Free Radical Substitution

When alkanes are mixed with chlorine or bromine under normal conditions, nothing happens. Alkanes are unreactive towards halogens in the dark. However, expose the mixture to ultraviolet (UV) light and a vigorous reaction occurs. A hydrogen atom on the alkane is replaced by a halogen atom — this is free radical substitution.

Understanding this mechanism is essential for Edexcel A-Level Chemistry. It is one of the first reaction mechanisms you will learn, and it introduces key concepts — radicals, homolytic fission, and chain reactions — that appear throughout organic chemistry.

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What Is a Free Radical?

A free radical is a species (atom or group of atoms) with an unpaired electron. Free radicals are represented with a dot to show the unpaired electron, for example Cl• or CH₃•.

Free radicals are:

• Highly reactive because the unpaired electron seeks to pair up
• Short-lived — they react almost immediately after formation
• Electrically neutral (they are not ions)

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Homolytic vs Heterolytic Fission

Understanding the difference between these two types of bond breaking is fundamental to organic mechanisms:

Homolytic fission: The bond breaks so that each atom takes one electron from the shared pair. This produces two free radicals. It is shown using single-headed curly arrows (fish-hook arrows), each showing the movement of one electron.

For the Cl-Cl bond: Cl-Cl → Cl• + Cl•

Heterolytic fission: The bond breaks so that one atom takes both electrons from the shared pair. This produces a cation and an anion. It is shown using double-headed curly arrows (standard curly arrows), showing the movement of an electron pair.

For H-Br: H-Br → H⁺ + Br⁻

Exam critical: Using the wrong type of curly arrow is a common mistake. Fish-hook arrows (half-headed) are ONLY for radical mechanisms. Double-headed curly arrows are for all ionic/polar mechanisms (electrophilic addition, nucleophilic substitution, etc.).

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The Three Stages of Free Radical Substitution

The mechanism is a chain reaction with three distinct stages: initiation, propagation, and termination.

Stage 1: Initiation

UV light provides the energy to break the Cl-Cl bond homolytically:

Cl₂ → 2Cl•

This is the only step that requires UV light. The energy of a UV photon is sufficient to break the relatively weak Cl-Cl bond (bond enthalpy 242 kJ mol⁻¹). Once chlorine radicals are produced, the chain reaction proceeds without further UV input.

Curly arrow description: Two fish-hook arrows are drawn from the Cl-Cl bond, one curving to each Cl atom. Each arrow represents one electron moving to the respective chlorine atom.

Stage 2: Propagation

Propagation steps are the "chain-carrying" reactions. Each step consumes one radical but produces another, sustaining the chain.

Propagation step 1: A chlorine radical abstracts a hydrogen atom from methane:

CH₄ + Cl• → CH₃• + HCl

Curly arrow description: A fish-hook arrow from the C-H bond to the H atom (one electron moves with the hydrogen). A second fish-hook arrow from the unpaired electron on Cl• to the space between H and Cl (forming the new H-Cl bond). The result: the C-H bond breaks homolytically, hydrogen pairs with the chlorine radical electron to form HCl, and the methyl group is left as a radical (CH₃•).

Propagation step 2: The methyl radical attacks a chlorine molecule:

CH₃• + Cl₂ → CH₃Cl + Cl•

Curly arrow description: A fish-hook arrow from the unpaired electron on CH₃• towards the nearer Cl atom (forming the new C-Cl bond). A fish-hook arrow from the Cl-Cl bond to the further Cl atom (that Cl leaves as a radical). The chlorine radical is regenerated — it can now participate in propagation step 1 again.

The overall equation is obtained by adding the two propagation steps:

CH₄ + Cl₂ → CH₃Cl + HCl

Notice that the radicals cancel out — they are intermediates, not in the overall equation.

Stage 3: Termination

Termination occurs when two radicals collide and their unpaired electrons pair up, forming a covalent bond. This removes radicals from the reaction mixture:

Cl• + Cl• → Cl₂ CH₃• + CH₃• → C₂H₆ CH₃• + Cl• → CH₃Cl

Termination steps are less likely than propagation because the concentration of radicals at any moment is very low, making radical-radical collisions rare. However, they do occur, which is why the reaction eventually stops.

Exam tip: The termination step CH₃• + CH₃• → C₂H₆ explains why traces of ethane are detected as a by-product. This is experimental evidence for the radical mechanism — ethane cannot be explained by any simple substitution of methane by chlorine.

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Bromination of Methane

The same mechanism applies to the reaction of methane with bromine in UV light:

Initiation: Br₂ → 2Br•

Propagation step 1: CH₄ + Br• → CH₃• + HBr

Propagation step 2: CH₃• + Br₂ → CH₃Br + Br•

Overall: CH₄ + Br₂ → CH₃Br + HBr

Bromination is slower than chlorination because the Br-H bond formed in propagation step 1 is weaker than the H-Cl bond. The first propagation step is endothermic for bromination (ΔH positive) but exothermic for chlorination (ΔH negative), making the activation energy higher for bromination.

