Mix methane and chlorine gas in a dark flask and nothing happens. Shine ultraviolet light on the same flask and two new compounds appear, chloromethane and hydrogen chloride: CH₄ + Cl₂ → CH₃Cl + HCl. One hydrogen on the methane has been swapped for a chlorine.
That swap is the one reaction of alkanes this course builds on. An alkane contains only C–C and C–H bonds, and both are close to nonpolar. It has no lone pair to donate, no empty orbital to accept electrons, and no partial charge for a nucleophile or an electrophile to aim at. None of the reactions you have met so far can touch it. That is why alkanes act as the inert frame that functional groups are attached to.
The swap is called radical halogenation. It needs a halogen plus heat or light, and its mechanism is unlike any you have seen: nothing in it carries a charge, and its arrows move single electrons instead of pairs. It also runs as a chain. Each step makes the species the next step needs, so once started, the reaction keeps itself going.
Homolysis, and the arrow that moves one electron
When H–Br dissolves in water, the bond breaks and both of its electrons leave with the bromine. The products are H⁺ and Br⁻, a cation and an anion. A break like this, where one atom takes both electrons, is called heterolysis. Almost every bond you have broken in this course so far broke this way, drawn with the ordinary curved arrow, which has a two-barbed head.
Shine light on Br–Br and the bond splits evenly instead: each bromine keeps one of the two electrons. This even split is homolysis. Its products are two neutral bromine atoms, each carrying one unpaired electron. Any atom or group with an unpaired electron is a radical, written with a dot, as in Br·. You first met homolysis in the bonding section of Foundations.
An arrow that moves a single electron has a single barb and is called a fishhook arrow. Drawing Molecules & Moving Electrons showed what it looks like. A bond that breaks homolytically needs two fishhooks, one for each of its electrons.
Three stages, and only one of them repeats
Follow methane and chlorine through the reaction. It runs in three stages, and each stage does something different to the number of radicals in the flask.
Initiation makes radicals from a molecule that had none. Light or heat splits Cl–Cl into two Cl·, so the radical count goes from zero to two. This step is slow and rare, but it only has to happen once in a while, and that is the whole trick of a chain reaction. A halogen–halogen bond is weak enough to break this way: breaking Cl–Cl takes 58 kcal/mol, and breaking Br–Br only 46.
When the starting materials have no weak bond, chemists add an initiator, a molecule built to fall apart into radicals on gentle warming. A peroxide, RO–OR, splits its weak O–O bond into two RO·. AIBN (azobisisobutyronitrile) breaks two C–N bonds instead, releasing N₂ gas and leaving two carbon radicals.
Propagation is two steps, and together they are the reaction. First, Cl· pulls a hydrogen atom off methane, leaving H–Cl and a methyl radical, ·CH₃. Removing a hydrogen atom this way is called hydrogen abstraction. Second, ·CH₃ takes one chlorine atom from Cl₂, which gives CH₃Cl and a fresh Cl·. Add the two steps and the radicals cancel, leaving CH₄ + Cl₂ → CH₃Cl + HCl. For any alkane R–H and halogen X₂ the net change is R–H + X₂ → R–X + H–X.
Each propagation step uses one radical and makes one, so the count never changes. The Cl· made in the second step goes on to start the first step again, which is why one initiation can turn over thousands of molecules before the chain stops.
Termination is any step in which two radicals meet and pair their electrons: Cl· + Cl·, ·CH₃ + Cl·, or ·CH₃ + ·CH₃. It uses two radicals and makes none, and that ends the chain. While the reaction runs, termination is rare, because radicals are always scarce and two of them seldom meet.
So the radical count names the stage. Up from zero is initiation, unchanged is propagation, and down to zero is termination.
The same four steps can be drawn with every electron shown. Where a bond breaks, one fishhook sends an electron to each atom. Where a bond forms, one fishhook comes from each partner, and the two meet between the atoms.
Which hydrogen gets taken
Methane has only one kind of hydrogen, but propane has two. Its six end hydrogens sit on carbons bonded to one other carbon, and its two middle hydrogens sit on a carbon bonded to two. Carbons, and the hydrogens on them, are called primary (1°), secondary (2°) or tertiary (3°) when that carbon is bonded to one, two or three other carbons. Pulling off an end hydrogen leaves a primary radical, and pulling off a middle one leaves a secondary radical.
A halogen removes a hydrogen faster when doing so leaves a more stable radical. The radical carbon has only seven valence electrons, one short of an octet, so anything that shares electron density with it helps. Neighboring C–H bonds do exactly that: each filled C–H σ orbital overlaps the half-filled p orbital and leaks a little of its electron density into it. This donation is hyperconjugation. It is the same effect often used to explain why staggered conformations are favored over eclipsed ones in Alkanes & Conformations. It also explains the carbocation order in How Reactions Happen. More alkyl groups on the radical carbon bring more C–H bonds that can donate, so the order is:
That is the carbocation order, for the same reason. A radical has one electron more than the matching cation and one fewer than the matching carbanion. Like the cation, it is short of an octet, so alkyl groups help it too. The difference is how much they help. Donation into an empty orbital is worth far more than donation into a half-filled one, so the spread from primary to tertiary is tens of kcal/mol for cations but only about 5 kcal/mol for radicals. The ranking is the same, only much flatter.
That 5 kcal/mol comes from bond dissociation energies, the energy needed to break a bond homolytically. The weaker a C–H bond, the more stable the radical left behind. A primary C–H takes about 101 kcal/mol to break, a secondary about 98, and a tertiary about 96.
How strongly the preference shows depends on the halogen. Each number below compares the rate at one hydrogen of each kind with the rate at one primary hydrogen.
| Relative rate per hydrogen | 3° | 2° | 1° |
|---|---|---|---|
| Chlorination (Cl₂) | ~5 | ~4 | 1 |
| Bromination (Br₂) | ~1600 | ~80 | 1 |
Chlorine prefers a tertiary hydrogen only about 5 to 1, while bromine prefers it about 1600 to 1.
Why bromine is choosy and chlorine is not
The explanation is the Hammond postulate, taught with energy diagrams in How Reactions Happen. A step's transition state is its highest-energy point, the top of the hill between reactants and products. The postulate says the transition state resembles whichever side of the step lies closer to it in energy. A downhill (exothermic) step therefore has an early, reactant-like transition state, and an uphill (endothermic) step has a late, product-like one.
An abstraction breaks a C–H bond and makes an H–X bond, so its enthalpy change, ΔH, is the energy of the bond broken minus the energy of the bond made:
| Abstraction | Bond broken | Bond made | ΔH |
|---|---|---|---|
| Cl· + R–H | C–H, ~98 | H–Cl, 103 | about −5 — downhill |
| Br· + R–H | C–H, ~98 | H–Br, 87.5 | about +11 — uphill |
Chlorine's abstraction runs downhill, so its transition state comes early. At that point the C–H bond has barely stretched and the radical has barely begun to form, so how stable the radical will be hardly affects the barrier. The barriers to a tertiary and a primary radical come out nearly level.
Bromine's abstraction runs uphill, so its transition state comes late. By the top of the hill the radical is almost fully formed, and nearly all of the 5 kcal/mol gap between a tertiary and a primary radical shows up in the barrier. A barrier difference that large makes one rate over a thousand times the other.
Why only chlorine and bromine
The same ΔH arithmetic decides which halogens are usable at all. H–F takes 136 kcal/mol to break, so fluorine's abstraction is downhill by nearly 40. That chain runs so hot that it is useless for picking one hydrogen over another, and it is often explosive. H–I takes only 71, so iodine's abstraction is uphill by about 27. That barrier is too high to cross at any practical temperature, and the overall reaction is uphill as well. Chlorine and bromine are the two halogens inside the workable window, one near each edge of it.
One more practical limit is that the product R–X still has C–H bonds, so it can be halogenated again. With chlorine, the chlorine already on the chain pulls electron density out of the nearby C–H bonds and makes them slightly harder to abstract, so a second chlorine tends to land farther along the chain. With bromine, the second bromine tends to land on the carbon next to the first, because the bromine already there helps stabilize a radical formed beside it. Either way, the product is a mixture of mono- and polyhalogenated compounds. Chemists therefore run radical halogenations with the alkane in large excess, so that a halogen radical is far more likely to meet starting material than product.
Count the hydrogens
The rates in the table are per hydrogen. Predicting a real product ratio also means counting how many of each kind of hydrogen the molecule has, and the count can outweigh the selectivity.
Start by sorting the hydrogens into sets of equivalent ones. Two hydrogens belong to the same set, and are called equivalent, when replacing either one with a halogen gives the same compound or its mirror image. In propane, replacing any of the six end hydrogens with chlorine gives 1-chloropropane every time, and replacing either middle hydrogen gives 2-chloropropane. So propane has two sets of hydrogens and gives two monochlorides. This replace-and-compare check is the substitution test from Stereochemistry. Mirror-image (enantiotopic) hydrogens react at the same rate with an achiral reagent such as Cl₂, so they count as one set.
Butane shows why you count sets of hydrogens, not carbons. It has four carbons but only two sets of hydrogens, the six on the ends and the four in the middle, so it gives exactly two monochlorides, not four. The two hydrogens on C2 are mirror-image partners (replacing one or the other gives the two enantiomers of 2-chlorobutane), but they react at the same rate, so they still belong to one set.
Chlorination of propane. Multiply each set's count by its per-hydrogen rate. Primary: 6 × 1 = 6. Secondary: 2 × 4 = 8. The total is 14, so the split is 6/14 and 8/14, about 43% 1-chloropropane and 57% 2-chloropropane. A four-to-one preference turns into a nearly even mixture, because there are three times as many primary hydrogens to find.
Bromination of 2-methylpropane. Primary: 9 × 1 = 9. Tertiary: 1 × 1600 = 1600. The total is 1609, so the tertiary bromide is 1600/1609, about 99%. A nine-to-one head start counts for nothing against a sixteen-hundred-fold preference.
The same substrate with chlorine. Primary: 9 × 1 = 9. Tertiary: 1 × 5 = 5. The total is 14, so the primary chloride is 9/14, about 64%. Here the major product comes from the less stable radical. On one molecule, bromine gives 99% of one isomer, while chlorine gives a roughly two-to-one mixture that favors the other.
The radical is flat
Brominate butane and most of the product comes from its secondary hydrogens, on C2 (C3 is equivalent to C2, so abstraction there gives the same product). The radical formed at C2 is flat. The carbon and its three groups, H, CH₃ and CH₂CH₃, all lie in one plane, and the unpaired electron sits in a p orbital standing at right angles to that plane, with one lobe above it and one below.
Br₂ can reach either lobe equally easily. Attack from the top face gives one 2-bromobutane, and attack from the bottom face gives its mirror image. C2 of the product is a new stereocenter, so the two products are enantiomers, and they form in equal amounts. A 50:50 mixture of two enantiomers is racemic, and it does not rotate plane-polarized light.
So whenever a radical halogenation creates a new stereocenter, the product is racemic. No choice of solvent or temperature changes this, because it comes from the shape of the radical. Stereochemistry made the same argument about any flat intermediate with two equivalent faces. Simple alkyl radicals are flat or very nearly so. A tertiary radical is slightly pyramidal, but it flips through the flat shape so fast that its two faces still behave alike.
Looking ahead: allylic bromination
This part is a preview. It needs alkenes, which arrive in Alkenes & Alkynes. It sits here because its mechanism is the chain you have just learned. Read it for the logic now, and come back once you have met double bonds properly.
One kind of hydrogen is far easier to abstract than any on an ordinary alkane: a hydrogen on an allylic carbon, the carbon right next to a C=C double bond. Take 1-butene, CH₂=CH–CH₂–CH₃. Its allylic carbon is C3. Removing a hydrogen from C3 leaves a radical whose unpaired electron is spread by resonance over C1 and C3. That extra stabilization makes the abstraction much easier than a secondary C–H alone would suggest. The carbon next to a benzene ring behaves the same way for the same reason, and Aromatic Chemistry calls that position benzylic.
The difficulty is practical. With Br₂ at ordinary concentration, addition across the double bond is faster than allylic substitution, so addition wins. The fix is NBS (N-bromosuccinimide). Each molecule of HBr the chain makes reacts with NBS to give one molecule of Br₂, so bromine is released slowly and its concentration stays low throughout. Addition needs a reasonable concentration of Br₂, but the radical chain does not. Starve the reaction of bromine and substitution wins.
The allylic radical from 1-butene has two resonance forms. In one, the double bond is C1=C2 and the unpaired electron is on C3. In the other, the double bond has moved to C2=C3 and the unpaired electron is on C1. Bromine can bond at either end, so NBS gives two products: 3-bromo-1-butene from the first form, and 1-bromo-2-butene, with its double bond moved, from the second. Drawing both resonance forms is how you predict the second product, and skipping the second form is how that product gets missed.
What carries forward
Radical chemistry comes back several times: in anti-Markovnikov addition of HBr, benzylic bromination, radical polymerization and mass spectrometry. The anti-Markovnikov addition of HBr to an alkene in the presence of peroxides, in Alkenes & Alkynes, runs on the same chain: initiation, two propagation steps and termination. There a bromine radical adds to the double bond first, not a proton, which is why the bromine ends up on the less substituted carbon, the reverse of the Markovnikov product. Mass spectrometry uses radicals too. The molecular ion is a radical cation, and it breaks into a cation and a neutral radical, favoring the breaks that give the most stable pieces.
The idea to carry away is smaller than the mechanism: the more reactive a reagent is, the less choosy it is. A very reactive species is unselective because it commits early, before the product has taken shape. A sluggish one is selective because it commits late, when it can already feel how stable its product will be.