Alkanes

Alkanes: saturated hydrocarbons—definition, general formula, naming and isomerism, plus key reactions (combustion and substitution).

  • SEC G3 Pure Chemistry 2027
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Learning objectives

  • describe the alkanes as a homologous series of saturated hydrocarbons with the general formula CnH2n+2
  • draw the structures of branched and unbranched alkanes, C1 to C4, and name the unbranched alkanes methane to butane
  • define isomerism and identify isomers
  • describe alkanes (exemplified by methane) as being generally unreactive except in terms of combustion and substitution by chlorine

Build this topic in a simple order: recognise the saturated carbon chain, use the general formula, compare physical trends, identify isomers, then apply combustion and UV substitution reactions.

1. Definition

A. Alkane

An alkane is a saturated hydrocarbon with only carbon-carbon single bonds, C-C. Alkanes form a homologous series with general formula CₙH₂ₙ₊₂.

B. Saturated hydrocarbon

A saturated hydrocarbon is a hydrocarbon with only C-C single bonds (no C = C).

2. Key Ideas

  • General formula: CₙH₂ₙ₊₂ (alkanes).
  • In a homologous series, successive members differ by CH₂.
  • As n increases, boiling point and viscosity generally increase.
  • Isomers: same molecular formula, different structural formulae.
  • Required alkane reaction: substitution by chlorine in UV light.
  • Combustion: complete → CO₂ + H₂O; incomplete → CO and/or soot.
Useful links

Homologous series basics: Introduction to Organic Chemistry
Unsaturated hydrocarbons + bromine water test: Alkenes
Combustion + air pollutants: Air pollutants

3. Detailed Explanations

Quick Recall (alkane essentials)
  • Alkanes are saturated hydrocarbons with only C-C single bonds; general formula CₙH₂ₙ₊₂.
  • Bromine water test: alkanes do not decolourise bromine water under normal conditions.
  • Key reaction tested: substitution of methane by chlorine under UV light: CH₄ + Cl₂ → [UV] CH₃Cl + HCl.
  • Combustion: complete → CO₂ + H₂O; incomplete → CO and/or soot (C). CO is toxic.
  • Isomers: same molecular formula, different structural formula (common with C₄H₁₀ and above).

A. General formula and the first few alkanes

The general formula for alkanes is CₙH₂ₙ₊₂.

NameMolecular formulaOne-line structural formulaState (at r.t.p.)
methaneCH₄CH₄gas
ethaneC₂H₆CH₃-CH₃gas
propaneC₃H₈CH₃-CH₂-CH₃gas

Required C4 alkane

NameMolecular formulaOne-line structural formulaState (at r.t.p.)
butaneC₄H₁₀CH₃-CH₂-CH₂-CH₃gas

B. Gradation in physical properties (do not overclaim)

As n increases (molecules get larger):

  • boiling point generally increases because intermolecular forces (forces between molecules) are stronger,
  • viscosity generally increases (liquids flow less easily).

Boiling Point Trend from Methane to Propane (Approx.)

Approximate boiling points for methane through propane plotted against carbon number, showing boiling point increases as chain length increases.

Scroll across the graph to read all labels.

Approximate boiling points for methane through propane plotted against carbon number, showing boiling point increases as chain length increases.Approximate boiling points for methane through propane plotted against carbon number, showing boiling point increases as chain length increases.
Approx values show the trend: as chain length increases, boiling point increases (stronger intermolecular forces).
Open full-size graph
View figure data
Values for Boiling Point Trend from Methane to Propane (Approx.)
Number of carbon atoms (unitless)Alkanes
1-162
2-89
3-42

C. Isomerism (what it is and why it matters)

Isomers have the same molecular formula but different structural formulae.

Example: C₄H₁₀ has two isomers:

  • butane (straight-chain): CH₃-CH₂-CH₂-CH₃
  • methylpropane (branched): CH₃-CH(CH₃)-CH₃

Recognise that these are different structures with the same molecular formula. You only need to recall the unbranched names methane to butane.

D. Combustion (complete vs incomplete)

Complete combustion happens in a plentiful supply of oxygen: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Incomplete combustion happens when oxygen is limited and can produce carbon monoxide and/or soot: 2C₂H₆(g) + 5O₂(g) → 4CO(g) + 6H₂O(l)

Carbon monoxide hazard

CO is colourless and odourless. Write: “forms carboxyhaemoglobin, so blood carries less oxygen”.

E. Substitution by chlorine (the key alkane reaction)

Methane is generally unreactive, but in UV light it reacts with chlorine. One hydrogen atom is replaced by a chlorine atom, so this is a substitution reaction.

Example (methane + chlorine): CH₄(g) + Cl₂(g) → [UV] CH₃Cl(g) + HCl(g)

4. Common Mistakes

  • Saying alkanes decolourise bromine water (they do not; that is alkenes).
  • Writing “addition reaction” for Cl₂ with alkanes (alkanes do substitution).
  • Forgetting the condition: UV light (in the dark, no reaction).
  • Writing incomplete combustion products as “only CO₂” (limited oxygen gives CO and/or soot).
  • Using “always” for physical-property trends (write “generally”).

5. Exam Tips

Substitution keywords

Write: “substitution reaction”, “UV light”, and “a hydrogen atom is replaced by a chlorine atom”.

Combustion keywords

Complete combustion: “CO₂ and H₂O”. Incomplete combustion: “CO and/or soot due to limited oxygen”.

