Alkanes & Conformations · Section 31 of 116

Radical halogenation

Practice this — interactive lesson

Alkanes are the one family in this course defined by what they will not do. Every bond is C–C or C–H, both close to nonpolar; there is no lone pair to donate, no empty orbital to accept, no partial charge for a nucleophile or an electrophile to aim at. Nothing in the first eight chapters attacks an alkane, and that is the point of them — they are the inert skeleton everything else hangs off.

There is one reaction. Given a halogen and either heat or ultraviolet light, an alkane will swap a hydrogen for a halogen. It runs by a mechanism unlike anything else here: no charges, no nucleophiles, no arrows pushing pairs of electrons. It moves single electrons instead, and it runs as a chain.

The condition is not decoration. Br₂ and an alkane in a dark flask at room temperature do nothing at all — which is exactly why bromine water is the classic test that tells an alkene from an alkane. The light or the heat is what starts the chain, and without it there is no reaction to have.

Homolysis, and the arrow that moves one electron

Every bond you have broken so far broke heterolytically: both electrons left with one atom, producing a cation and an anion, drawn with the ordinary double-barbed curved arrow. A halogen molecule under UV light does the opposite. The bond breaks homolytically — one electron to each atom — and the products are two neutral species each carrying an unpaired electron. Those are radicals.

Because the arrow now carries one electron rather than two, it is drawn with a single barb, and is called a fishhook arrow. Two fishhooks are needed to break one bond, one for each electron. Mixing the two arrow types in a single step is the most common way a radical mechanism is drawn wrongly: if a step produces a charge, it was not a radical step.

Three stages, and only one of them repeats

INITIATIONX–Xhv or heatX• + X•radicals: 0 → 2happens rarely, andonly has to happen oncePROPAGATIONX• + R–H → R• + H–XR• + X–X → R–X + X•each step uses one radical and makes oneradicals: 2 → 2, so the cycle never stops itselfnet: R–H + X₂ → R–X + H–XTERMINATIONX• + X• → X–XR• + X• → R–XR• + R• → R–Rradicals: 2 → 0rare while it runs —two radicals have to meet
Sort the steps by what each does to the number of radicals in the flask, and the three stages name themselves. Up from zero is initiation, unchanged is propagation, down to zero is termination.The middle panel is the reaction; the other two only start and stop it. Because each propagation step consumes one radical and produces one, the pair runs as a loop, and a single initiation event can turn over thousands of molecules before two radicals happen to collide and end it. That is also why the second propagation step is not termination even though the product appears there — the product is not what distinguishes the stages, the radical count is.

Initiation makes radicals out of a molecule that had none. Light or heat splits X–X into two X·. This happens rarely — it is the slow, expensive step — and it does not need to happen often, which is the whole trick of a chain reaction.

Propagation is two steps, and together they are the reaction. A halogen radical abstracts a hydrogen from the alkane, giving H–X and a carbon radical. That carbon radical then attacks X₂, taking one halogen and releasing the other as a fresh X·. Add the two together and the radicals cancel out: R–H + X₂ → R–X + H–X. Look at what each step does to the radical count — it consumes one and produces one. The chain feeds itself, and a single initiation event can turn thousands of alkane molecules over before it stops.

Termination is any step where two radicals find each other and pair up: X· + X·, R· + X·, or R· + R·. These consume two radicals and make none, which is what ends the chain. They are rare while the reaction runs, because the radical concentration is always tiny — two radicals have to collide, and there are very few of them in the flask at any moment.

Common slip: labelling the second propagation step as termination because it "makes the product." Sort the steps by what happens to the radical count, not by whether a product appears. Up from zero is initiation, unchanged is propagation, down to zero is termination.

Which hydrogen gets taken

An alkane with more than one kind of hydrogen can give more than one product, and which one dominates is decided by the stability of the carbon radical formed when that hydrogen is abstracted. Radicals follow the ordering you already know from carbocations, and for the same reason — hyperconjugation from neighboring C–H bonds into the half-filled orbital:

3° > 2° > 1° > methyl

What is genuinely surprising is how much that ordering matters, and that the answer depends on which halogen you picked.

Relative rate per hydrogen
Chlorination (Cl₂)~5~41
Bromination (Br₂)~1600~801

Bromine is fussy and chlorine is not, and the reason is the Hammond postulate again. Abstraction by the very reactive chlorine radical is exothermic, so its transition state comes early and looks like the reactants — at which point the radical has barely formed and its stability hardly matters. Abstraction by the much less reactive bromine radical is endothermic, so its transition state comes late and looks like the radical itself, and every bit of that radical's stability shows up in the rate.

Practically: if a synthesis needs one clean product from a substrate with a tertiary C–H, use bromine. Chlorination of the same substrate gives a mixture you would have to separate. "Bromine is selective, chlorine is indiscriminate" is worth memorizing in exactly those terms, because exam questions are usually asking which halogen to choose rather than asking you to draw anything.

Two corrections to the arithmetic

The table is per hydrogen, so predicting a product ratio means multiplying each rate by how many equivalent hydrogens there are. Propane has six primary hydrogens and only two secondary ones, so even with the secondary position four times more reactive toward chlorine, the two products come out close to even — the statistics nearly cancel the selectivity.

And the carbon radical is essentially planar, with its unpaired electron in a p orbital, so a halogen can arrive on either face with equal ease. If the reaction creates a new stereocenter, the product is racemic. A radical mechanism cannot deliver a single enantiomer, and that is a structural fact about the intermediate rather than a limitation of the conditions.

Allylic and benzylic bromination

One position is far easier to halogenate than any ordinary alkane carbon: the one next door to a double bond or a benzene ring. Abstracting a hydrogen there gives a radical whose unpaired electron is delocalized by resonance across the pi system, and that extra stabilization makes the abstraction much easier than the substitution pattern alone would suggest.

The difficulty is practical. Adding Br₂ to an alkene does not give allylic substitution — it gives addition across the double bond, which is faster. The solution is NBS (N-bromosuccinimide), a reagent that releases bromine slowly and keeps its concentration permanently low. Addition needs a reasonable concentration of Br₂; the radical chain does not. Starve the reaction of bromine and substitution wins.

Worth noticing

Because the allylic radical is a resonance hybrid, the bromine can end up at either end of that delocalized system, and allylic bromination frequently gives two products with the double bond in two different places. Drawing both resonance forms of the radical is how you predict the second one, and forgetting it is how the second one gets missed.

What carries forward

Radical chemistry reappears twice in this course. The anti-Markovnikov addition of HBr in the presence of peroxides, in the Markovnikov section, is the same chain logic applied to an alkene — initiation, two propagation steps, termination — which is why the regiochemistry flips: a bromine radical adds first, not a proton. And the fragmentation patterns in mass spectrometry are radical chemistry too, since the molecular ion is itself a radical cation that breaks apart to leave the most stable radical it can.

The transferable idea is smaller than the mechanism: a reaction's selectivity tracks how reactive its attacking species is. A very reactive species is unselective because it commits early; a sluggish one is selective because it commits late and gets to see what it is making. That is the same Hammond reasoning that explains carbocation rearrangements, and it will explain the difference between kinetic and thermodynamic enolates in Module 11.