Alkanes and Free-radical Substitution

Learn and apply Alkanes and Free-radical Substitution in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Alkanes and Free-Radical Substitution: Orientation

Free-radical substitution is the one core A Level organic mechanism that uses radical (fishhook) thinking: UV light → homolytic fission → chain reaction steps. This lesson trains you to write initiation/propagation/termination cleanly and explain why mixtures form.

This topic gets cleaner if you cross-check Organic Mechanisms: Curly Arrows, Electrophiles, Nucleophiles while navigating from the Organic Chemistry hub.

Definitions (Must Know)

A. Alkane

An alkane is a hydrocarbon with only single C–C bonds (saturated). Acyclic alkanes have general formula CₙH₂ₙ₊₂.

B. Free-radical substitution

Free-radical substitution is a reaction where a hydrogen atom on an alkane is replaced by a halogen atom, via radical intermediates.

Example overall equation (chlorination): CH₄ + Cl₂ → CH₃Cl + HCl

C. Free radical

A free radical is a species with an unpaired electron (shown as a dot, e.g. Cl.).

Detailed Explanations

A. Why alkanes are relatively unreactive

Alkanes mainly contain strong, non-polar σ bonds. Because there is no strongly polar bond (no permanent δ + centre) and no π bond, therefore they do not react readily with typical nucleophiles/electrophiles under mild conditions.

B. Conditions and what they imply

  • Conditions: Cl₂ or Br₂ with UV light.
  • UV implies homolytic fission: a bond splits evenly to form two radicals.

C. Workflow: writing a free-radical substitution mechanism (exam method)

  1. Write the initiation step: X₂ → [UV] 2X..
  2. Write 2 propagation steps that show (i) H atom removal and (ii) halogen addition.
  3. Write one termination step (any two radicals combine).
  4. Check: every radical has a dot, and atoms are conserved in each step.

Mini example (chlorination of methane):

  • Initiation: Cl₂ → [UV] 2Cl.
  • Propagation: Cl. + CH₄ → HCl + CH₃. then CH₃. + Cl₂ → CH₃Cl + Cl.

D. Mechanism stages (example: chlorination)

1) Initiation

Homolytic fission produces radicals: Cl₂ → [UV] 2Cl.

2) Propagation

Radicals are used and re-formed (this is why it’s a chain reaction): Cl. + CH₄ → HCl + CH₃. CH₃. + Cl₂ → CH₃Cl + Cl.

3) Termination

Two radicals collide to form a non-radical product (chain stops): Cl. + Cl. → Cl₂ CH₃. + Cl. → CH₃Cl CH₃. + CH₃. → C₂H₆

E. Why mixtures form (and why bromine is “more selective”)

  • If an alkane has more than one type of hydrogen (e.g. primary vs secondary), substitution can occur at multiple positions → isomer mixture.
  • Multiple substitutions can occur (e.g. CH₃Cl then CH₂Cl₂ etc.), especially for methane.
  • Bromination is slower but tends to give a higher proportion of the “more stable radical” product (often more selective than chlorination).
Chlorination vs Bromination Selectivity (Propane Example)Illustrative: using typical relative H-abstraction reactivities and H counts in propane, bromination gives a much higher proportion of 2-halopropane than chlorination.Chlorination vs Bromination Selectivity (Propane Example)ProductPredicted proportion (%)KeyChlorinationChlorinationBrominationBromination
Illustrative: using typical relative H-abstraction reactivities and H counts in propane, bromination gives a much higher proportion of 2-halopropane than chlorination.
Data table
ProductChlorinationBromination
1-halopropane444
2-halopropane5697

F. Combustion, pollutants and the atmosphere

Complete combustion in excess oxygen forms carbon dioxide and water. Limited oxygen can produce carbon monoxide and soot:

CH₄ + 2O₂ → CO₂ + 2H₂O 2CH₄ + 3O₂ → 2CO + 4H₂O

In vehicle engines, high temperatures allow nitrogen and oxygen from air to form nitrogen oxides. Sulfur impurities in fuel form sulfur dioxide, and unburnt hydrocarbons can escape. Their effects differ: carbon monoxide is toxic, nitrogen oxides contribute to photochemical smog and acid deposition, sulfur dioxide contributes to acid deposition, particulates harm health, and carbon dioxide strengthens the greenhouse effect.

