# Alkanes

> Chemistry · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9701-u13-alkanes/

Alkanes are the simplest homologous series of saturated hydrocarbons, derived from crude oil. This guide covers their bonding, structure, physical properties, combustion reactions and free radical substitution mechanism, core content for CIE A-Level Chemistry.

**Prerequisites:** [Introduction to organic chemistry and homologous series](https://www.owlsprep.com/study/cie-9701-u12-introduction-to-organic-chemistry/); [Intermolecular forces](https://www.owlsprep.com/study/cie-9701-u3-chemical-bonding-intermolecular-forces/)

## Learning objectives

- Describe the general formula, structure and bonding of alkanes
- Explain trends in physical properties of alkanes in terms of intermolecular forces
- Write balanced equations for complete and incomplete combustion of alkanes
- Outline the three stages of free radical substitution of alkanes with halogens

## Structure and Bonding

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

*Notation:* General formula: $C_nH_{2n+2}$ (acyclic)

*Example:* Methane ($CH_4$), ethane ($C_2H_6$), propane ($C_3H_8$)

All carbon atoms in alkanes are $sp^3$ 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 $C_nH_{2n+2}$, 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 $C_4H_{10}$.

> **Exam tip:** Remember cyclic alkanes have the general formula $C_nH_{2n}$, not $C_nH_{2n+2}$, due to an extra internal C-C bond reducing hydrogen count by 2.

## Physical Properties

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 $C_5H_{12}$ 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. $$\text{2,2-dimethylpropane} < \text{2-methylbutane} < \text{pentane}$$

> **tip**
>
> For alkanes with the same number of carbon atoms: more branching = lower boiling point. This is a very common exam question.

## Combustion Reactions

**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:* $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ (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 ($C_6H_{14}$).

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 $CO_2$ on right:
4. $$C_6H_{14} + O_2 \rightarrow 6CO_2 + H_2O$$
5. Balance hydrogen next: 14 H on left, so 7 $H_2O$ 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.

## Free Radical Substitution Mechanism

**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. 2. Propagation (first step): A chlorine free radical abstracts a hydrogen from methane:
4. $$Cl^{\bullet} + CH_4 \rightarrow CH_3^{\bullet} + HCl$$
5. 3. 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. 4. Termination (example step): Two free radicals combine to form a stable molecule, ending the chain:
8. $$CH_3^{\bullet} + Cl^{\bullet} \rightarrow CH_3Cl$$

> **info**
>
> Multiple propagation steps produce a mixture of products (e.g. dichloromethane, trichloromethane) not just chloromethane, which is a common exam point.

## Common pitfalls

- **Wrong:** Using the general formula $C_nH_{2n+2}$ for cycloalkanes
  - Why it fails: Cycloalkanes have one extra C-C bond, so they have two fewer hydrogen atoms than acyclic alkanes
  - Correct: Use $C_nH_{2n}$ for cycloalkanes, reserve $C_nH_{2n+2}$ for acyclic alkanes
- **Wrong:** Claiming branched alkanes have higher boiling points than straight chain isomers
  - Why it fails: Branched alkanes have a compact shape that reduces surface area for intermolecular interactions, weakening London forces
  - Correct: For alkanes of the same molecular formula, increasing branching decreases boiling point
- **Wrong:** Leaving a half-integer coefficient for oxygen in balanced combustion equations
  - Why it fails: CIE examiners require whole number coefficients for full marks
  - Correct: Multiply all coefficients by 2 to eliminate any fractions after balancing C and H
- **Wrong:** Showing heterolytic fission for the initiation step of free radical substitution
  - Why it fails: Free radicals form only from homolytic fission, where each atom gets one electron from the broken bond
  - Correct: Use single-headed fishhook curly arrows to show homolytic fission and movement of single electrons

## Cheatsheet

| Property | Key Fact |
| --- | --- |
| General formula (acyclic alkanes) | $C_nH_{2n+2}$ |
| General formula (cycloalkanes) | $C_nH_{2n}$ |
| Intermolecular force | London dispersion forces only |
| Boiling point trend | Increases with chain length, decreases with branching |
| Complete combustion products | $CO_2 + H_2O$ (excess $O_2$) |
| Incomplete combustion products | $CO/C + H_2O$ (limited $O_2$) |
| Reaction with halogens | Free radical substitution, requires UV light |
| Mechanism stages | Initiation → Propagation → Termination |

## 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.

- [Alkenes](https://www.owlsprep.com/study/cie-9701-u13-alkenes/)
- [Halogenoalkanes](https://www.owlsprep.com/study/cie-9701-u14-halogenoalkanes/)
- [Halogen derivatives and alcohols](https://www.owlsprep.com/study/cie-9701-u14-overview/)

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