OCR A Level Chemistry A: Alkanes (H432) – Revision Notes

Use these notes to revise Alkanes, section 4.1.2 of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. All seven outcomes of this Module 4 (Core organic chemistry) section, 4.1.2(a) to (g), are condensed here. Module 4 is examined in Unified chemistry (03) and in Synthesis and analytical techniques (02), according to the specification. H432 is the full A level; this section carries no PAG reference.

Start with the Alkanes study guide if you need each idea explained from scratch. Then test yourself with the Alkanes practice questions. For other topics, open the OCR chemistry hub; record progress on the printable checklist and look for wider gaps with the free diagnostics.

Definitions to learn

Word	What it means
Hydrocarbon	Compound of carbon and hydrogen only
Saturated	Contains single carbon–carbon bonds only
Alkane general formula	CₙH₂ₙ₊₂ (open-chain)
σ-bond	Bond from direct overlap of orbitals between the two bonding atoms
London forces	Induced dipole–dipole interactions between molecules
Radical	Species with an unpaired electron, written with a dot (Cl•)
Homolytic fission	Bond breaks so each atom keeps one electron of the pair, giving two radicals
Radical substitution	Replacement of H by a halogen atom through a radical chain

Names of the first ten straight-chain alkanes: methane, ethane, propane, butane, pentane, hexane, heptane, octane, nonane, decane (C₁ to C₁₀).

4.1.2(a): σ-bonds and free rotation

• Every C–C and C–H bond in an alkane is a σ-bond.
• The overlap is directly between the nuclei, on the line joining them.
• Rotating one end does not change that overlap, so there is free rotation about each σ-bond.
• Rotated arrangements of the same molecule are not isomers.
• Hybridisation is not required.

4.1.2(b): Shape around each carbon

• Four bonding pairs, no lone pairs, around every carbon.
• Pairs repel and get as far apart as possible: tetrahedral, 109.5°.
• In a 3-D drawing: two bonds as lines in the plane, one solid wedge (towards you), one dashed wedge (away).

4.1.2(c): Boiling points

Method in steps: chain-length answer

1. Name the force: London forces (induced dipole–dipole interactions) between molecules.
2. Longer chain: extra electrons and a larger contact area.
3. So stronger London forces.
4. More energy needed to overcome them, so a higher boiling point.

Method in steps: branching answer

1. Isomers have the same number of electrons.
2. Branched molecules are more compact, with fewer points of contact.
3. So weaker London forces.
4. Less energy needed to overcome them, so a lower boiling point.

Straight-chain alkanes from methane to butane are gases at room temperature; pentane to decane are liquids.

4.1.2(d): Low reactivity

Two points, both needed:

• High bond enthalpy: C–C and C–H σ-bonds are strong and need a lot of energy to break.
• Very low polarity: C and H have very similar electronegativities, so the σ-bonds have little or no dipole and nothing attracts nucleophiles or electrophiles.

4.1.2(e): Combustion

Type	Oxygen supply	Products
Complete	Plentiful	CO₂ and H₂O
Incomplete	Limited	CO and/or C (soot), plus H₂O

Alkanes are used as fuels because complete combustion releases a lot of energy. Incomplete combustion happens wherever oxygen is short, for example in a blocked flue or a badly ventilated room.

Method in steps: balancing a combustion equation

1. Carbon: one CO₂ (or CO, or C) per carbon atom.
2. Hydrogen: half as many H₂O as there are H atoms.
3. Add up the O atoms in the products; divide by two for the number of O₂.
4. If you get a half, double every coefficient.

Worked reminder, butane in limited oxygen to carbon monoxide:

• 4 CO and 5 H₂O need 4 + 5 = 9 O atoms, so 4½ O₂.
• 2C₄H₁₀ + 9O₂ → 8CO + 10H₂O

Why CO is dangerous: it is toxic because it binds to haemoglobin and reduces oxygen transport in the blood; it is colourless and odourless, so it is not noticed without a detector.

4.1.2(f): Radical substitution

Conditions: chlorine or bromine with UV radiation.

Method in steps: writing the mechanism

1. Initiation: homolytic fission of the halogen. Cl₂ → 2Cl•
2. Propagation 1: halogen radical takes an H atom, giving HX and an alkyl radical.
3. Propagation 2: alkyl radical takes a halogen atom from X₂, giving the haloalkane and X• again.
4. Termination: any two radicals combine.

Worked reminder, propane and chlorine forming 2-chloropropane:

• Initiation: Cl₂ → 2Cl•
• Propagation: CH₃CH₂CH₃ + Cl• → CH₃CH•CH₃ + HCl
• Propagation: CH₃CH•CH₃ + Cl₂ → CH₃CHClCH₃ + Cl•
• Termination: Cl• + Cl• → Cl₂; CH₃CH•CH₃ + Cl• → CH₃CHClCH₃; 2CH₃CH•CH₃ → (CH₃)₂CHCH(CH₃)₂

Checks: one radical on each side of a propagation step; two radicals to none in termination. Half curly arrows are not required.