Catalytic converters reduce emissions of carbon monoxide, nitrogen oxides and unburnt hydrocarbons, but they do not remove the carbon dioxide formed by burning fuel. A complete environmental answer names the pollutant, its source and its effect rather than calling every emission “greenhouse gas”.

Worked Examples

Modelled example 1

Bromination of Methane

Core

Problem

Write the initiation and two propagation steps for the bromination of methane.
Study the worked solution
  1. Initiate the chain

    Method

    Split bromine homolytically under UV light.

    Reason

    Each bromine atom takes one bonding electron, producing the radicals needed to start the chain.

    Working

    Br₂ → [UV] 2Br.
  2. Remove a hydrogen atom

    Method

    Use a bromine radical to form hydrogen bromide and a methyl radical.

    Reason

    This propagation step consumes one radical but creates another.

    Working

    Br. + CH₄ → HBr + CH₃.
  3. Regenerate the chain carrier

    Method

    React the methyl radical with another bromine molecule.

    Reason

    The desired organic product forms while a bromine radical is regenerated, so the chain can continue.

    Working

    CH₃. + Br₂ → CH₃Br + Br.

Common misconception 2

What UV Light Does

Find and correct the mistake

Learner claim

A learner says UV light starts chlorination by breaking a C–H bond in methane. Correct the claim and explain why UV is needed.

Identify the bond and fission type

Bond broken first
Fission type

View solution step by step
  1. Correct the first bond change

    Method

    Break the Cl-Cl bond, not a methane C–H bond.

    Reason

    UV supplies energy for homolytic fission of the halogen molecule.

    Working

    Cl₂ → [UV] 2Cl.
  2. Connect initiation to the chain

    Method

    Use the chlorine radicals to initiate propagation.

    Reason

    Without the first radicals, no chain carrier exists to abstract hydrogen from methane.

    Working

    UV → chlorine radicals → chain mechanism starts.

Challenge 3

Major Product from Propane

Minimal support

Selectivity transfer

Chlorination of propane gives mainly 2-chloropropane rather than 1-chloropropane. Explain this major-product preference.

Compare the radical pathways

Favoured intermediate
Major product

Hints

Hint 1: intermediates
Substitution at carbon 1 and carbon 2 forms different propyl radicals.
Hint 2: stability
Compare the stability of a primary radical with a secondary radical.
View solution step by step
  1. Compare the two intermediates

    Method

    Identify the secondary propyl radical as more stable than the primary propyl radical.

    Reason

    The two substitution sites therefore do not give equally favoured radical pathways.

    Working

    Secondary radical stability > primary radical stability.
  2. Link pathway to product

    Method

    State that the favoured secondary-radical pathway produces 2-chloropropane.

    Reason

    Greater intermediate stability makes that pathway more favourable, although both products may form.

    Working

    Major product: 2-chloropropane; 1-chloropropane remains a possible minor product.

Mind Stretchers

Mind stretcher 1Extension

Why does methane chlorination often give a mixture including CH₂Cl₂ and CHCl₃?

Show Hint

Identify where each radical is consumed and whether the chain continues or stops.

Show Answer

Mark scheme:

  • After CH₃Cl forms, it can undergo further substitution because it still contains C–H bonds.
  • The same chain mechanism can replace additional H atoms with Cl atoms, giving multiple chloro-products.

Mind stretcher 2: Explaining product mixturesExtension

Question. Chlorination of propane gives more than one monochlorinated product. Explain why without calculating a product ratio.

Show Hint

Propane contains hydrogen atoms in two different carbon environments.

Show Answer

A chlorine radical can replace a hydrogen on an end carbon or on the central carbon. These substitutions give 1-chloropropane and 2-chloropropane. Further substitution can also occur, so radical chlorination does not normally produce one pure product.