Last updated: 21 August 2026
Alkanes are often introduced as the simplest family of organic compounds, but they establish several ideas that keep returning throughout A-Level Organic Chemistry. Their chemical behaviour becomes much easier to understand once we connect the structure of the molecule with the strength and polarity of the bonds inside it.
We will use ethane, CH₃CH₃, as the main example. By looking at its sp³-hybridised carbon atoms, σ bonds and lack of strongly polar reactive sites, we can explain both why alkanes are normally rather unreactive and why particular conditions are needed when they do react.
Try these five questions before reading the chapter. Use the structures and reaction schemes carefully — Organic Chemistry often becomes much easier once you learn to read what is happening to the molecule.
- 5 diagnostic questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
1. Structure and Bonding in Ethane
Alkanes are saturated hydrocarbons. A hydrocarbon contains only carbon and hydrogen, while the word saturated means that all carbon–carbon bonds are single bonds. An alkane therefore contains no C=C or C≡C bonds.
Ethane has molecular formula C₂H₆ and condensed structural formula CH₃CH₃. Each carbon atom forms three bonds to hydrogen atoms and one bond to the other carbon atom.
Alkanes contain only single covalent bonds. Every C–C and C–H bond in ethane is therefore a σ bond.
Hybridisation of carbon in ethane
Each carbon atom in ethane is sp³ hybridised. One 2s orbital and three 2p orbitals are combined to form four equivalent sp³ hybrid orbitals.
These four orbitals point approximately towards the corners of a tetrahedron, allowing each carbon atom to form four σ bonds.
C–C bond: formed by head-on overlap between an sp³ orbital on one carbon atom and an sp³ orbital on the other carbon atom.
C–H bond: formed by head-on overlap between an sp³ orbital on carbon and the 1s orbital of hydrogen.
Shape and bond angles around each carbon atom
Because each carbon atom has four bonding regions, the arrangement around each carbon is tetrahedral. The bond angles are approximately 109.5°.
In an examination answer, it is more precise to say that the geometry around each carbon atom in ethane is tetrahedral rather than simply saying that “ethane is tetrahedral”.
2. Why Are Alkanes Relatively Unreactive?
Alkanes do not readily react with many common reagents under ordinary conditions. This relatively low reactivity can be explained directly from their bonding.
Two ideas are important. First, considerable energy is required to break the strong σ bonds. Second, the molecules do not contain a strongly polar functional group that can readily attract polar reagents.
Strong C–C and C–H σ bonds
Both the C–C and C–H bonds in an alkane are strong σ bonds. A reaction involving cleavage of these bonds therefore has a significant energy requirement.
Lack of strongly polar reactive sites
A C–C bond is non-polar because the two bonded atoms have the same electronegativity. C–H bonds have only a small bond polarity. As a result, an alkane does not contain the strongly δ⁺ or δ⁻ centres commonly found in more reactive organic functional groups.
Many nucleophiles and electrophiles therefore have no obvious reactive centre to attack.
Relatively unreactive does not mean unable to react. Alkanes undergo combustion and can react with halogens when enough energy is supplied to generate highly reactive radicals.
3. Combustion of Alkanes
Alkanes are widely used as fuels. Although they are relatively unreactive towards many reagents, combustion can proceed rapidly once sufficient activation energy is supplied.
Complete combustion
When oxygen is plentiful, the carbon atoms are fully oxidised to carbon dioxide and the hydrogen atoms form water.
The products of complete combustion of an alkane are therefore carbon dioxide and water.
Incomplete combustion
If the supply of oxygen is limited, complete oxidation of the carbon atoms may not occur. Carbon monoxide, CO, and sometimes carbon particles, C, can be formed.
| Combustion | Oxygen supply | Products |
|---|---|---|
| Complete combustion | Plentiful oxygen | CO₂ and H₂O |
| Incomplete combustion | Limited oxygen | CO and H₂O may form |
| Very limited oxygen | Severely oxygen-deficient | C and H₂O may form |
Carbon monoxide is particularly dangerous because it binds strongly to haemoglobin, reducing the ability of the blood to transport oxygen.
4. Free-Radical Substitution of Ethane
A characteristic reaction of alkanes is their reaction with halogens such as chlorine or bromine in the presence of ultraviolet light.
A hydrogen atom in the alkane is replaced by a halogen atom. Since one atom is substituted for another, the reaction is called free-radical substitution.
The mechanism occurs as a chain reaction involving three types of step: initiation, propagation and termination.
Step 1: Initiation
UV light supplies the energy required to break the Cl–Cl bond.
The bond undergoes homolytic fission. Each chlorine atom takes one electron from the shared pair, producing two chlorine radicals.
A free radical is a species containing an unpaired electron. The dot in Cl• represents this unpaired electron.
Step 2: Propagation
A chlorine radical is highly reactive. It removes a hydrogen atom from ethane:
An ethyl radical, CH₃CH₂•, is produced. This radical then reacts with another chlorine molecule:
Notice that a chlorine radical is regenerated. This new Cl• radical can react with another ethane molecule, so the sequence can repeat.
