Alkanes

Chemistry· Unit 13: Hydrocarbons, Sub-topic 1: Alkanes· 20 min read

1. Structure and Bonding★☆☆☆☆⏱ 5 min

📘 Definition

Alkanes

General formula: $C_{n} H_{2 n + 2}$ (acyclic)

Saturated hydrocarbons where all carbon atoms form four single sigma bonds, with no multiple bonds between carbons.

Example:

Methane ( $C H_{4}$ ), ethane ( $C_{2} H_{6}$ ), propane ( $C_{3} H_{8}$ )

All carbon atoms in alkanes are $s p^{3}$ hybridised, with a tetrahedral geometry around each carbon and approximate bond angles of 109.5°. Rotation around C-C single bonds is free, so alkane chains can adopt multiple conformations.

📐 Worked Example

Find the molecular formula and draw the displayed formula for straight chain butane (4 carbon acyclic alkane).

1
Use the general formula for acyclic alkanes $C_{n} H_{2 n + 2}$ , substitute n=4:
2
$C_4H_{(2 \times 4) + 2} = C_4H_{10}$
3
Connect 4 carbon atoms with single C-C bonds, then add hydrogen atoms to satisfy the 4-bond rule for each carbon:
4
$\begin{array}{r} H H H H \\ | | | | \\ H-C-C-C-C-H \\ | | | | \\ H H H H \end{array}$
5
Final molecular formula is $C_{4} H_{10}$ .

Exam tip:

Remember cyclic alkanes have the general formula $C_{n} H_{2 n}$ , not $C_{n} H_{2 n + 2}$ , due to an extra internal C-C bond reducing hydrogen count by 2.

2. Physical Properties★★☆☆☆⏱ 5 min

Alkanes are non-polar because the electronegativity of carbon and hydrogen is nearly identical. The only intermolecular forces between alkane molecules are weak London dispersion (instantaneous dipole-induced dipole) forces.

Boiling point increases with increasing chain length: longer chains have larger molecular surface area and higher relative mass, leading to stronger London forces. For isomers of the same molecular formula, branching reduces boiling point.

📐 Worked Example

Arrange these $C_{5} H_{12}$ isomers in order of increasing boiling point: pentane (straight chain), 2-methylbutane (single branch), 2,2-dimethylpropane (two branches).

1
All three have the same molecular mass, so differences depend on branching only.
2
More branching produces a more compact molecular shape, which reduces surface area for intermolecular interactions, leading to weaker London forces and lower boiling point.
3
Order of increasing branching: pentane < 2-methylbutane < 2,2-dimethylpropane
4
Final order of increasing boiling point:
5
$\text{2,2-dimethylpropane} < \text{2-methylbutane} < \text{pentane}$

3. Combustion Reactions★★☆☆☆⏱ 6 min

📘 Definition

Complete Combustion

Combustion of alkanes in excess oxygen, producing only carbon dioxide and water as products. It is highly exothermic, so alkanes are widely used as fuels.

Example:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ (complete combustion of methane)

Incomplete combustion occurs when oxygen is limited. Products include toxic carbon monoxide and/or solid soot (carbon), in addition to water. It releases less energy than complete combustion.

📐 Worked Example

Write a fully balanced equation for the complete combustion of hexane ( $C_{6} H_{14}$ ).

1
Write the unbalanced equation:
2
$C_6H_{14} + O_2 \rightarrow CO_2 + H_2O$
3
Balance carbon first: 6 C on left, so 6 $C O_{2}$ on right:
4
$C_6H_{14} + O_2 \rightarrow 6CO_2 + H_2O$
5
Balance hydrogen next: 14 H on left, so 7 $H_{2} O$ on right:
6
$C_6H_{14} + O_2 \rightarrow 6CO_2 + 7H_2O$
7
Balance oxygen: right side has (6×2)+(7×1) = 19 O atoms, so $\frac{19}{2} O_{2}$ on left, then multiply all coefficients by 2 to get whole numbers:
8
$2C_6H_{14} + 19O_2 \rightarrow 12CO_2 + 14H_2O$

Exam tip:

Examiners always penalise balanced equations with half-integer coefficients. Always multiply through to get whole numbers.

