OCR A Level Chemistry A: Alkanes (H432)

Alkanes is section 4.1.2 of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, and this guide works through it from the start. Seven lettered outcomes, 4.1.2(a) to (g), make up the section within Module 4, Core organic chemistry; each is taught below with worked examples. The specification is for the full A level. Module 4 is examined, the specification says, in Unified chemistry (03) and Synthesis and analytical techniques (02). The section carries no PAG reference.

Other H432 topics are reached from the OCR chemistry hub; the printable checklist lets you tick off each outcome as it becomes secure.

Outcomes in 4.1.2

Outcome	Your task	Specification’s additional guidance
4.1.2(a)	Describe alkanes as saturated hydrocarbons with C–C and C–H σ-bonds; explain free rotation about a σ-bond	Hybridisation not required; HSW1 (orbital overlap model)
4.1.2(b)	Explain the tetrahedral shape and bond angle around each carbon by electron pair repulsion	M4.1, M4.2; draw 3-D diagrams; links to 2.2.2 g–h
4.1.2(c)	Explain boiling point trends with chain length and branching using London forces	M3.1; links to 2.2.2 k
4.1.2(d)	Explain the low reactivity of alkanes using bond enthalpy and polarity	HSW1; links to 2.2.2 j
4.1.2(e)	Write equations for complete combustion of alkanes as fuels and incomplete combustion in limited oxygen; explain the danger from CO	None listed
4.1.2(f)	Describe radical substitution with chlorine and bromine in UV, with a mechanism of initiation, propagation and termination	Half curly arrows not required; a single dot shows each radical; links to 4.1.1 f–g
4.1.2(g)	Explain why radical substitution is limited in synthesis	Mixtures from further substitution and from reaction at different chain positions

You also need the names of the first ten straight-chain alkanes, from 4.1.1(a): methane, ethane, propane, butane, pentane, hexane, heptane, octane, nonane and decane.

4.1.2(a): Saturated hydrocarbons and σ-bonds

An alkane is a hydrocarbon (it contains carbon and hydrogen only) and is saturated (every carbon–carbon bond is a single bond). Open-chain alkanes have the general formula CₙH₂ₙ₊₂.

Every bond in an alkane is a σ-bond. A σ-bond forms when two orbitals overlap directly between the two bonding nuclei, along the line joining them. The shared electron pair sits on that axis.

Because the electron density lies along the bond axis, turning one end of the molecule relative to the other does not reduce the overlap. So there is free rotation about a σ-bond. In butane the two ends of the chain spin about the central C–C bond. The different arrangements are not isomers: they interconvert freely.

Worked example 1. 1-bromo-2-chloroethane, CH₂BrCH₂Cl, can be drawn with the Br and Cl atoms on the same side or on opposite sides. Explain why it exists as a single compound.

• The C–C bond is a σ-bond, formed by direct overlap of orbitals between the two carbon nuclei.
• Rotation about that axis keeps the overlap the same, so the bond does not break.
• Every arrangement of the CH₂Br and CH₂Cl ends converts into every other, so only one compound exists.

4.1.2(b): Shape and bond angle

Each carbon atom in an alkane has four bonding pairs of electrons and no lone pairs. Electron pairs repel one another and spread out as far apart as possible. Four pairs are furthest apart in a tetrahedral arrangement, with a bond angle of 109.5°.

This applies to every carbon, so a long chain is a zig-zag, not a straight line.

Drawing a 3-D diagram. Put two bonds in the plane of the paper as plain lines, one bond coming towards you as a solid wedge, and one going away as a dashed wedge.

Worked example 2. State and explain the H–C–H angle at the central carbon of propane, CH₃CH₂CH₃.

• The central carbon forms four bonds (two C–C, two C–H): four bonding pairs, no lone pairs.
• The pairs repel equally and move as far apart as possible.
• The shape around that carbon is tetrahedral, so the angle is 109.5°.

