In simple terms
A friendly intro before the formal notes — no formulas yet.
The Chain of Lonely Electrons
Most bonds break by handing both shared electrons to one atom, making ions. But a bond can also split fairly, giving one electron to each atom. Each fragment is then left with a single unpaired electron — a free radical — and radicals are so reactive that they set off a chain reaction that keeps making more of themselves.
Think of a shared bond as two people holding one rope between them. Heterolytic fission is one person snatching the whole rope and walking off (they become 'charged', an ion). Homolytic fission is the rope tearing exactly in half, so each person walks away holding a frayed end — an unpaired, reactive strand. Those frayed ends are radicals: they grab at anything to pair up, and in doing so they create a new frayed end somewhere else, which is why the reaction runs as a chain.
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A bond must first be broken evenly (homolytically) to make the first radicals — this needs energy, usually from UV light. This is INITIATION.
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A radical attacks a stable molecule, makes a product, and generates a brand-new radical. The number of radicals is conserved, so the cycle repeats over and over. This is PROPAGATION.
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Occasionally two radicals collide and pair up their lone electrons, forming a stable molecule and removing radicals from the mixture. This is TERMINATION.
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Because radicals attack somewhat indiscriminately and can substitute more than once, the reaction produces a MIXTURE of products, not a single clean one.
Explore the concept
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Full topic notes
Formal explanation with the rigour you need for the exam.
Two ways to break a bond: homolytic vs heterolytic fission
A covalent bond is a shared pair of electrons. When the bond breaks, those two electrons have to go somewhere, and there are exactly two possibilities. In HETEROLYTIC fission the bond breaks unevenly: one atom keeps BOTH electrons, so it becomes a negative ion, while the other atom leaves with none and becomes a positive ion. In HOMOLYTIC fission the bond breaks evenly: each atom keeps ONE of the shared electrons, so each fragment leaves with a single unpaired electron. A fragment with an unpaired electron is a free radical, written with a dot, such as Cl• or •CH₃.
Heterolytic fission: X–Y → X⁺ + :Y⁻. Both electrons go to one atom → two IONS. Common for polar bonds and reactions in solution.
Homolytic fission: X–X → X• + X•. One electron to each atom → two RADICALS. Common for non-polar bonds energised by UV light or heat.
A free radical has an unpaired electron (shown by a dot) and is highly reactive; it seeks another electron to pair with.
Radicals carry a DOT and no charge; ions carry a CHARGE and no dot. Keep the two notations strictly apart.
Free radicals and the role of UV light
Making the first radicals costs energy: you have to break a bond homolytically, and that does not happen on its own at room temperature in the dark. The energy is supplied by ULTRAVIOLET (UV) light. A UV photon carries enough energy to break a relatively weak, non-polar bond such as Cl–Cl evenly, generating two chlorine radicals. This is why the condition attached to free-radical substitution is always UV light (sometimes written as 'sunlight' or hv). Alkanes and halogens can be mixed in the dark almost indefinitely; shine UV on them and the chain reaction begins.
UV light provides the energy to break a bond homolytically and create radicals.
The condition 'UV light' (or hv / sunlight) is required and must be stated in the initiation step.
Only the initiation step needs the UV input; once radicals exist, the propagation chain sustains itself.
The single-headed fish-hook arrow represents the movement of ONE electron (a full curly arrow represents a pair).
Free-radical substitution: the three stages
The reaction of an alkane with a halogen in UV light replaces a hydrogen atom with a halogen atom — a substitution — and it proceeds as a radical chain in three named stages. We use methane and chlorine as the standard example. The overall reaction for the first substitution is CH₄ + Cl₂ →(UV) CH₃Cl + HCl, but that overall equation hides the mechanism, which is what examiners test.
Read the propagation steps carefully: the first one consumes a Cl• radical but produces a •CH₃ radical; the second consumes the •CH₃ radical but produces a Cl• radical again. The chlorine radical is regenerated, so the two steps cycle round and round, converting many molecules of methane and chlorine for every radical made in initiation. That is what 'chain reaction' means.
Initiation — radicals are CREATED by homolytic fission of the halogen using UV light: Cl₂ →(UV) 2Cl•.
Propagation — the chain-carrying steps; each USES one radical and MAKES another, so the radical count stays constant: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•.
Termination — radicals are DESTROYED when two of them combine, pairing their lone electrons: Cl• + Cl• → Cl₂, •CH₃ + Cl• → CH₃Cl, •CH₃ + •CH₃ → C₂H₆.
Each single electron shifting in these steps is drawn with a fish-hook (single-headed) arrow.
Why a mixture of products forms
Free-radical substitution is not clean. The chlorine radical is indiscriminate: it will abstract a hydrogen atom from a product molecule just as happily as from methane itself. So once CH₃Cl has formed, a Cl• radical can turn it into •CH₂Cl, which reacts with chlorine to give CH₂Cl₂ — and the process repeats to give CHCl₃ and finally CCl₄. On top of this, termination joins radicals in several combinations, so even ethane (C₂H₆, from two methyl radicals) appears. The result is always a MIXTURE, and this is a key limitation of the reaction. Using a large excess of the alkane favours the mono-substituted product but never delivers it pure.