4. Free Radical Substitution Mechanism★★★☆☆⏱ 7 min

📘 Definition

Free Radical

An uncharged species with an unpaired electron, formed by homolytic fission of a covalent bond.

Alkanes react with halogens (chlorine, bromine) under UV light to form halogenoalkanes via free radical substitution. The mechanism has three distinct stages: initiation, propagation and termination.

📐 Worked Example

Write the key steps for the formation of chloromethane from methane and chlorine via free radical substitution.

1
1. Initiation: UV light provides energy for homolytic fission of the Cl-Cl bond:
2
$Cl_2 \xrightarrow{UV} 2Cl^{\bullet}$
3
1. Propagation (first step): A chlorine free radical abstracts a hydrogen from methane:
4
$Cl^{\bullet} + CH_4 \rightarrow CH_3^{\bullet} + HCl$
5
1. Propagation (second step): A methyl free radical reacts with a chlorine molecule to form chloromethane and regenerate a chlorine free radical (chain reaction):
6
$CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}$
7
1. Termination (example step): Two free radicals combine to form a stable molecule, ending the chain:
8
$CH_3^{\bullet} + Cl^{\bullet} \rightarrow CH_3Cl$

5. Common Pitfalls

Wrong move:

Using the general formula $C_{n} H_{2 n + 2}$ for cycloalkanes

Why:

Cycloalkanes have one extra C-C bond, so they have two fewer hydrogen atoms than acyclic alkanes

Correct move:

Use $C_{n} H_{2 n}$ for cycloalkanes, reserve $C_{n} H_{2 n + 2}$ for acyclic alkanes

Wrong move:

Claiming branched alkanes have higher boiling points than straight chain isomers

Why:

Branched alkanes have a compact shape that reduces surface area for intermolecular interactions, weakening London forces

Correct move:

For alkanes of the same molecular formula, increasing branching decreases boiling point

Wrong move:

Leaving a half-integer coefficient for oxygen in balanced combustion equations

Why:

CIE examiners require whole number coefficients for full marks

Correct move:

Multiply all coefficients by 2 to eliminate any fractions after balancing C and H

Wrong move:

Showing heterolytic fission for the initiation step of free radical substitution

Why:

Free radicals form only from homolytic fission, where each atom gets one electron from the broken bond

Correct move:

Use single-headed fishhook curly arrows to show homolytic fission and movement of single electrons

6. Quick Reference Cheatsheet

Property	Key Fact
General formula (acyclic alkanes)	$C_{n} H_{2 n + 2}$
General formula (cycloalkanes)	$C_{n} H_{2 n}$
Intermolecular force	London dispersion forces only
Boiling point trend	Increases with chain length, decreases with branching
Complete combustion products	$C O_{2} + H_{2} O$ (excess $O_{2}$ )
Incomplete combustion products	$C O / C + H_{2} O$ (limited $O_{2}$ )
Reaction with halogens	Free radical substitution, requires UV light
Mechanism stages	Initiation → Propagation → Termination

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · Paper 1
Multiple choice: boiling point trends
2023 · Paper 2
Free radical chlorination mechanism
2024 · Paper 1
Combustion product identification

Going deeper

syllabusCIE 9701 2025-2026 SyllabusCheck page 42 for alkanes learning outcomes

What's Next

Alkanes are the foundation of organic chemistry for CIE A-Level, and their reactions and properties underpin all subsequent topics in hydrocarbon chemistry. Understanding alkane structure, physical property trends and free radical substitution here will also help you recognise and compare other mechanism types across different organic functional groups later in your course. Alkanes are a common topic in both multiple choice and structured questions, so mastering core skills like balancing combustion equations and drawing mechanism steps will earn you consistent, easy marks in your exam. Next, you will build on this foundation to study unsaturated hydrocarbons, their distinct structures and more reactive addition reactions.