4.1.2(c): Boiling points and London forces

Alkane molecules are non-polar. The only forces between them are induced dipole–dipole interactions (London forces). An uneven electron cloud at any instant gives an instantaneous dipole, which induces a dipole in a neighbouring molecule.

Chain length. Each extra CH₂ adds electrons and surface area. More electrons give larger instantaneous and induced dipoles, and longer molecules touch at more points. Overcoming these stronger attractions needs more energy, which raises the boiling point.

Alkane	Boiling point / °C
methane	−162
ethane	−89
propane	−42
butane	−1
pentane	36
hexane	69
heptane	98
octane	126
nonane	151
decane	174

The first four are gases at room temperature. Plotted as a graph (M3.1), the rise is steep at first and then gentler, because each CH₂ adds a smaller fraction to a larger molecule.

Branching. Isomers have the same number of electrons, but branched molecules are more compact and closer to spherical. They have fewer points of contact with neighbours, so their London forces are weaker and they boil at a lower temperature.

Worked example 3. Three isomers of C₇H₁₆ have these boiling points: heptane 98 °C, 2-methylhexane 90 °C, 2,2-dimethylpentane 79 °C. Explain the order.

• All three have the same molecular formula, so the same number of electrons.
• Heptane is unbranched, so its molecules pack closely with the most surface contact.
• Each extra branch makes the molecule more compact, so there are fewer points of contact.
• Fewer contacts mean weaker London forces, less energy to overcome them, and a lower boiling point.

Always say that intermolecular forces are overcome on boiling. No C–C or C–H bonds break.

4.1.2(d): Why alkanes are unreactive

Alkanes do not react with many common reagents, such as acids, bases and most oxidising agents at room temperature. Two reasons:

• High bond enthalpy. C–C and C–H σ-bonds are strong. Typical average bond enthalpies are about +347 kJ mol⁻¹ for C–C and +413 kJ mol⁻¹ for C–H, so a lot of energy is needed to break them.
• Very low polarity. Carbon and hydrogen have very similar electronegativities (2.55 and 2.20 on the Pauling scale). The C–H bond has almost no dipole and the C–C bond has none, so there are no δ+ or δ− sites to attract nucleophiles or electrophiles.

Alkanes do react with oxygen when ignited, and with halogens in UV radiation.

4.1.2(e): Combustion

Complete combustion happens in a plentiful supply of oxygen. The products are carbon dioxide and water, and a lot of energy is released, which is why alkanes are used as fuels.

Worked example 4. Write the equation for the complete combustion of decane, C₁₀H₂₂.

• Carbon: 10 CO₂. Hydrogen: 22 H atoms make 11 H₂O.
• Oxygen atoms needed: 20 + 11 = 31, so 15½ O₂.
• C₁₀H₂₂ + 15½O₂ → 10CO₂ + 11H₂O, or doubled: 2C₁₀H₂₂ + 31O₂ → 20CO₂ + 22H₂O.

Incomplete combustion happens in a limited supply of oxygen, for example a gas heater in a poorly ventilated room or an engine. The carbon ends up as carbon monoxide or as soot (carbon). For pentane:

• to carbon monoxide: 2C₅H₁₂ + 11O₂ → 10CO + 12H₂O
• to carbon: C₅H₁₂ + 3O₂ → 5C + 6H₂O

Why CO is dangerous. Carbon monoxide is toxic. It binds to haemoglobin in red blood cells, which reduces the blood’s ability to carry oxygen. It is colourless and odourless, so you cannot detect it without a CO detector.

Worked example 5. 2.88 g of pentane burns completely. Calculate the volume of oxygen used at RTP and the mass of CO₂ formed. Use M(C₅H₁₂) = 72.0 g mol⁻¹, M(CO₂) = 44.0 g mol⁻¹ and 24.0 dm³ mol⁻¹ at RTP.