This is why the process is called a chain reaction. Initiation creates the first radicals, while propagation repeatedly consumes one radical and produces another.
Step 3: Termination
The chain stops when two radicals collide and combine. Because the product contains no unpaired electron, no new radical is available to continue that particular chain.
| Stage | What happens? | How to recognise it |
|---|---|---|
| Initiation | Cl₂ undergoes homolytic fission | Radicals are first generated |
| Propagation | A radical reacts and another radical is formed | The radical chain continues |
| Termination | Two radicals combine | No radical is produced |
5. Further Substitution Can Occur
The formation of chloroethane does not necessarily stop the reaction. Chloroethane still contains C–H bonds, so another hydrogen atom can be substituted by chlorine.
Further substitution can continue, giving increasingly chlorinated products. This means that free-radical chlorination can produce a mixture of products rather than a single pure compound.
Free-radical substitution is not usually a good method for preparing a single pure halogenoalkane because further substitution can occur.
6. Alkanes and the Environment
Because alkanes are widely used as fuels, their combustion has important environmental consequences. Different products create different problems, so their effects should be kept separate.
Carbon dioxide and the greenhouse effect
Complete combustion produces carbon dioxide. CO₂ is a greenhouse gas that absorbs infrared radiation emitted by the Earth and contributes to the greenhouse effect.
Carbon monoxide from incomplete combustion
Carbon monoxide may be produced when oxygen is limited. CO is toxic because it binds strongly to haemoglobin and therefore reduces the ability of the blood to transport oxygen.
Unburnt hydrocarbons
Some hydrocarbons may pass through an engine without being completely burned. These unburnt hydrocarbons can participate in atmospheric reactions that contribute to the formation of photochemical smog.
Keep the effects distinct: CO₂ contributes to the greenhouse effect, CO is toxic, while unburnt hydrocarbons can contribute to photochemical smog.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Calling the whole ethane molecule “tetrahedral” | The tetrahedral description refers specifically to the arrangement around each sp³ carbon atom. | State that the geometry around each carbon atom is tetrahedral, with bond angles of approximately 109.5°. |
| Writing Cl⁻ instead of Cl• during initiation | Homolytic fission produces neutral radicals, not chloride ions. | Write Cl₂ → 2Cl• and remember that each atom takes one electron from the shared pair. |
| Calling UV light a catalyst | UV light supplies energy to break the Cl–Cl bond; it is not a chemical catalyst. | State that UV light provides energy for homolytic fission during initiation. |
| Assuming chlorination produces only chloroethane | Further hydrogen atoms can also be substituted, producing more highly chlorinated compounds. | Recognise that free-radical substitution can give a mixture of products. |
What you should be able to do
Describe an alkane
Explain what is meant by a saturated hydrocarbon and identify the bonding present in ethane.
Explain the bonding in ethane
Describe sp³ hybridisation, σ-bond formation and tetrahedral geometry around each carbon atom.
Explain alkane stability
Relate their relatively low reactivity to strong σ bonds and low bond polarity.
Write combustion equations
Distinguish complete and incomplete combustion and identify their products.
Write a free-radical substitution mechanism
Give correct initiation, propagation and termination equations for the chlorination of ethane.
Recognise further substitution and environmental effects
Explain why mixtures of halogenated products may form and relate combustion products to their environmental or health effects.
These ten questions test whether you can apply the chapter to structures, reactions and unfamiliar examples.
- 10 mastery questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about alkanes
Most of the facts about alkanes become easier to remember once you relate them back to their strong σ bonds and relatively non-polar structure.
Why are alkanes described as saturated?
Alkanes contain only carbon–carbon single bonds. There are no C=C or C≡C bonds in the carbon skeleton, so they are described as saturated hydrocarbons.
Why is each carbon atom in ethane sp³ hybridised?
Each carbon atom forms four σ bonds. In the hybridisation model, one 2s orbital and three 2p orbitals combine to form four equivalent sp³ hybrid orbitals directed approximately tetrahedrally.
If alkanes are relatively unreactive, why can they burn?
Relatively unreactive does not mean that a reaction is impossible. Combustion requires sufficient activation energy to begin, after which formation of stable products such as CO₂ and H₂O makes the overall process highly exothermic.
Why is UV light required for the reaction between ethane and chlorine?
UV light provides the energy required for homolytic fission of the Cl–Cl bond. This produces Cl• radicals, which initiate the free-radical chain reaction.
Why does free-radical chlorination often give a mixture of products?
The first substitution product still contains C–H bonds. These can undergo further substitution, so compounds containing two or more chlorine atoms can also form. This makes free-radical chlorination relatively unselective when the aim is to prepare one pure product.
Ethane captures the essential chemistry of the alkanes: sp³-hybridised carbon atoms form strong σ bonds, making the molecules relatively stable, yet combustion and free-radical substitution become possible when the appropriate conditions are provided. Keeping that link between structure, bonding and reactivity in mind will make the reactions of more reactive organic functional groups much easier to understand.