Why are alkanes useful fuels and how do they react?
Alkanes as saturated hydrocarbons from crude oil, fractional distillation, cracking. Combustion of alkanes and the formation of pollutants. Free-radical substitution of alkanes by halogens, including initiation, propagation and termination.
A focused answer to the AQA A-Level Chemistry 3.3.2 specification points on alkanes. Covers fractional distillation and cracking of crude oil, complete and incomplete combustion, pollutants and catalytic converters, and the free-radical substitution mechanism.
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What this dot point is asking
AQA wants you to explain how alkanes are separated from crude oil and cracked, write equations for complete and incomplete combustion, describe the pollutants formed and how catalytic converters remove them, and give the full free-radical substitution mechanism.
Crude oil, fractional distillation and cracking
Crude oil is separated by fractional distillation: it is heated and the vapour rises up a column with a temperature gradient, so each fraction condenses at its boiling point. Boiling point rises with chain length because of stronger van der Waals (London) forces between larger molecules.
Cracking breaks long-chain alkanes into shorter, more useful molecules.
- Thermal cracking: high temperature and pressure, produces a high proportion of alkenes.
- Catalytic cracking: zeolite catalyst, slight pressure, lower temperature, produces branched alkanes, cycloalkanes and aromatics for motor fuels.
Combustion
Incomplete combustion (limited oxygen) gives toxic carbon monoxide () and carbon (soot).
Alkanes are saturated (only single and bonds), non-polar and have only weak van der Waals forces between molecules, so they are generally unreactive. They burn readily and react with halogens only under ultraviolet light. Their lack of polarity means they are not attacked by nucleophiles or electrophiles, which is why their main reactions are combustion and radical substitution.
Free-radical substitution
The reaction of an alkane with a halogen needs UV light to start it, and proceeds through three stages with single-electron (fishhook) movement.
Because each propagation step regenerates a radical, a single initiation event can trigger thousands of product molecules. The reaction is hard to control: further substitution gives a mixture of , , and , and termination of different radicals gives traces of ethane and other products. This poor selectivity is why radical substitution is rarely used to make a single pure product in industry.
Try this
Q1. Write an equation for the incomplete combustion of propane forming carbon monoxide. [2 marks]
- Cue. .
Q2. Name the three stages of free-radical substitution. [3 marks]
- Cue. Initiation, propagation, termination.
Exam-style practice questions
Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
AQA 20185 marksMethane reacts with chlorine in the presence of ultraviolet light to form chloromethane. Outline the free-radical substitution mechanism, naming the three stages and giving an equation for each.Show worked answer β
Name and give an equation for each stage:
Initiation: UV light causes homolytic fission of the chlorine molecule, , forming two chlorine radicals.
Propagation (two steps): , then . The chlorine radical is regenerated, so the chain continues.
Termination: two radicals combine, e.g. (or ).
Markers reward naming each stage, the UV/homolytic fission point, both propagation steps with the regenerated radical, and a valid termination step.
AQA 20204 marksOctane, , undergoes complete and incomplete combustion in a car engine. Write balanced equations for the complete combustion of octane and for incomplete combustion producing carbon monoxide only, and explain why nitrogen oxides also form.Show worked answer β
Complete combustion: .
Incomplete combustion (carbon monoxide only): .
Nitrogen oxides form because the high temperature and pressure inside the engine give nitrogen and oxygen from the air enough energy to react, , which is otherwise unreactive at ordinary temperatures.
Markers reward both balanced equations (correctly balanced oxygen) and the explanation that the high engine temperature lets atmospheric nitrogen react with oxygen.
Related dot points
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Sources & how we know this
- AQA A-level Chemistry (7405) specification β AQA (2015)