6. Worked Examples

Modelled example 1

Identify an alkane from formula

Core

Problem

Is C₄H₁₀ an alkane? Justify your answer using the general formula.
Study the worked solution
  1. Substitute the carbon number

    Method

    Set n = 4 in CₙH₂ₙ₊₂.

    Reason

    The subscript on carbon gives the value of n.

    Working

    2n + 2 = 2(4) + 2 = 10.
  2. Compare the formula

    Method

    Compare the predicted and given hydrogen subscripts.

    Reason

    A matching molecular formula is consistent with the alkane homologous series.

    Working

    C₄H₁₀ fits CₙH₂ₙ₊₂; therefore it is an alkane.

Guided practice 2

Complete combustion equation (balancing)

About 6 min

Problem

Write a balanced equation for the complete combustion of propane, C₃H₈.

Balance C, then H, then O

CO2 coefficient
H2O coefficient
O2 coefficient

Hints

Hint 1: products
Complete combustion produces only carbon dioxide and water.
Hint 2: order
Balance carbon, then hydrogen, and count all oxygen atoms on the product side last.
View solution step by step
  1. Balance carbon and hydrogen

    Method

    Use three CO₂ and four H₂O.

    Reason

    These coefficients account for three C atoms and eight H atoms in propane.

    Working

    C₃H₈ + O₂ → 3CO₂ + 4H₂O.
  2. Balance oxygen

    Method

    Use five oxygen molecules.

    Reason

    The products contain 3(2) + 4(1) = 10 O atoms.

    Working

    C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).

Common misconception 3

Correct an isomer claim

Find and correct the mistake

Learner claim

Butane and methylpropane have different names, so a learner says they cannot be isomers. Correct the claim.

Compare molecular and structural formulae

Molecular formula
Structural formula

View solution step by step
  1. Compare atom counts

    Method

    Show that both compounds have molecular formula C₄H₁₀.

    Reason

    Isomers must contain the same numbers of each type of atom.

    Working

    Butane: C₄H₁₀; methylpropane: C₄H₁₀.
  2. Compare connectivity

    Method

    State that their structural formulae are different.

    Reason

    Butane is unbranched while methylpropane has a branched carbon skeleton.

    Working

    Same molecular formula + different structural formulae → structural isomers.

Examiner practice 4

Substitution (condition + product)

3 marks

Examination question

Methane reacts with chlorine to form chloromethane. State the condition and write the equation for replacing one hydrogen atom. [3 marks]

Give the condition, products and balanced equation

View solution step by step
  1. State the condition

    1 mark

    Method

    State UV light.

    Reason

    UV light initiates the alkane–halogen substitution reaction.

    Working

    Condition: UV light.
  2. Form the substituted products

    1 mark

    Method

    Replace one H atom in methane with Cl and pair the removed H with the other Cl.

    Reason

    A substitution replaces a hydrogen atom; it does not add chlorine across a double bond.

    Working

    Products: CH₃Cl and HCl.
  3. Write the balanced equation

    1 mark

    Method

    Use a 1:1:1:1 coefficient ratio.

    Reason

    The equation then conserves C, H and Br atoms.

    Working

    CH₄(g) + Cl₂(g) → [UV] CH₃Cl(g) + HCl(g).

Challenge 5

Draw the two C4 alkane isomers

Minimal support

Structure transfer

Draw the two structural isomers of C₄H₁₀. Name the unbranched isomer and explain why the pair are isomers.

Keep the formula while changing the carbon skeleton

Number of C4H10 structures
Why they are isomers

Hints

Hint 1: straight then branched
Start with four carbons in a row, then make a three-carbon chain with a one-carbon branch.
Hint 2: count atoms
Both drawings must contain four carbon and ten hydrogen atoms.
View solution step by step
  1. Draw and name the straight chain

    Method

    Place all four carbon atoms in one continuous chain.

    Reason

    The required unbranched name for C₄H₁₀ is butane.

    Working

    butane: CH₃-CH₂-CH₂-CH₃.
  2. Draw the branched structure

    Method

    Use a three-carbon chain with the fourth carbon attached to the middle carbon.

    Reason

    The atoms are connected differently while the molecular formula remains C₄H₁₀.

    Working

    CH₃-CH(CH₃)-CH₃; this is the branched isomer.

7. Mind Stretchers

Mind stretcher 1: Spot the wrong testExtension

Question: A student says: “Propane decolourises bromine water because it is a hydrocarbon.” Explain why this is wrong.

Show Answer

Decolourising bromine water is a test for a C = C double bond (unsaturation).

Propane is an alkane and has only C-C single bonds (saturated), so it does not decolourise bromine water under normal conditions.

Final: Bromine water tests for C = C; propane has none.

Mind stretcher 2: CO vs CO2 (exam trap)Extension

Question: A heater produces fumes. A student writes: “The dangerous gas is CO₂ from incomplete combustion.” Correct the gas and give the mark-scheme reason it is dangerous.

Show Answer

The dangerous gas is carbon monoxide, CO, produced in incomplete combustion.

Reason: CO forms carboxyhaemoglobin, so blood carries less oxygen.

Final: CO; reduces oxygen transport in blood (carboxyhaemoglobin).

8. Quiz

Quiz Time!

Test general formula checks, isomerism, trends, combustion products, and UV substitution.