Further substitution — when chlorine is in excess
Nothing in the mechanism stops at one substitution. Chloromethane still has three hydrogens, so a Cl• can attack it exactly as it attacked methane — the same pair of propagation steps, swapping one more H for Cl:
Cl• + CH3Cl → •CH2Cl + HCl
•CH2Cl + Cl2 → CH2Cl2 + Cl•
With chlorine in excess that keeps happening, one hydrogen at a time, until none are left:
CH4 → CH3Cl → CH2Cl2 → CHCl3 → CCl4
Each arrow is another pair of propagation steps, so the reaction delivers a mixture — chloromethane, dichloromethane, trichloromethane and tetrachloromethane, all halogenoalkanes. With excess methane instead, a Cl• is far more likely to meet CH4 than CH3Cl, so monosubstitution to CH3Cl is favoured.
Substitution along a longer chain
Methane’s four hydrogens are all identical. In a longer alkane they are not: every carbon along the chain carries hydrogens, and a Cl• abstracts whichever one it happens to hit — the mechanism has no way of aiming. So even monosubstitution of butane gives two different products:
This is why free-radical substitution is a poor way to make one clean product:
- No control over position — substitution can happen at any carbon along the chain, so longer alkanes give a mixture of isomers (here 1-chlorobutane and 2-chlorobutane).
- No control over how many substitutions — every product still has hydrogens, so di- and tri-substituted molecules build up alongside the mono-products.
The result is a mixture that must be separated afterwards — by fractional distillation, since the products have different boiling points.
- State the UV light. Examiner reports are explicit: free-radical substitution needs UV light for initiation — forget it and you can lose the mark even if the rest is right.
- Propagation vs termination. A propagation step has a radical on both sides (one in, one out); a termination step has radicals only on the left (two combine).
- Radicals get a dot, not a curly arrow — and the product is a mixture, for two reasons: substitution runs on with excess Cl2 (CH3Cl → CH2Cl2 → CHCl3 → CCl4), and beyond methane it can hit any position along the chain.
🧪 Exam-style questions
An excess of methane reacts with chlorine in the presence of ultraviolet radiation. What are the main products? Tick (✓) one box.
Which is a propagation step in the chlorination of methane? Tick (✓) one box.
2-Bromopropane reacts with bromine to form 2,2-dibromopropane. What is the name of the mechanism of this reaction? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Capstone: one barrel of crude oil
The whole topic is one journey. A barrel of crude oil is fractionally distilled into fractions by boiling point; the surplus long-chain fractions are cracked into the short-chain fuels and alkenes the market wants; those fuels are burned — ideally completely to CO2 and water, but in practice with a tail of pollutants that catalytic converters and flue-gas desulfurisation have to clean up; and the alkanes themselves, given UV light, undergo free-radical substitution to become the halogenoalkanes the next topic builds on.
The mechanism checklist that protects the marks: state the UV condition; show radicals with a dot; give an initiation, a propagation (radical in, radical out) and a termination (two radicals combine) step; and remember the product is a mixture.
Interactive — build the mechanism
Tap four tiles in order — initiation, both propagations, then one termination (any of the three counts). Two of the tiles are traps.
- Alkanes are saturated hydrocarbons, CnH2n+2. Crude oil is a mixture separated by fractional distillation (a physical separation) into fractions — groups of hydrocarbons with similar boiling points.
- Cracking breaks C–C bonds into smaller, more useful molecules (an alkane + alkene[s]). Thermal: high temperature + high pressure → many alkenes. Catalytic: high temperature, slight pressure, zeolite catalyst → motor fuels + aromatics. Balance every C and H.
- Complete combustion → CO2 + H2O; incomplete (limited O2) → CO and/or C (soot) + H2O.
- Engine pollutants: CO, C/particulates, unburned hydrocarbons, NOx (N2 + O2 from the air react at the engine’s high temperature), and SO2 (from sulfur impurities).
- Catalytic converters remove CO, NOx and unburned hydrocarbons (e.g. 2CO + 2NO → 2CO2 + N2). SO2 is removed from flue gases by CaO or CaCO3 (bases that neutralise it) — not by a catalytic converter.
- Free-radical substitution (methane + chlorine, UV light): initiation Cl2 → 2Cl•; propagation Cl• + CH4 → •CH3 + HCl, then •CH3 + Cl2 → CH3Cl + Cl•; termination two radicals combine. The product is a mixture: with excess Cl2 substitution runs on (CH3Cl → CH2Cl2 → CHCl3 → CCl4), and longer alkanes substitute at any position along the chain (butane → 1- and 2-chlorobutane) — so the products need separating. Excess methane favours CH3Cl.