• Equation: C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O
• n(C₅H₁₂) = 2.88 ÷ 72.0 = 0.0400 mol
• n(O₂) = 8 × 0.0400 = 0.320 mol; volume = 0.320 × 24.0 = 7.68 dm³
• n(CO₂) = 5 × 0.0400 = 0.200 mol; mass = 0.200 × 44.0 = 8.80 g

4.1.2(f): Radical substitution with chlorine and bromine

In ultraviolet radiation, alkanes react with chlorine or bromine. A hydrogen atom is replaced by a halogen atom, so this is substitution, and it goes through radicals (species with an unpaired electron), so it is radical substitution. Each radical is written with a single dot, for example Br•.

Worked example 6. Butane reacts with bromine in UV light to form 1-bromobutane, among other products. Write the mechanism.

Initiation. UV radiation breaks the Br–Br bond by homolytic fission: each bromine atom takes one electron from the bonded pair.

Br₂ → 2Br•

Propagation. A radical reacts with a molecule to give a new radical. Two steps form a chain:

CH₃CH₂CH₂CH₃ + Br• → CH₃CH₂CH₂CH₂• + HBr

CH₃CH₂CH₂CH₂• + Br₂ → CH₃CH₂CH₂CH₂Br + Br•

Step two hands back a Br• for step one, so the chain repeats many times after a single initiation.

Termination. Two radicals combine and the chain stops. Possible steps:

Br• + Br• → Br₂

CH₃CH₂CH₂CH₂• + Br• → CH₃CH₂CH₂CH₂Br

CH₃CH₂CH₂CH₂• + CH₃CH₂CH₂CH₂• → C₈H₁₈ (octane)

Check each propagation step: the number of radicals is one on each side. Check each termination step: two radicals become none. Half curly arrows are not required for this mechanism; the specification shows radical mechanisms as a sequence of equations.

Chlorine reacts in exactly the same way, starting with Cl₂ → 2Cl•.

4.1.2(g): Limits of radical substitution in synthesis

Radical substitution is a poor way to make one pure haloalkane, because it gives a mixture of organic products.

Further substitution. The product still has C–H bonds, so Br• can remove another hydrogen from it. Repeating the propagation steps on 1-bromobutane gives dibromobutanes, then tribromobutanes and so on.

C₄H₉Br + Br• → C₄H₈Br• + HBr, then C₄H₈Br• + Br₂ → C₄H₈Br₂ + Br•

Reaction at different positions. Br• can remove a hydrogen from any carbon. In butane, removing H from an end carbon leads to 1-bromobutane; removing H from carbon 2 or 3 leads to 2-bromobutane.

Worked example 7. How many different monobromo products can form from butane, and what does that mean for a chemist who wants only 1-bromobutane?

• Butane has two kinds of carbon environment: the two end CH₃ groups and the two middle CH₂ groups.
• So there are two monobromo isomers: 1-bromobutane and 2-bromobutane.
• Termination also makes octane, and further substitution makes dibromo and higher products.
• 1-bromobutane must be separated from all of these, so its yield is low.

Count products by grouping equivalent carbons. Do not count carbon 1 and carbon 4 of butane as giving different products: both give 1-bromobutane.

Common errors

• Describing σ overlap as sideways. Sideways overlap is the π-bond in alkenes (4.1.3).
• Explaining boiling points with “bonds breaking” or with permanent dipoles. Only London forces act between alkane molecules.
• For branching, saying “fewer electrons”. Isomers have the same number; the difference is surface contact.
• Explaining low reactivity by bond strength only. You need polarity as well.
• Leaving out the dots on radicals, or putting a dot on HBr or the haloalkane product.
• Writing H• instead of HBr in the first propagation step. Br• removes a hydrogen atom to form HBr.

Where to go next

• Condensed recall and a quick self-test: Alkanes revision notes
• Exam-style questions with worked answers: Alkanes practice questions
• Background: Bonding and structure and Amount of substance
• Check yourself across the course: free diagnostics

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.2 Alkanes, learning outcomes (a) to (g). Published by OCR.