Chlorine radicals substitute repeatedly: CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄.
Termination gives extra products, including ethane from •CH₃ + •CH₃ → C₂H₆.
A single pure product is not obtainable; excess alkane only shifts the balance toward mono-substitution.
Relevance: CFCs and the ozone layer
Free-radical chains are not a laboratory curiosity — they shape the atmosphere. Chlorofluorocarbons (CFCs), once widely used as refrigerants and propellants, are so unreactive that they drift intact into the stratosphere. There the intense UV light breaks a C–Cl bond homolytically, releasing a chlorine radical, Cl•. That radical then destroys ozone in a propagation-like cycle: Cl• + O₃ → ClO• + O₂, followed by ClO• + O → Cl• + O₂. The chlorine radical is regenerated each cycle, so one Cl• can destroy thousands of ozone molecules before a termination step removes it. It is exactly the initiation–propagation logic of alkane halogenation, playing out on a planetary scale — and it is why CFCs were phased out under the Montreal Protocol.
Model answer — marked the way our engine marks it
Mechanism questions in Paper 2 are marked analytically: each mark is tied to one specific correct step or condition. A method mark (M) is earned for a chemically valid step in the right stage; an answer mark (A) is earned for the correct species — with the radical dots present. Error-carried-forward (ECF) means that if you wrote a slightly wrong radical earlier but then used it correctly in the next step, the follow-on mark can still be awarded. Study how each mark below is pinned to a single line.
Common mistakes examiners penalise
Confusing homolytic and heterolytic fission — homolytic (even split) makes radicals with dots; heterolytic (uneven split) makes ions with charges. Don't describe a radical reaction as making ions.
Dropping the radical dot — Cl and Cl•, or CH₃ and •CH₃, are different species. A missing dot changes the species and loses the answer mark.
Writing an ionic step in a radical mechanism — free-radical substitution never has a step like Cl• + CH₄ → CH₃⁺ + ...; ionic products score zero even when balanced.
Forgetting to state the UV condition — the initiation step is meaningless without it, and it is a mark in its own right.
Mislabelling the stages — initiation CREATES radicals (needs UV), propagation USES and REMAKES a radical (chain-carrying), termination DESTROYS radicals (two combine). Putting a step in the wrong stage loses the mark.
Claiming a single pure product — chlorine radicals substitute repeatedly and termination joins radicals, so a MIXTURE (CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄, plus C₂H₆) always forms.
Not balancing HCl in propagation — the hydrogen abstracted from the alkane must appear as HCl, not as a free H atom or as H₂.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
For each process, state whether the bond fission is homolytic or heterolytic, and identify the species produced. (a) Cl₂ → 2Cl. (b) H–Br → H + Br. [2]
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(a) The Cl–Cl bond splits so that each chlorine atom keeps one electron, giving two identical fragments each with an unpaired electron: 2Cl•. Because the electrons are shared out evenly and radicals are formed, this is homolytic fission. [1]
Ethane, C₂H₆, reacts with chlorine, Cl₂, in the presence of UV light. Write one equation for each of the initiation, propagation (two steps) and termination stages of the free-radical substitution, using ethyl radicals where appropriate. [5]
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Initiation — homolytic fission of chlorine by UV light: [1]
Write equations for the initiation and the two propagation steps in the free-radical substitution of methane by chlorine, and state the condition required. [4]
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Model answer.
How it all connects
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Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Try to recall each definition before you reveal it.
Quick check
Answer in your head first — then tap to check. No pressure.
Revision flashcards
Flip the card. Test yourself before the exam.
Free radical
A species with an unpaired electron, shown with a dot (e.g. Cl•, •CH₃). The unpaired electron makes radicals extremely reactive. The dot is not optional — an examiner reads Cl and Cl• as different species.
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
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Heterolytic fission: X–Y → X⁺ + :Y⁻. Both electrons go to one atom → two IONS. Common for polar bonds and reactions in solution.
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Homolytic fission: X–X → X• + X•. One electron to each atom → two RADICALS. Common for non-polar bonds energised by UV light or heat.
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A free radical has an unpaired electron (shown by a dot) and is highly reactive; it seeks another electron to pair with.
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Radicals carry a DOT and no charge; ions carry a CHARGE and no dot. Keep the two notations strictly apart.
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Get a Paper 2 mechanism marked: write the initiation, propagation and termination steps for a free-radical substitution with full radical notation
Get a Paper 2 mechanism marked: write the initiation, propagation and termination steps for a free-radical substitution with full radical notation
Extra simulations & links
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Frequently asked
Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do Get a Paper 2 mechanism marked: write the initiation, propagation and termination steps for a free-radical substitution with full radical notation on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.