IB Chemistry SLTopic 6 — Electron SharingPaper 1 & 2Organic~13 min read
Halogenation of Alkanes
Alkanes are the most unreactive family in organic chemistry. Nothing polar can get a grip on them — so the only thing that will attack one is a species so desperate that it does not care: a radical.
📚 What you need to know
Alkanes are unreactive because their C–C and C–H bonds are strong and nonpolar.
With no δ+ or δ– regions, neither nucleophiles nor electrophiles have anything to attack.
Free-radical substitution replaces a hydrogen with a halogen, and requires UV light.
The mechanism has three stages: initiation, propagation and termination.
Initiation makes the radicals; propagation uses one up and makes another; termination destroys two at once.
Propagation is a chain: one halogen radical can convert thousands of alkane molecules.
The reaction gives a mixture of substitution products, so it is a poor way to make a specific compound.
Why nothing attacks an alkane
Two things protect an alkane. The C–C and C–H bonds are strong, so a lot of energy is needed to break them. And carbon and hydrogen have almost identical electronegativities, so the electrons in every bond are shared very nearly equally.
Compare this with a halogenoalkane, where the electronegative halogen creates a δ+ carbon that nucleophiles home in on. An alkane offers no such target.
A radical does not need a target. It is not attracted to a charge, because it has none — it simply collides with whatever is nearby and grabs an atom to pair up its lone electron. That indifference is exactly why radicals can attack a molecule that everything else ignores.
The evidence for UV light
The classic demonstration takes two identical tubes of hexane and bromine water. One goes on a sunny windowsill, the other into a dark cupboard.
A controlled experiment in the proper sense: only one variable differs between the tubes, so the light must be what makes the difference.
The three stages
The mechanism has a beginning, a middle and an end, and each stage does something different to the number of radicals in the flask. Keeping that in mind is the fastest way to remember which equation belongs where.
The middle stage is drawn as a loop for a reason. Neither propagation step uses up a radical overall — each one consumes a radical and creates another, so the cycle turns until two happen to meet.
Stage
What it does to the radical count
Trigger or outcome
Initiation
0 → 2, radicals are created
needs UV light to break the halogen bond
Propagation
1 → 1, the count is unchanged
the chain keeps turning; this is where the product forms
The error examiners look for. The first propagation step is not CH4 + Cl• → CH3Cl + H•. A hydrogen radical is never formed. The chlorine radical takes the hydrogen atom, giving HCl and leaving a methyl radical behind.
Why it is a poor way to make anything
The product, chloromethane, still has three hydrogens. Chlorine radicals cannot tell it apart from methane, so they attack it too, and the substitution keeps going.
It does not stop
CH4 → CH3Cl → CH2Cl2 → CHCl3 → CCl4
The flask ends up holding a mixture of all of them, plus HCl and the odd trace of ethane from termination. Separating one product from that mixture is difficult and the yield of any single compound is low — which is why free-radical substitution is a mechanism to understand rather than a method to use.
WORKED EXAMPLE
Write the full mechanism for the reaction of ethane with chlorine in UV light to form chloroethane, labelling each stage.
InitiationCl₂ → 2Cl• (UV light)Propagation, step 1CH₃CH₃ + Cl• → •CH₂CH₃ + HClPropagation, step 2•CH₂CH₃ + Cl₂ → CH₃CH₂Cl + Cl•Termination, any one ofCl• + Cl• → Cl₂•CH₂CH₃ + Cl• → CH₃CH₂Cl•CH₂CH₃ + •CH₂CH₃ → CH₃CH₂CH₂CH₃the last one gives butaneCheck each propagation step: a radical goes in and a radical comes out. If one of your steps has no radical on a side, it is wrong.
WORKED EXAMPLE
A student writes the following as a propagation step: CH4 + Cl• → CH3Cl + H• Explain what is wrong with it.
What the student has doneThey have swapped the chlorine straight onto the carbon and released a hydrogen radical.Why it does not happenThe chlorine radical removes a hydrogen atom to form the very stable H—Cl bond. A free hydrogen radical would be extremely high in energy and is not produced.the correct step is CH₄ + Cl• → •CH₃ + HClThe chlorine ends up on the carbon in the second propagation step, from a chlorine molecule, not from the radical.
WORKED EXAMPLE
Explain why the chlorination of methane is unsuitable for preparing a pure sample of chloromethane, and suggest one way of improving the proportion of it in the mixture.
The problemChloromethane still has three hydrogens, and a chlorine radical attacks it as readily as it attacks methane.a mixture of CH₃Cl, CH₂Cl₂, CHCl₃ and CCl₄ formsMaking it worseTermination also produces small amounts of ethane, and HCl is formed throughout.An improvementuse a large excess of methaneA radical is then far more likely to meet an unreacted methane molecule than a chloromethane one, so further substitution is less likely. The yield is still not clean.
💡 Exam tip
Label the three stages by name. Marks are usually awarded for the labels as well as the equations.
Write UV light above the arrow in the initiation step.
Check that every propagation step has one radical on each side.
Never produce H•. The halogen radical takes the hydrogen and forms HX.
Give a termination step that matches the radicals present in your mechanism.
Mention the mixture of products when asked to evaluate the reaction as a synthesis.
⚠️ Common mix-up
Forming a hydrogen radical in propagation. The single most-penalised error in this topic.
Putting Cl2 in the first propagation step and the radical in the second. It is the other way round.
Writing Cl– instead of Cl•, which turns a radical mechanism into an ionic one.
Calling the initiation step a propagation step because it also involves chlorine.
Giving only one product and claiming a clean reaction.
Saying the alkane is attacked because it is polar. It is attacked precisely because a radical does not need polarity.
That completes electron sharing, and with it the radical mechanisms. Up next: Electron-Pair Sharing Reactions — where a lone pair attacks rather than a lone electron, and every arrow goes back to being double-headed.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
