IB Chemistry HLTopic 6 — Electron SharingPaper 1 & 2Organic~12 min read
Halogenation of Alkanes
Alkanes are the dullest molecules in organic chemistry. Strong bonds, no charge anywhere, nothing for a nucleophile or an electrophile to grab hold of. Radicals are the one thing aggressive enough to attack them — and once they start, they set off a chain reaction that is hard to stop.
📚 What you need to know
Alkanes are unreactive because C−C and C−H bonds are strong and the molecule is non-polar.
The reaction is free-radical substitution: a halogen atom replaces a hydrogen.
UV light is essential. In the dark, nothing happens.
Three stages: initiation, propagation, termination.
Initiation breaks the halogen bond, not the alkane. Propagation uses a radical and makes a new one. Termination joins two radicals together.
The reaction gives a mixture of products, so it is a poor way to make one specific halogenoalkane.
Propagation never produces a hydrogen radical. This is the classic trap.
Why alkanes need such extreme conditions
Strong bonds. C−H is 414 kJ mol−1 and C−C is 346 kJ mol−1. Both take a lot of breaking.
No polarity. Carbon has an electronegativity of 2.6 and hydrogen 2.2 — a difference of only 0.4. The electrons are shared almost equally.
Nothing to attack. With no δ+ region, nucleophiles have no target. With no δ− region, electrophiles have none either.
This is why alkanes make good fuels and good lubricants: they sit there and do nothing until you set fire to them. Their unreactivity is a feature, not a flaw. It also means the only two reactions you need for them are combustion and radical substitution.
The evidence that UV light is essential
A clean controlled experiment: same hexane, same bromine, same temperature. Light is the only variable, so light must be the cause.
The three stages
Use the radical count to identify a step you are unsure about. Zero in and two out is initiation; one in and one out is propagation; two in and none out is termination.
Initiation
UV light breaks the halogen−halogen bond homolytically. It is the weakest bond present, which is why it goes first rather than a C−H bond.
Initiation
Cl2UV → 2Cl•
Propagation
Two steps that feed each other. The first uses up a chlorine radical and makes a methyl radical; the second uses up the methyl radical and hands back a chlorine radical, ready to start again.
Check every propagation step this way: exactly one radical on the left, exactly one on the right. If your step has two radicals on one side, it is a termination step. If it has none, it is not a radical step at all.
Termination
Two radicals meet and pair up their electrons. Both radicals vanish, so the chain stops. Several combinations are possible, which is one reason the product mixture is messy.
Three possible terminations
•CH3 + Cl• → CH3Cl
•CH3 + •CH3 → C2H6
Cl• + Cl• → Cl2
The problem with this reaction
As a way of making one specific halogenoalkane, free-radical substitution is poor. The chlorine radicals cannot tell the difference between the starting alkane and the product you have just made, so substitution keeps going:
Stage
What forms
Why it does not stop
First substitution
CH3Cl
The product still has three C−H bonds left to attack
Second substitution
CH2Cl2
Radicals attack whichever molecule they meet first
Keeps going
CHCl3, then CCl4
With excess halogen, every hydrogen is eventually replaced
Termination products
C2H6 and others
Radicals combining with each other add extra by-products
Two more things make the mixture worse. A longer alkane has hydrogens in different positions, so you get isomers as well — propane gives both 1-chloropropane and 2-chloropropane. And separating that lot by fractional distillation is expensive. This is why industry uses other routes when it wants one clean product.
Worked examples
WORKED EXAMPLE
Write the initiation, propagation and one termination step for the reaction of ethane with chlorine in UV light.
Step 1: Initiation breaks the halogen bondCl2 → 2Cl•Step 2: First propagation — the radical takes a hydrogenCH3CH3 + Cl• → •CH2CH3 + HClStep 3: Second propagation — the chlorine radical comes back•CH2CH3 + Cl2 → CH3CH2Cl + Cl•Step 4: Termination joins any two radicals•CH2CH3 + Cl• → CH3CH2Clcheck each propagation step has one radical each side — both do
WORKED EXAMPLE
A student writes this propagation step: CH3CH3 + Cl• → CH3CH2Cl + H•. Explain what is wrong with it.
Step 1: Check the radical countOne radical in, one radical out. That part looks fine, which is exactly why the error is so easy to miss.Step 2: Look at which bonds would have to breakThis would need the strong C−H bond broken and a hydrogen radical released, which is very unfavourable.Step 3: State what actually happensThe chlorine radical takes the hydrogen atom away with it, forming stable HCl.Wrong: it should be CH3CH3 + Cl• → •CH2CH3 + HCla hydrogen radical is never a propagation product — examiners look for this every year
WORKED EXAMPLE
Explain why the chlorination of methane produces a mixture of products rather than pure chloromethane.
Step 1: Consider what the product still hasCH3Cl still contains three C−H bonds.Step 2: Ask whether radicals can be selectiveThey are extremely reactive and attack whatever they collide with, product or reactant.CH3Cl → CH2Cl2 → CHCl3 → CCl4Step 3: Add the termination productsRadicals also combine with each other, giving ethane among other things.Further substitution plus termination by-products give a mixtureusing excess methane reduces further substitution but never eliminates it
💡 Exam tip
Always label your steps as initiation, propagation or termination. Unlabelled equations often score nothing.
Write UV or sunlight above the arrow in the initiation step.
Check the one-radical-in, one-radical-out rule on every propagation step before moving on.
Never write H• as a propagation product. It is the single most penalised error in this topic.
Give both propagation steps if the question asks for propagation. One is only half the answer.
For “why a mixture”, mention further substitution and termination by-products, and isomers for longer chains.
⚠ Common mix-up
Breaking the alkane in initiation. UV breaks the halogen bond, which is much weaker than C−H.
Producing a hydrogen radical. The hydrogen leaves attached to the halogen as HCl or HBr.
Giving only one propagation step. The cycle needs both to keep running.
Calling a termination step propagation. Two radicals reacting together is always termination.
Forgetting the dots. An equation without them does not show a radical mechanism at all.
Saying the alkane is attacked because it is polar. It is not polar — that is precisely why only radicals will touch it.
Up next: Electron-Pair Sharing Reactions — back to electrons moving in pairs, but now with molecules that do have δ+ and δ− regions. That is where nucleophiles and electrophiles finally get something to attack.
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