Exploring Free-Radical Mechanisms Revision Notes | A Level AQA Chemistry

1. Why Radicals Matter in Organic Chemistry

Alkanes are famously unreactive — they resist acids, bases, and most reagents at room temperature. Yet shine UV light on a mixture of methane and chlorine, and a rapid, exothermic reaction takes place. The reason is free-radical substitution, and understanding its mechanism is a core AQA A Level requirement.

This lesson unpacks the three stages of the mechanism, shows you how to write balanced equations for each step, and explains why the reaction produces a mixture of products — a fact examiners love to probe.

2. Homolytic Fission and the Birth of a Radical

Every radical mechanism begins with bond breaking. There are two ways a covalent bond can break: heterolytic fission, where both electrons go to one atom (producing ions), and homolytic fission, where one electron goes to each atom (producing radicals). Free-radical mechanisms always involve homolytic fission.

The dot notation (Cl•) is the AQA-standard way to show a radical. Each dot represents the single unpaired electron that makes the species so reactive — it will grab an electron from almost anything nearby to achieve a stable pair.

3. The Three Stages: Initiation, Propagation, Termination

Free-radical substitution always follows the same three-stage pattern. Each stage has a distinct role, and AQA requires you to write a balanced equation for every step. The example used throughout this lesson is the reaction of methane with chlorine under UV light.

The overall reaction is: $CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)$. Now let's see how that happens at the molecular level, one stage at a time.

3a. Initiation

Initiation is the step that generates the first radicals. UV light provides the energy needed to break the Cl–Cl bond homolytically. Without this energy input, the reaction does not start — which is why the mixture of methane and chlorine is stable in the dark.

The balanced equation for initiation is: $Cl_2(g) \xrightarrow{UV} 2Cl\bullet(g)$. Notice that one molecule of $Cl_2$ produces two chlorine radicals. The UV light is shown above the arrow as the condition, not as a reactant.

3b. Propagation

Propagation is where the chain reaction runs. There are always two propagation steps, and together they regenerate a radical at the end — which is why the chain keeps going without needing more UV light.

• Propagation step 1: $Cl\bullet(g) + CH_4(g) \rightarrow CH_3\bullet(g) + HCl(g)$
• Propagation step 2: $CH_3\bullet(g) + Cl_2(g) \rightarrow CH_3Cl(g) + Cl\bullet(g)$

In step 1, the chlorine radical abstracts a hydrogen atom from methane, forming HCl and a methyl radical ($CH_3\bullet$). In step 2, the methyl radical attacks a chlorine molecule, forming chloromethane and regenerating a chlorine radical. That regenerated $Cl\bullet$ feeds straight back into step 1 — hence the term chain reaction.

Each propagation step must be balanced: check that atom counts and radical counts are consistent on both sides. A radical is consumed and a new radical is produced in each step — the total number of unpaired electrons is conserved within each propagation equation.

3c. Termination

Termination occurs when two radicals collide and combine, destroying both unpaired electrons and forming a stable, closed-shell molecule. There is no new radical produced, so the chain ends.

• $Cl\bullet(g) + Cl\bullet(g) \rightarrow Cl_2(g)$
• $CH_3\bullet(g) + Cl\bullet(g) \rightarrow CH_3Cl(g)$
• $CH_3\bullet(g) + CH_3\bullet(g) \rightarrow C_2H_6(g)$

All three termination equations above are valid. The third one — two methyl radicals combining to form ethane — explains why trace amounts of ethane are always detected as a by-product of this reaction. This is a classic exam question: examiners ask you to suggest why ethane is found in the products.

4. Writing Balanced Equations for Each Step

AQA specifically requires you to write balanced equations for the steps in a free-radical mechanism. This means atom counts must balance on both sides, and radical dots must be shown correctly. The worked example below takes you through the full mechanism for methane and chlorine, step by step.

Worked example: Writing the Full Mechanism for CH4 + Cl2

Write balanced equations for all steps in the free-radical substitution of methane with chlorine to form chloromethane.