Energetics Comparison

	Chlorination Step 1	Bromination Step 1
Bond broken	C-H (412 kJ mol⁻¹)	C-H (412 kJ mol⁻¹)
Bond formed	H-Cl (431 kJ mol⁻¹)	H-Br (366 kJ mol⁻¹)
ΔH for step	−19 kJ mol⁻¹ (exothermic)	+46 kJ mol⁻¹ (endothermic)

This energetic difference explains why chlorination is faster: the first propagation step is more thermodynamically favourable.

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Limitations and Problems

Multiple Substitution

Once chloromethane (CH₃Cl) is formed, it can itself undergo further substitution:

CH₃Cl + Cl• → CH₂Cl• + HCl CH₂Cl• + Cl₂ → CH₂Cl₂ + Cl•

This continues to give trichloromethane (CHCl₃) and tetrachloromethane (CCl₄). The product is always a mixture of mono-, di-, tri-, and tetra-substituted products.

Lack of Selectivity

For alkanes with different types of C-H bond (e.g., propane has both primary and secondary C-H bonds), the chlorine radical is not very selective. It attacks primary and secondary C-H bonds with similar ease, producing a mixture of position isomers.

However, bromine radicals are much more selective than chlorine radicals. Bromine preferentially attacks the weaker secondary and tertiary C-H bonds, giving a higher proportion of the secondary/tertiary substitution product. This selectivity difference is because the endothermic first propagation step in bromination has a late transition state that resembles the radical product, so the stability of the carbon radical matters more.

Difficult to Control

Because of multiple substitution and lack of selectivity, free radical substitution is not a good synthetic method when a single, pure product is desired. The yield of any one product is low.

To maximise the yield of the monosubstituted product, use a large excess of the alkane relative to the halogen. This ensures that chlorine radicals are statistically more likely to encounter an unreacted alkane molecule than an already-substituted one.

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Why UV Light Is Required

The Cl-Cl bond does not break at room temperature because the bond enthalpy (242 kJ mol⁻¹) is too high to be provided by thermal energy alone at 25 °C. UV photons carry sufficient energy to cause homolytic fission. The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the light. UV light has a high enough frequency to provide the required energy.

Once the chain reaction begins, the propagation steps are self-sustaining and exothermic (for chlorination), so no further UV input is needed.

If the UV light source is removed, no new radicals are generated in the initiation step, but existing chain reactions continue until termination occurs. The reaction rate decreases and eventually stops.

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Why Not Fluorination or Iodination?

Fluorination: The reaction of fluorine with alkanes is too vigorous — it is explosive. Both propagation steps are highly exothermic because F forms very strong bonds (H-F bond enthalpy 562 kJ mol⁻¹). The reaction cannot be controlled.

Iodination: The reaction is too slow to be useful. The first propagation step (R-H + I• → R• + HI) is highly endothermic because the H-I bond is weak (297 kJ mol⁻¹), giving a very high activation energy. The reaction essentially does not proceed.

Only chlorination and bromination are practically useful.

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Summary of Key Points

• Free radicals have an unpaired electron and are highly reactive
• Homolytic fission produces radicals; shown with single-headed (fish-hook) curly arrows
• Heterolytic fission produces ions; shown with double-headed curly arrows
• The mechanism has three stages: initiation (UV breaks halogen bond), propagation (chain-carrying steps), termination (radicals combine)
• Products are always a mixture due to multiple substitution
• Large excess of alkane favours monosubstitution
• Ethane detected as a by-product is evidence for the radical mechanism
• Chlorination is faster but less selective than bromination

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A-Level Deep Dive: Free Radical Substitution

Spec mapping

Edexcel 9CH0 specification Topic 6 — Organic Chemistry I, sub-topic 6.3 covers the mechanism of free-radical substitution: homolytic fission of the halogen molecule under UV light to generate radicals, the three-stage mechanism (initiation, propagation, termination), the formation of polyhalogenated by-products as a consequence of multiple substitution, and the limited industrial value of this route as a result (refer to the official specification document for exact wording). Sub-topic 6.3 is examined directly on Paper 2 (mechanism with curly arrows) and on Paper 3 (atmospheric chemistry, where the same mechanism explains stratospheric ozone destruction by chlorofluorocarbons). The data booklet does not list bond enthalpies for halogens specifically; common values: F-F (158), Cl-Cl (242), Br-Br (193), I-I (151) kJ mol$^{-1}$−1. The Edexcel formula booklet does not provide this — candidates use mean bond enthalpies given in question.

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Worked example with full mark scheme

Question (8 marks):

(a) Write the full mechanism for the free-radical substitution of methane by chlorine to form chloromethane ($CH_3Cl$CH3​Cl). Label each step as initiation, propagation, or termination. (5)

(b) Explain, with reference to the mechanism, why a mixture of $CH_3Cl$CH3​Cl, $CH_2Cl_2$CH2​Cl2​, $CHCl_3$CHCl3​ and $CCl_4$CCl4​ is obtained, and why this reaction is not industrially useful for the selective synthesis of chloromethane. (3)

Solution with mark scheme:

(a) Step 1 — initiation. Cl2 absorbs a UV photon and undergoes homolytic fission (each atom takes one electron):

$Cl_2 \xrightarrow{UV} 2Cl^{\bullet}$Cl2​UV​2Cl∙

M1 — initiation correct: Cl2 → 2Cl•, with UV (or hν) condition stated.

