THE CONCEPT
Why Alkanes Need Radicals to React
Alkanes are famously unreactive toward ionic reagents — there's no pi bond, no polar bond, and no lone pair anywhere for an acid or nucleophile to grab onto. The one reaction that reliably gets an alkane to do anything under mild conditions is radical halogenation: exposing the alkane to a halogen (Cl₂ or Br₂, typically) and either UV light or heat, which supplies enough energy to homolytically cleave the weak halogen-halogen bond and kick off a self-sustaining radical chain reaction.
This isn't a polar, electron-pushing mechanism at all — it's a chain reaction built from single, unpaired electrons reacting one at a time, and it always proceeds through the same three-stage pattern: initiation, propagation, and termination.
💡 Memory Trick
The hub's trick anchors the whole reaction to its simplest example: CH₄ + Cl₂ → CH₃Cl (under UV light), via a free radical chain mechanism. Hold onto that one concrete reaction as your template, and remember the three-word stage sequence — initiation, propagation, termination — as the skeleton every radical halogenation mechanism you draw will follow, no matter how big the alkane gets.
THE THREE STAGES
Walking Through the Full Chain Mechanism
1
Initiation
UV light (or heat) supplies enough energy to homolytically cleave the weak Cl-Cl (or Br-Br) bond, splitting it evenly into two chlorine radicals. This step consumes energy but produces the reactive radicals that drive everything after it.Cl₂ --UV--> 2 Cl•
2
Propagation (two repeating steps)
A halogen radical abstracts a hydrogen from the alkane, forming H-X and a carbon radical. That carbon radical then reacts with another molecule of X₂, forming the alkyl halide product and regenerating a fresh halogen radical to keep the chain going. These two steps repeat over and over, which is why a single initiation event can convert enormous numbers of alkane molecules.Cl• + CH₄ → HCl + •CH₃, then •CH₃ + Cl₂ → CH₃Cl + Cl•
3
Termination
Any two radicals colliding and combining removes both from the chain, ending that particular chain (though the overall reaction keeps going elsewhere as long as other chains are still propagating). Termination steps are relatively rare compared to propagation steps, simply because radical concentrations stay low throughout the reaction.Cl• + Cl• → Cl₂, or •CH₃ + Cl• → CH₃Cl, or •CH₃ + •CH₃ → CH₃CH₃
SELECTIVITY ACROSS THE HALOGENS
Why Bromine Is the Chemist's Favorite
The four halogens behave very differently in this reaction. Fluorine is so reactive that the reaction is essentially uncontrollable and explosively exothermic — not useful synthetically. Chlorine reacts readily but with poor selectivity, often giving a messy mixture of every possible mono- and poly-chlorinated product. Bromine reacts more slowly but with excellent selectivity, strongly favoring substitution at the most substituted (and therefore most stable radical-forming) carbon — which makes it the halogen of choice whenever you actually care about which carbon gets halogenated. Iodine is essentially unreactive in this reaction — the process is endothermic and far too slow to be useful.
🧪 Lab Application
You need to selectively brominate the single tertiary C-H bond in 2-methylbutane without generating a messy mixture of products.
1
Choose the right halogen for the job. Reach for Br₂, not Cl₂ — bromine's much greater selectivity for the more substituted carbon makes this exact kind of clean, targeted substitution possible.
2
Initiate with light or heat. Irradiate with UV light (or apply gentle heat) to homolytically cleave Br₂ into bromine radicals and start the chain.
3
Let propagation do the selecting. Bromine radical abstraction strongly favors the hydrogen on the tertiary carbon, because the resulting tertiary radical is far more stable than a primary or secondary radical would be — so the reaction naturally funnels toward the product you want.
4
Expect a much cleaner product mixture than chlorination would give. If you'd used Cl₂ instead, poor selectivity would have produced a substantial mix of primary-, secondary-, and tertiary-substituted products all together, complicating your purification.
📌 Exam Application
Exam questions frequently ask you to predict or compare the product distribution from chlorination versus bromination of the same alkane — the expected answer hinges entirely on remembering that bromine is far more selective for the more-substituted (more stable radical) position, while chlorine gives a statistical, much less selective mixture.
⚠️ Most Common Free Radical Halogenation of Alkanes Mistakes
A common mistake is forgetting that even bromination isn't perfectly selective — it strongly favors the more substituted carbon, but minor amounts of other products can still form. The other frequent trap is mixing up which halogen is which: students often reverse chlorine's poor selectivity and bromine's high selectivity when writing exam answers under time pressure.
✓ Quick Self-Test
1) Name the three stages of a radical chain mechanism in order. 2) Why is fluorine impractical for controlled radical halogenation? 3) Why does bromination give cleaner, more selective products than chlorination? 4) What breaks the Cl-Cl bond in the initiation step? 5) Why is iodine essentially unreactive in this reaction?
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