Why do examiners insist on initiation, propagation, and termination?

Because the marks are testing whether you understand the logic of a chain reaction, not just the final product. In IB Chemistry, “propagation” is the idea that radicals are both used up and remade, so the reaction can repeat many times. Writing the stages makes your answer easy to mark and hard to misinterpret. It also forces you to show where the radicals come from (initiation), which is where UV light matters. Termination then proves you know how the reaction stops and why radical concentration doesn’t rise forever. If you format your answer by stages, you usually gain method marks even if you slip on a minor formula.

Why do alkanes need UV light, but alkenes often don’t?

Alkanes are saturated and relatively non-polar, so they lack an electron-rich region that attracts reagents. In IB Chemistry, this is why alkanes tend to react via radical pathways only after radicals are generated by UV light. Alkenes have a π bond with exposed electron density, which makes them reactive toward electrophiles without needing radical initiation. That difference is the reason alkenes are commonly tested with addition mechanisms, while alkanes show up with radical substitution. If a question includes UV and a halogen, think chain reaction and radicals. If it features C=C and HX, think electrophilic addition.

Why does multiple substitution happen so easily?

Because the first substitution product is still a hydrocarbon with remaining C-H bonds. In the same UV conditions, halogen radicals don’t “know” they’ve already substituted once; they can abstract another hydrogen and continue the chain. In IB Chemistry, this is why you’re often told that chlorination of methane produces a mixture, not a single pure compound. The mixture grows as more hydrogens are replaced, producing di-, tri-, and tetra-halogenated products. This is also why controlling conditions (like limiting halogen supply or time) matters in real chemistry. In an exam, your safest phrasing is usually: “further substitution is possible, so a mixture forms.”

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IB Chemistry: Radical Substitution Explained

IB Chemistry

RevisionDojo Team

•Last updated: 19 Jul 2026•7 min read

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IB Chemistry: Radical Substitution Explained

Radical substitution feels like one of those IB Chemistry topics that should be simple: shine UV light, swap a hydrogen, move on.

But then an exam question asks you for every step, why the light matters, why the products multiply, and why bromine behaves differently from chlorine. Suddenly, it’s not “one reaction” anymore. It’s a story about tiny, unstable particles that can’t stop bumping into things.

In IB Chemistry, radical substitution is where organic mechanisms stop being vocabulary and start being cause-and-effect. Once you see the logic, the marks become predictable.

Radical substitution in IB Chemistry (definition)

Radical substitution is a reaction in which a halogen atom replaces a hydrogen atom in an alkane via a chain reaction involving free radicals.

Three ideas to keep straight in IB Chemistry:

Radical: a species with an unpaired electron (written as a dot, like Cl·)
Substitution: one atom/group is swapped for another
Chain reaction: once radicals exist, they can regenerate themselves and repeat the cycle

This mechanism is most associated with alkanes, which are otherwise stubbornly unreactive because they’re saturated and lack a C=C bond.

For syllabus-aligned notes on this exact content, use Substitution reactions with alkanes (R3.3.3).

Quick exam checklist (before you write the mechanism)

When a question screams “radical substitution” in IB Chemistry, check:

Reactants include an alkane + Cl₂ or Br₂
Condition includes UV light (or sunlight)
You’re expected to show: initiation, propagation, termination
You may be asked about mixtures of products (multiple substitutions)
You may be asked about chlorine vs bromine (reactivity vs selectivity)

If you want a broader organic map, IB Chemistry Organic Chemistry is a useful hub for connecting mechanisms.

Why UV light matters in IB Chemistry radical substitution

Alkanes don’t politely react with halogens in the dark. The C-H bonds are strong and non-polar overall, so there’s no easy “handle” for reaction.

UV light provides the kick-start: it breaks the halogen bond evenly to form radicals.

That phrase “breaks evenly” is the part examiners love.

Homolytic fission: each atom takes one electron from the bond
Result: two radicals, each with one unpaired electron

The three stages of radical substitution (initiation, propagation, termination)

In IB Chemistry, you’re rewarded for structure. Write the headings, then write the steps.

Initiation (make radicals)

Example with chlorine:

Cl₂ → 2Cl·

Key phrase for marks: UV light causes homolytic fission of Cl₂.

Propagation (keep the chain going)

Propagation usually appears as two repeating steps.

Step 1: radical attacks the alkane

Cl· + CH₄ → HCl + CH₃·

A methyl radical forms (CH₃·). One radical is consumed, another radical is created.

Step 2: radical reacts with halogen molecule

CH₃· + Cl₂ → CH₃Cl + Cl·

Notice what happened: you formed a product (chloromethane) and regenerated Cl·. That regeneration is the “chain” in chain reaction.

If you want to drill these as exam-style prompts, practise via R3.3 Electron sharing reactions Questionbank and track weak spots using RevisionDojo’s Grading tools and AI Chat feedback.

Termination (remove radicals)

Termination happens when two radicals collide and form a stable molecule. Examples:

Cl· + Cl· → Cl₂
CH₃· + Cl· → CH₃Cl
CH₃· + CH₃· → C₂H₆

In IB Chemistry, the definition-level explanation is enough: termination removes radicals, stopping the chain.

Products and why mixtures form

A classic IB Chemistry trap is assuming you only get one substitution.

After forming CH₃Cl, the molecule still contains C-H bonds. Under the same UV conditions, it can keep reacting to form further substituted products:

CH₃Cl
CH₂Cl₂
CHCl₃
CCl₄

So the realistic answer is usually: a mixture of products, unless conditions are carefully controlled.

For quick structural recognition (and to avoid mixing up reaction families), IB Chemistry: How to Tell Alkanes vs Alkenes Fast is a helpful refresher.

Chlorine vs bromine (reactivity vs selectivity)

In IB Chemistry, you can often earn marks with a simple contrast:

Chlorine: more reactive, faster reaction, less selective, more by-products
Bromine: less reactive, slower reaction, more selective, fewer by-products

When exam questions say “explain why a mixture forms” or “compare chlorine and bromine,” they’re asking you to connect rate and product spread.

To keep mechanism types distinct, it also helps to compare with Electrophilic Addition Explained and Nucleophilic Substitution Explained.

Conclusion: make radical substitution a predictable IB Chemistry mark

Radical substitution in IB Chemistry is a small mechanism with a big payoff: it teaches you how UV-driven homolytic fission creates radicals, how propagation sustains a chain reaction, and how termination ends it.

If you want this to feel automatic before exams, use RevisionDojo as your routine: read the Functional group chemistry notes, practise with the Questionbank, then lock it in with Flashcards, Study Notes, Mock Exams, Predicted Papers, and targeted support from Tutors and the Coursework Library. That’s how IB Chemistry stops being memorised steps and becomes a mechanism you can explain under pressure.