4.1.2(g): Why it is limited in synthesis

• Further substitution: the haloalkane still has C–H bonds, so it reacts again to give di-, tri- and more substituted products.
• Different positions: the halogen radical can remove H from any carbon, giving structural isomers.
• Termination also gives longer alkanes.
• Result: a mixture; after separation only a small amount of each product remains.

Method in steps: counting monosubstituted isomers

1. Draw the skeleton.
2. Group carbons that are equivalent by symmetry.
3. Count only groups that carry at least one H.
4. Each group gives one monohalo isomer.

Worked reminder, 2,2-dimethylbutane, CH₃C(CH₃)₂CH₂CH₃:

• The three CH₃ groups on carbon 2 are equivalent: one group.
• Carbon 2 has no hydrogen: no product.
• The CH₂ (carbon 3) is one group; the end CH₃ (carbon 4) is another.
• So three monochloro isomers, before you count any further substitution.

Links to other sections

• 2.2.2 g–h (shapes from electron pair repulsion) supports 4.1.2(b).
• 2.2.2 j (electronegativity and bond polarity) supports 4.1.2(d).
• 2.2.2 k (induced dipole–dipole interactions) supports 4.1.2(c).
• 4.1.1 f–g (homolytic fission, radicals and dot notation) supports 4.1.2(f).
• Synoptic questions can join these with moles and gas volumes from Module 2, so practise combustion calculations alongside the Amount of substance revision notes, and revise intermolecular forces with the Bonding and structure revision notes.

Must-know distinctions

• σ-bond vs π-bond. σ: direct overlap between nuclei, free rotation. π: sideways p-orbital overlap in alkenes (4.1.3), no rotation.
• Intermolecular vs covalent. Boiling overcomes London forces between molecules; it does not break C–H or C–C bonds.
• Homolytic vs heterolytic fission. Homolytic: one electron to each atom, two radicals. Heterolytic: both electrons to one atom, ions.
• Propagation vs termination. Propagation keeps the number of radicals the same; termination removes two.
• Complete vs incomplete combustion. Same fuel; the difference is the oxygen supply.

Quick self-test

1. Give the molecular formula of nonane.
2. State what “saturated” means for a hydrocarbon.
3. State the H–C–C bond angle in ethane.
4. Octane boils at a higher temperature than its isomer 2,2,4-trimethylpentane. Give the reason in one sentence.
5. Balance the complete combustion of tridecane, C₁₃H₂₈.
6. Calculate the mass of water formed when 0.0100 mol of decane burns completely. Use M(H₂O) = 18.0 g mol⁻¹.
7. Write the initiation step for bromine and name the type of bond fission.
8. Name the type of step: CH₃CH₂• + Br₂ → CH₃CH₂Br + Br•
9. How many monochloro isomers can form from hexane? Name them.
10. Give the two reasons alkanes are unreactive towards most reagents.
11. Two propyl radicals, CH₃CH₂CH₂•, combine. Name the product.

Answers

1. C₉H₂₀. CₙH₂ₙ₊₂ with n = 9.
2. It contains only single carbon–carbon bonds.
3. 109.5°. Each carbon has four bonding pairs, tetrahedral.
4. Straight-chain octane has more surface contact, so stronger London forces to overcome than the compact branched isomer.
5. C₁₃H₂₈ + 20O₂ → 13CO₂ + 14H₂O. Oxygen atoms: 13 × 2 + 14 = 40, so no half to clear.
6. Each C₁₀H₂₂ gives 11 H₂O, so 0.110 mol; 0.110 × 18.0 = 1.98 g.
7. Br₂ → 2Br•, homolytic fission, using UV radiation.
8. Propagation: one radical in, one radical out.
9. Three: 1-chlorohexane, 2-chlorohexane and 3-chlorohexane. Carbons 1 and 6, 2 and 5, and 3 and 4 are equivalent pairs.
10. High bond enthalpy of C–C and C–H bonds, and very low polarity of those bonds.
11. Hexane, CH₃CH₂CH₂CH₂CH₂CH₃ (a termination step).

Where marks are usually lost

• Calling the σ-bond “sideways overlap”, or saying the bond “cannot rotate”.
• Explaining the tetrahedral shape without stating four bonding pairs and repulsion.
• Naming permanent dipoles, or just “intermolecular forces”, instead of London forces (induced dipole–dipole interactions), or saying bonds break on boiling.
• For branching, claiming the branched isomer has fewer electrons.
• Giving only one reason for low reactivity; bond enthalpy and polarity are separate points.
• Leaving a half in a combustion equation when whole numbers are asked for, or writing CO₂ for incomplete combustion.
• Missing dots on radicals, or adding a dot to a molecule such as HCl.
• Putting H• in propagation step 1 instead of HX.
• Omitting UV radiation from the conditions.
• Counting equivalent positions twice, for example treating 1-chlorohexane and 6-chlorohexane as different.

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.