Part 2: Alkane fuels and radical substitution

Connect crude-oil processing and combustion to bond changes, then build a radical chain mechanism and explain its limits as a synthetic route.

Saturation and the general formula

An acyclic alkane has formula CₙH₂ₙ₊₂. The two end carbons in an unbranched chain are CH₃; internal carbons are CH₂. Branching changes the arrangement but not the formula. A monocyclic cycloalkane has CₙH₂ₙ because closing the ring replaces two C–H bonds with a C–C bond. Both contain only single carbon–carbon bonds.

C–C and C–H bonds are relatively strong and have little polarity, so alkanes do not readily attract nucleophiles or electrophiles at room temperature. They still burn and can undergo radical substitution when suitable activation energy is supplied. Unreactive does not mean incapable of reaction.

Distillation, cracking and reforming solve different problems

Crude oil is a mixture, mainly of hydrocarbons. In fractional distillation, heated vapour enters a column that is hotter at the bottom and cooler at the top. Fractions condense over different boiling ranges. Larger molecules usually have stronger London forces and higher boiling temperatures, so they condense lower in the column. Fractions are mixtures of similar boiling range, not individual compounds.

Distillation is a physical separation: intermolecular attractions are overcome and re-formed without breaking carbon chains. Cracking is a chemical change that breaks C–C bonds in larger hydrocarbons, using heat and often a catalyst. It supplies smaller fuel molecules and alkene feedstocks where demand is greater. Always balance the proposed cracking equation: more than one combination of products is possible.

Reforming converts straight-chain hydrocarbons into branched-chain alkanes and cyclic hydrocarbons suitable for more efficient engine combustion. A branch changes connectivity while retaining the molecular formula; ring formation can release hydrogen. Reforming is not simply another name for fractional distillation.

Balance combustion by following carbon, then hydrogen

With enough oxygen, complete combustion produces CO₂ and H₂O. For one mole of CₙH₂ₙ₊₂, nCO₂ accounts for carbon and (n + 1)H₂O accounts for hydrogen; oxygen then requires (3n + 1)/2 O₂. A fractional oxygen coefficient is valid for a one-mole fuel equation; multiply every coefficient by two if integers are required.

Worked example: heptane gives seven CO₂ and eight H₂O. These contain 14 + 8 = 22 oxygen atoms, so eleven O₂ are required. With insufficient oxygen, CO and/or solid carbon can form instead. In a real engine several combustion routes operate simultaneously, so a single incomplete-combustion equation is a model of one route.

Explain a pollutant by its origin and effect

Incomplete combustion makes carbon monoxide. CO binds strongly to haemoglobin, reducing the blood’s ability to transport oxygen. Carbon particulates and unburned hydrocarbons also arise from incomplete combustion or fuel escaping reaction. Carbon dioxide arises even in complete combustion and contributes to climate change.

High engine temperatures allow nitrogen and oxygen from air to react, producing nitrogen oxides. Sulfur-containing fuel impurities produce sulfur oxides when burned. Oxides such as NO₂ and SO₂ contribute to acidity when they react in the atmosphere and dissolve in water. Avoid saying every nitrogen oxide is itself acidic: NO can be oxidised further to NO₂.

A catalytic converter provides a high-surface-area catalyst for reactions converting CO and unburned hydrocarbons to CO₂ and H₂O, and nitrogen oxides to N₂. In 2CO + 2NO → 2CO₂ + N₂, carbon is oxidised and nitrogen reduced. The converter reduces several harmful emissions but does not remove the CO₂ climate impact, and it works effectively only when warm.

Compare the whole fuel life cycle

Crude oil is non-renewable on a human timescale. Ethanol can be produced by fermentation of plant sugars, and biodiesel can be made from oils or fats. Growing replacement crops can replenish those resources and photosynthesis removes CO₂ while the crop grows. The feedstock origin matters: ethanol made from fossil-derived ethene is not automatically renewable.

A balanced comparison includes cultivation, fertiliser manufacture, processing, distillation, transport and any land-use change, as well as combustion. These stages use energy and can release greenhouse gases. Biofuel production can compete with food production, land, biodiversity and water. “Renewable” therefore does not prove “zero net emissions”. Compare energy delivered per unit mass or per journey as well as the fuel’s price.

Worked evaluation: if a biofuel supplies 25 MJ kg⁻¹ and a fossil fuel 40 MJ kg⁻¹, delivering 200 MJ ideally takes 8.0 kg and 5.0 kg respectively. Comparing emissions per kilogram alone would miss the extra mass needed. Real engines also differ in efficiency, so use measured efficiency data where supplied.

A radical chain has three distinct stages

A radical is a species with an unpaired electron, shown by a dot. Homolytic fission splits a covalent bond so each fragment takes one bonding electron. In alkane halogenation, ultraviolet radiation initiates homolysis of Cl–Cl or Br–Br. This differs from heterolytic fission, where both bonding electrons go to one fragment and ions form.

For chlorination of methane, initiation produces radicals; propagation consumes one radical and produces another, allowing the chain to continue; termination removes radicals by joining two of them. The initiation equation is not the overall substitution equation. Edexcel does not require curly half-arrows for this radical mechanism, but the radical dots and species must be correct.

Why radical substitution rarely makes one pure product

Chloromethane still has C–H bonds, so further substitution can form CH₂Cl₂, CHCl₃ and CCl₄. A longer alkane also has chemically different hydrogen positions. Propane can form 1-chloropropane and 2-chloropropane. Radical combinations add other impurities. A mixture means lower selectivity and a separation stage even if the starting alkane reacts extensively.

Worked transfer: to show formation of 2-bromopropane, write Br• + CH₃CH₂CH₃ → HBr + CH₃C•HCH₃, followed by CH₃C•HCH₃ + Br₂ → CH₃CHBrCH₃ + Br•. The dot is on the central carbon that lost H. Writing the terminal radical would describe the other substitution product.

Excess alkane can reduce further substitution by making a radical more likely to encounter unreacted alkane than a halogenated product, but does not guarantee one product. In a synthesis answer, discuss both positional isomers and repeated substitution where they apply.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources