Part 3: Free-radical substitution and product mixtures

Follow chlorine radicals through a chain reaction, then apply the same bookkeeping to different substitution positions and further chlorination.

Substitution of methane by chlorine

Under UV irradiation, methane and chlorine react by free-radical substitution. A chlorine atom replaces a hydrogen atom in the organic molecule; that hydrogen forms HCl. Chloromethane is an initial product, but further substitution and radical combination create a mixture.

A mechanism is a sequence of steps explaining the overall change. Do not present the overall equation as though it were an initiation or propagation step. Keep a radical dot on every species with an unpaired electron.

Initiation creates radicals

UV light supplies energy for homolytic fission of the Cl–Cl bond. Each chlorine atom receives one electron from the shared pair, giving two chlorine radicals. The bond being split at this initiation stage is Cl–Cl, not C–H.

Propagation regenerates the chain carrier

First, a chlorine radical removes a hydrogen atom from methane. HCl and a methyl radical form. Next, the methyl radical reacts with a chlorine molecule, forming chloromethane and a new chlorine radical. Each of these steps consumes one radical and produces one radical.

Add the two equations and cancel CH₃• and Cl•. What remains is the overall substitution equation. This cancellation is a useful check that the radical cycle is complete; the regenerated chlorine radical can attack another methane molecule.

Termination removes radicals from the chain

Two radicals combine to form a product with no unpaired electrons. Any one of these collisions removes chain carriers. Ethane can therefore appear even though it is not the desired chlorination product.

Further substitution reduces selectivity

Chloromethane still contains C–H bonds, so it can undergo another substitution to form dichloromethane. Continued replacement can give trichloromethane and tetrachloromethane. Excess methane relative to chlorine favours monosubstitution but does not guarantee a pure product.

The next pair of propagation equations gives dichloromethane. Adding them gives CH₃Cl + Cl₂ → CH₂Cl₂ + HCl. For replacement of all four original H atoms, the net equation uses four chlorine molecules and produces four HCl molecules.

Apply the pattern to another carbon position

Propane can form 1-chloropropane or 2-chloropropane because terminal and central hydrogen positions are different. To show the route to 2-chloropropane, remove the central hydrogen and place the radical dot on that carbon. Molecular formula C₃H₇• alone conceals the position.

Use CH₃C•HCH₃ to show the central radical, then form CH₃CHClCH₃. For terminal substitution, the intermediate is CH₃CH₂CH₂• and the product CH₃CH₂CH₂Cl. Do not assume equal product amounts: the number of equivalent H atoms and the rates of their abstraction both affect the mixture.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources