OCR A Chemistry H032 / H432 · Year 12 / AS · 4.1.2

Part 1: Structure, properties and combustion

All 2 parts available. Reviewed 6 October 2026.

Relate sigma bonding to alkane shape and low reactivity, then balance complete and incomplete combustion.

Saturated hydrocarbons and sigma bonds

Acyclic alkanes have general formula CnH2n+2; a saturated single-ring cycloalkane has CnH2n. Both have only single carbon–carbon bonds. A sigma (σ) bond forms by direct, head-on orbital overlap along the line between the nuclei. Rotation about an isolated C–C sigma bond is possible without breaking that overlap.

Each saturated carbon has four bonding pairs arranged approximately tetrahedrally at 109.5°. Strong C–C/C–H bonds and their low polarity explain why alkanes do not react readily with many polar reagents. Low reactivity does not mean they cannot burn or undergo radical substitution.

Molecular size and branching

Along the unbranched homologous series, larger molecules have more electrons, more polarisable clouds and greater surface contact, giving stronger London forces and higher boiling points.

Among isomers with the same formula, branching tends to make molecules more compact, reducing contact and weakening overall London attractions, so boiling points tend to fall. Do not explain boiling by breaking C–C bonds or invoke hydrogen bonding in a pure alkane.

Oxygen supply changes the carbon products

Complete combustion in sufficient oxygen forms CO₂ and water. Balance carbon first, hydrogen second and oxygen last. For C₆H₁₄, products are 6CO₂ + 7H₂O, requiring 19 oxygen atoms, or 9½O₂. Integer coefficients can be obtained by doubling the whole equation.

Limited oxygen can produce CO and/or soot, usually as a mixture with other products, releasing less energy per mole than complete combustion. CO binds strongly to haemoglobin and interferes with oxygen transport. Do not rely on smell to detect it. Combustion releases CO₂, connecting fuel use to the infrared absorption discussed in Analytical Techniques.

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
2CH₄ + 3O₂ → 2CO + 4H₂O
CH₄ + O₂ → C + 2H₂O

Choose the right comparison before explaining boiling

Compare hexane with pentane to study increasing chain length: hexane has more electrons and a larger region of intermolecular contact, so its London attractions are stronger. Compare pentane with 2,2-dimethylpropane to study branching: they have the same electron count, so the relevant difference is molecular shape and contact, not “the branched molecule has fewer electrons”.

A complete explanation reaches an energy conclusion: more energy must be supplied to overcome stronger attractions, so boiling occurs at a higher temperature at the same pressure. The molecules remain covalently intact in the vapour. Do not automatically transfer a boiling-point trend to melting, where crystal packing also matters.

An alkane is insoluble in water to a good approximation because mixing does not supply strong alkane–water attractions to compensate for disruption of water’s hydrogen bonding. The hydrocarbon has C–H bonds, but that does not make it a hydrogen-bond donor.

A general combustion equation and a limiting-oxygen check

For CnH2n+2, balance to nCO₂ and (n+1)H₂O. Those products contain 3n+1 oxygen atoms, giving an O₂ coefficient of (3n+1)/2. For butane, n = 4 gives 6.5O₂ per mole of C₄H₁₀. If integer coefficients are requested, multiply the entire equation by two.

For gases at the same temperature and pressure, volumes follow the balanced mole ratio. Therefore 20.0 cm³ of butane needs 130 cm³ of O₂ for complete combustion. Do not use this gas-volume shortcut for liquid water or compare volumes measured at different temperatures without an appropriate gas calculation.

C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Worked example: compare carbon dioxide per unit of energy

Use these supplied illustrative data: ethane has ΔcH = −1560 kJ mol⁻¹, and M(CO₂) = 44.0 g mol⁻¹. Burning one mole of ethane forms two moles of CO₂, or 88.0 g, while releasing 1560 kJ. For 1.00 MJ = 1000 kJ released, the mass is (1000/1560) × 88.0 = 56.4 g CO₂ to three significant figures.

Use the positive magnitude of the released energy to find the amount burned; a negative mass has no meaning. Do not use q = mcΔT unless the problem actually supplies a mass being heated and a temperature rise. Here the molar combustion enthalpy is the bridge from energy to moles.

This comparison concerns emissions from combustion per energy released. It is not a complete environmental comparison: energy delivered by an appliance also depends on efficiency, and production and transport can add emissions.

Quick checks

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

Q1. Give the formula of an acyclic alkane with seven carbons.Show answer

C₇H₁₆, from CnH2n+2.

Q2. Why can a sigma bond rotate?Show answer

Head-on overlap along the internuclear axis remains when one end rotates relative to the other.

Q3. Which usually boils higher: pentane or 2,2-dimethylpropane?Show answer

Pentane: its less compact shape provides greater intermolecular contact and stronger London interactions.

Q4. Balance complete combustion of propane.Show answer

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

Q5. Why is carbon monoxide dangerous?Show answer

It binds strongly to haemoglobin, reducing oxygen transport; it cannot be reliably detected by smell.

Q6. Why is “2,2-dimethylpropane has fewer electrons than pentane” a poor explanation of its lower boiling point?Show answer

They are isomers with the same molecular formula and electron count. Branching makes 2,2-dimethylpropane more compact, reducing contact and weakening the overall London attractions between its molecules.

Q7. Calculate the O₂ volume required for 15.0 cm³ gaseous ethane at the same temperature and pressure.Show answer

C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O. The gas-volume ratio is 1:3.5, so O₂ volume = 15.0 × 3.5 = 52.5 cm³.

Q8. Using a supplied methane combustion enthalpy of −890 kJ mol⁻¹ and M(CO₂) = 44.0, find the CO₂ mass per 1.00 MJ released.Show answer

One mole of methane forms one mole of CO₂. n(CH₄) = 1000/890 mol, so m(CO₂) = (1000/890) × 44.0 = 49.4 g, using the supplied data to three significant figures.

Q9. Explain why a fuel may be hard to ignite but burn strongly once lit.Show answer

Strong bonds and an activation-energy barrier make reaction slow initially. Ignition supplies energy to reach the barrier. Subsequent bond formation can release more energy than was needed to break reactant bonds, so combustion is exothermic.

Sources

Sources and examiner guidance (reviewed 6 October 2026)

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