R3.3 Electron sharing reactions | IB DP Chemistry

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🎯What you need to be able to do

Identify and represent radicals, and explain why they are highly reactive.
Explain, including with equations, the homolytic fission of halogens in the presence of ultraviolet light or heat, and recognise it as the initiation step of a chain reaction.
Use a single-barbed arrow (fish hook) to show the movement of a single electron.
Explain, using equations, the propagation and termination steps in the reaction between an alkane and a halogen.
Explain why radical substitution produces a mixture of products.
Refer to the stability of alkanes in terms of the strengths of the C–C and C–H bonds and their essentially non-polar nature.

📚The chemistry

Radicals

A radical is a molecular entity that has an unpaired electron. It is represented by a dot beside the symbol: \( \mathrm{Cl\cdot} \), \( \mathrm{\cdot CH_3} \), \( \mathrm{\cdot OH} \), \( \mathrm{\cdot H} \).

Radicals are highly reactive precisely because of that unpaired electron: pairing it up by forming a bond releases energy, so a radical will react with almost anything it meets. Their lifetimes are correspondingly short.

Note that a radical may be an atom (\( \mathrm{Cl\cdot} \)), a molecule (\( \mathrm{NO} \), \( \mathrm{NO_2} \), and \( \mathrm{O_2} \), which has two unpaired electrons), a cation or an anion — the defining feature is only the unpaired electron, not the charge. This is also why radicals are exceptions to the octet rule (S2.2): an odd number of electrons cannot be arranged in pairs.

Homolytic and heterolytic fission

Homolytic fission
the bonding pair splits evenly, one electron to each fragment
→ produces two radicals
shown with single-barbed arrows (fish hooks), one electron each

Heterolytic fission
the bonding pair goes entirely to one atom
→ produces two ions
shown with a double-barbed curly arrow, an electron pair
see R3.4

Homolytic fission of a non-polar bond such as Cl–Cl is favoured because there is no reason for the pair to prefer one atom over the other. It requires energy — ultraviolet light or heat — and the Cl–Cl bond enthalpy is low enough that ordinary sunlight suffices.

Why alkanes are unreactive

Alkanes are remarkably inert towards acids, bases, oxidising agents and reducing agents at room temperature. Two reasons, and the syllabus asks for both:

the C–C and C–H bonds are strong, so a great deal of energy is needed to break them;
those bonds are essentially non-polar, because carbon and hydrogen have similar electronegativities. With no \( \delta+ \) or \( \delta- \) site anywhere in the molecule, there is nothing for a nucleophile or an electrophile to attack.

That is why the only reactions of alkanes on this course are combustion (R1.3) and radical substitution: a radical is reactive enough not to need a polar site to attack. Note also that alkanes are described as kinetically stable but thermodynamically unstable — combustion is strongly exothermic, so they should burn, and at room temperature they do not, because the activation energy is high.

Radical substitution, step by step

The reaction of methane with chlorine in ultraviolet light proceeds by a chain reaction in three stages.

1. Initiation — a radical is created where there was none. UV light supplies the energy for homolytic fission of the halogen:

\[ \mathrm{Cl_2} \xrightarrow{\ \mathrm{UV}\ } 2\,\mathrm{Cl\cdot} \]

Note that it is the Cl–Cl bond that breaks, not the C–H bond: at 242 kJ mol⁻¹ it is much the weaker of the two, C–H being about 414.

2. Propagation — a radical reacts to produce another radical, so the chain continues. There are always two propagation steps, and they must add to the overall equation:

\[ \mathrm{Cl\cdot} + \mathrm{CH_4} \rightarrow \mathrm{HCl} + \mathrm{\cdot CH_3} \]

\[ \mathrm{\cdot CH_3} + \mathrm{Cl_2} \rightarrow \mathrm{CH_3Cl} + \mathrm{Cl\cdot} \]

The chlorine radical consumed in the first step is regenerated in the second, so a single initiation event can lead to thousands of product molecules. That is what makes it a chain reaction.