1. $Cl_2(g) \xrightarrow{UV} 2Cl\bullet(g)$ — One Cl-Cl bond breaks homolytically under UV light. One molecule produces two radicals. Check: 2 Cl atoms on each side. [B1]
2. $Cl\bullet(g) + CH_4(g) \rightarrow CH_3\bullet(g) + HCl(g)$ — The Cl radical abstracts one H from methane. A methyl radical is produced. Check: 1 Cl, 1 C, 4 H on each side. One radical in, one radical out. [M1]
3. $CH_3\bullet(g) + Cl_2(g) \rightarrow CH_3Cl(g) + Cl\bullet(g)$ — The methyl radical attacks Cl2, taking one Cl atom and regenerating a Cl radical. Check: 1 C, 3 H, 2 Cl on each side. One radical in, one radical out. [M1]
4. $CH_3\bullet(g) + Cl\bullet(g) \rightarrow CH_3Cl(g)$ — Two radicals combine. No new radical is produced, so the chain ends. Check: 1 C, 3 H, 1 Cl on each side. [B1]
5. Step 1 + Step 2: $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$ — Adding the two propagation equations and cancelling the radical intermediates ($Cl\bullet$ and $CH_3\bullet$) gives the overall equation. This confirms the mechanism is internally consistent. [A1]

Answer: Initiation: $Cl_2 \xrightarrow{UV} 2Cl\bullet$; Propagation 1: $Cl\bullet + CH_4 \rightarrow CH_3\bullet + HCl$; Propagation 2: $CH_3\bullet + Cl_2 \rightarrow CH_3Cl + Cl\bullet$; Termination (example): $CH_3\bullet + Cl\bullet \rightarrow CH_3Cl$

The sanity check in step 5 is a powerful self-verification tool. If you add your two propagation equations and the radical intermediates do not cancel cleanly, at least one equation is wrong. Use this in the exam whenever you have time.

5. Why Free-Radical Reactions Produce Mixtures

Here is where the chemistry gets genuinely interesting — and where examiners probe deeper understanding. The reaction of methane with chlorine does not stop at chloromethane. The $CH_3Cl$ product still contains C–H bonds, and those bonds can be attacked by further $Cl\bullet$ radicals.

This leads to a cascade of further substitutions: $CH_3Cl \rightarrow CH_2Cl_2 \rightarrow CHCl_3 \rightarrow CCl_4$. All four chlorinated products form simultaneously, making the reaction poor for selective synthesis. The ratio of products depends on the relative amounts of $CH_4$ and $Cl_2$ present — excess methane favours monosubstitution, but complete selectivity is impossible.

With longer-chain alkanes, the problem compounds further. Propane, for example, has two distinct types of hydrogen — primary (on $CH_3$ groups) and secondary (on the $CH_2$ group). Radicals can abstract either type, producing a mixture of 1-chloropropane and 2-chloropropane. The reaction has no mechanism for distinguishing between them.

6. Comparing Free-Radical Mechanisms with Other Mechanism Types

AQA expects you to compare free-radical substitution with other mechanisms. For example, nucleophilic substitution may proceed by SN1 via a carbocation intermediate or by SN2 via a single transition state, whereas free-radical substitution proceeds through radicals generated by homolytic fission.

The table above highlights a critical AQA specification point: curly arrows are not required for radical mechanisms at A Level. You are expected to write balanced equations for each step instead. For nucleophilic substitution and electrophilic addition, full curly arrow mechanisms are required — so do not confuse the two conventions.

7. Exam Technique and Common Mistakes

Free-radical mechanism questions are reliable mark-earners once you know the conventions. The mistakes that cost marks are almost always the same ones — and they are all avoidable.

Exam questions on this topic frequently use the command word explain, which requires cause and effect — not just a description. For example, 'Explain why the reaction produces a mixture of products' demands that you link the lack of selectivity of radical attack to the structural features of the alkane, not merely state that a mixture forms.

Free-radical substitution is a foundational mechanism that introduces the logic of chain reactions, balanced radical equations, and the concept of non-selective reactivity — all of which reappear in later topics across organic chemistry.