Step 2 — propagation, step 1. A chlorine radical attacks methane, abstracting an H atom to give HCl and a methyl radical:

$Cl^{\bullet} + CH_4 \rightarrow HCl + CH_3^{\bullet}$Cl∙+CH4​→HCl+CH3∙​

M1 — propagation step 1 correct.

Step 3 — propagation, step 2. The methyl radical attacks Cl2, abstracting a Cl atom to give CH3Cl and regenerate a chlorine radical:

$CH_3^{\bullet} + Cl_2 \rightarrow CH_3Cl + Cl^{\bullet}$CH3∙​+Cl2​→CH3​Cl+Cl∙

M1 — propagation step 2 correct; the regenerated Cl• re-enters the propagation cycle, sustaining the chain.

Step 4 — termination (any one). Two radicals combine to terminate the chain:

$Cl^{\bullet} + Cl^{\bullet} \rightarrow Cl_2$Cl∙+Cl∙→Cl2​

$CH_3^{\bullet} + CH_3^{\bullet} \rightarrow CH_3CH_3 \text{ (ethane)}$CH3∙​+CH3∙​→CH3​CH3​ (ethane)

$CH_3^{\bullet} + Cl^{\bullet} \rightarrow CH_3Cl$CH3∙​+Cl∙→CH3​Cl

M1 — at least one termination step correct.

A1 — all three stages clearly labelled as initiation / propagation / termination, with arrows balanced.

(b) Step 1 — chain propagation continues beyond first substitution. Once $CH_3Cl$CH3​Cl is formed, a Cl• radical can abstract another H from $CH_3Cl$CH3​Cl:

$Cl^{\bullet} + CH_3Cl \rightarrow HCl + CH_2Cl^{\bullet}$Cl∙+CH3​Cl→HCl+CH2​Cl∙

The $CH_2Cl^{\bullet}$CH2​Cl∙ radical reacts with Cl2 to give $CH_2Cl_2$CH2​Cl2​, which is then attacked again, producing $CHCl_3$CHCl3​ and finally $CCl_4$CCl4​.

M1 — explanation of further substitution: each successive chloromethane molecule is itself a substrate for further H-abstraction.

A1 — therefore the product is a mixture of $CH_3Cl$CH3​Cl, $CH_2Cl_2$CH2​Cl2​, $CHCl_3$CHCl3​ and $CCl_4$CCl4​, in ratios depending on Cl2:CH4 stoichiometry.

A1 — this is not industrially useful for the selective synthesis of $CH_3Cl$CH3​Cl because:

• yields of any single product are low (typically 30–50%);
• separation of the four products by distillation is energy-intensive;
• there is poor atom economy and significant HCl waste.

Total: 8 marks (M5 A3).

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Specimen question modelled on the Edexcel 9CH0 Paper 3 format

Question (6 marks): Chlorofluorocarbons (CFCs) such as CCl2F2 (R-12) were used as refrigerants until the discovery that they damage the ozone layer.

(a) Write the equation for the photolysis of CCl2F2 in the stratosphere, and explain why C-Cl is preferentially broken rather than C-F. (3)

(b) Show how the chlorine radical produced catalyses the destruction of ozone, identifying the propagation steps and the regenerated radical. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1) — $CCl_2F_2 \xrightarrow{UV} CClF_2^{\bullet} + Cl^{\bullet}$CCl2​F2​UV​CClF2∙​+Cl∙.
• A1 (AO2.1) — C-Cl bond enthalpy (~340 kJ mol$^{-1}$−1) is lower than C-F (~485 kJ mol$^{-1}$−1), so C-Cl is preferentially photolysed.
• A1 (AO2.1) — UV photons in the stratosphere have energies sufficient to break C-Cl but not C-F, making C-Cl the kinetically accessible bond.

(b)

• M1 (AO1.1) — propagation step 1: $Cl^{\bullet} + O_3 \rightarrow ClO^{\bullet} + O_2$Cl∙+O3​→ClO∙+O2​.
• M1 (AO1.1) — propagation step 2: $ClO^{\bullet} + O_3 \rightarrow Cl^{\bullet} + 2O_2$ClO∙+O3​→Cl∙+2O2​ (or $ClO^{\bullet} + O \rightarrow Cl^{\bullet} + O_2$ClO∙+O→Cl∙+O2​).
• A1 (AO2.1) — net: $2O_3 \rightarrow 3O_2$2O3​→3O2​, with Cl• regenerated; Cl• acts as a catalyst, allowing one Cl atom to destroy thousands of ozone molecules before being removed.