Part 2: Radical substitution and product mixtures

Build initiation, propagation and termination equations, and explain why alkane halogenation is a poor route to one pure product.

UV supplies energy for homolysis

Alkanes react with chlorine or bromine in UV light by radical substitution: a hydrogen atom is replaced by a halogen atom. For methane chlorination, the overall first substitution is CH₄ + Cl₂ → CH₃Cl + HCl. UV splits the Cl–Cl bond homolytically to generate radicals.

Consume one radical and regenerate another

First Cl• removes a hydrogen atom from methane, making HCl and a methyl radical. The methyl radical then reacts with Cl₂ to form chloromethane and regenerate Cl•. Adding the two equations gives the overall substitution; radicals cancel.

For propane, abstraction from an end carbon gives CH₃CH₂CH₂•, while abstraction from the middle gives CH₃CH•CH₃. Keep the dot at the carbon missing H. Reaction with Br₂ gives respectively 1-bromopropane or 2-bromopropane plus Br•.

Two radicals combine

Termination removes radicals without producing a new radical. Possible methane-chlorination endings include chlorine recombination, product formation and coupling to ethane. A step with a radical on both sides is propagation, not termination.

Different positions and further substitution

Two distinct reasons create mixtures: hydrogen can be replaced at inequivalent positions, and already halogenated products can react again. Methane gives no positional isomers at first substitution, but CH₃Cl can continue to CH₂Cl₂, CHCl₃ and CCl₄. Excess alkane relative to halogen helps favour monosubstitution but does not guarantee a pure product.

The side product is hydrogen halide, not H₂. Verify conservation in every equation, especially when a carbon radical couples with another. There is no fixed universal product ratio: relative pathways depend on the alkane, halogen and conditions.

Why propagation sustains a chain

Add the two ethane chlorination steps below. The ethyl radical appears once as a product and once as a reactant; chlorine radical does too. Cancelling them leaves the overall substitution. A radical produced in the second step can start the first step again with a new ethane molecule.

UV is needed to initiate radical formation; it is not written as a material reactant consumed in the overall equation. The chain eventually ends when radicals are removed, including through termination collisions. A reaction equation and a mechanism answer different questions: the overall equation alone does not show these intermediate steps.

Count unique positions using symmetry

In butane, the two terminal CH₃ groups are equivalent and the two internal CH₂ groups are equivalent. Monochlorination can therefore form two structural isomers: 1-chlorobutane and 2-chlorobutane. Drawing the chain in the opposite direction does not add 3- and 4-chlorobutane as new structures.

In 2-methylpropane, the three CH₃ groups are equivalent to one another, but the central C–H is different. Replacing a terminal H gives 1-chloro-2-methylpropane; replacing the central H gives 2-chloro-2-methylpropane. Count distinct environments before writing radicals or products.

The number of H atoms in an environment is not enough to predict exact product percentages. Relative abstraction rates differ with the halogen and environment. If a question supplies selectivity data, use it; otherwise explain the possible products without inventing a ratio.

Diagnose an incorrect propagation step

The proposed equation CH₃• + HCl → CH₃Cl + H• is not the usual second propagation step for methane chlorination. The reacting halogen source in that step is Cl₂, and Cl• is regenerated. Memorising the correct two-step pattern avoids changing reagents to force the desired product.

For substitution at an internal carbon, remove exactly one H there and place the radical dot on that carbon. For example CH₃CH₂CH₂CH₃ gives CH₃CH•CH₂CH₃ for the route to 2-chlorobutane. Do not put both a full positive charge and a radical dot on an ordinary neutral alkyl radical in this mechanism.

Termination products are found by joining the atoms bearing unpaired electrons. Two ethyl radicals make a four-carbon chain; an ethyl radical and a chlorine radical make chloroethane. All atoms must remain accounted for.

Quick checks

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

Sources