Alkanes and Free-radical Substitution
Learn and apply Alkanes and Free-radical Substitution in the published Chemistry course sequence.
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The core idea
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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)
- Write the initiation step: X₂ → [UV] 2X..
- Write 2 propagation steps that show (i) H atom removal and (ii) halogen addition.
- Write one termination step (any two radicals combine).
- 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).
Data table
| Product | Chlorination | Bromination |
|---|---|---|
| 1-halopropane | 44 | 4 |
| 2-halopropane | 56 | 97 |
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
Problem
Study the worked solution
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.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.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
Learner claim
Identify the bond and fission type
View solution step by step
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.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
Selectivity transfer
Compare the radical pathways
Hints
Hint 1: intermediates
Hint 2: stability
View solution step by step
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.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.