3. Termination — two radicals combine, removing radicals from the system and breaking the chain. Any two radicals present may combine, so there are several possible steps:

\[ \mathrm{Cl\cdot} + \mathrm{Cl\cdot} \rightarrow \mathrm{Cl_2} \qquad \mathrm{\cdot CH_3} + \mathrm{Cl\cdot} \rightarrow \mathrm{CH_3Cl} \qquad \mathrm{\cdot CH_3} + \mathrm{\cdot CH_3} \rightarrow \mathrm{C_2H_6} \]

The appearance of ethane among the products is direct evidence for the mechanism: there is no way to get a C–C bond from methane and chlorine except by two methyl radicals meeting.

Writing a propagation step that consumes two radicals or produces two. The test is simple and mechanical: a propagation step has exactly one radical on each side. One radical in, one radical out. A step with radicals on the left and none on the right is termination; a step with none on the left and two on the right is initiation. Label each step by counting radicals, not by where it appears in your answer.

Why a mixture of products

Radical substitution is synthetically poor, and you should be able to say why in two distinct ways:

Further substitution. The product \( \mathrm{CH_3Cl} \) still contains C–H bonds, and a chlorine radical cannot tell it apart from methane. So it is attacked in turn, giving \( \mathrm{CH_2Cl_2} \), then \( \mathrm{CHCl_3} \), then \( \mathrm{CCl_4} \). Using a large excess of methane reduces this, but does not eliminate it.
Substitution at different positions. In a larger alkane the radical may abstract a hydrogen from any carbon, so propane gives both 1-chloropropane and 2-chloropropane.
Termination products. Ethane and other coupling products appear as contaminants.

The consequence is that radical substitution is used industrially for making bulk chlorinated solvents, where a mixture can be separated by fractional distillation and every fraction sold, but it is a poor route to a single pure compound in the laboratory.

✏️Worked example

Ethane reacts with bromine in the presence of ultraviolet light.
(a) Write equations for the initiation, both propagation steps, and two possible termination steps, labelling each.
(b) State the overall equation, and show that the propagation steps sum to it.
(c) Explain why the reaction does not occur in the dark at room temperature, and why the products include bromoethane, 1,2-dibromoethane and butane.
(d) State the type of bond fission involved and how it is represented.

(a) Initiation — UV light causes homolytic fission of the weaker Br–Br bond:

\[ \mathrm{Br_2} \xrightarrow{\ \mathrm{UV}\ } 2\,\mathrm{Br\cdot} \]

Propagation — one radical in, one radical out, in each step:

\[ \mathrm{Br\cdot} + \mathrm{C_2H_6} \rightarrow \mathrm{HBr} + \mathrm{\cdot C_2H_5} \]

\[ \mathrm{\cdot C_2H_5} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_5Br} + \mathrm{Br\cdot} \]

Termination — any two radicals combine (any two of these):

\[ \mathrm{Br\cdot} + \mathrm{Br\cdot} \rightarrow \mathrm{Br_2} \qquad \mathrm{\cdot C_2H_5} + \mathrm{Br\cdot} \rightarrow \mathrm{C_2H_5Br} \qquad \mathrm{\cdot C_2H_5} + \mathrm{\cdot C_2H_5} \rightarrow \mathrm{C_4H_{10}} \]

(b) Overall:

\[ \mathrm{C_2H_6} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_5Br} + \mathrm{HBr} \]

Adding the two propagation steps, the \( \mathrm{Br\cdot} \) and \( \mathrm{\cdot C_2H_5} \) appear once on each side and cancel, leaving exactly the overall equation — which is the check that the mechanism is consistent with the stoichiometry.

(c) In the dark at room temperature there is no source of energy to break the Br–Br bond homolytically, so no radicals are formed and the initiation step does not occur. Without an initiator there is nothing to start the chain. Ethane is otherwise unreactive because its C–C and C–H bonds are strong and essentially non-polar, offering no site for attack.

The products are mixed for three separate reasons. Bromoethane is the intended substitution product. 1,2-dibromoethane arises because bromoethane still has C–H bonds and is attacked by a bromine radical in the same way, giving further substitution. Butane arises from a termination step in which two ethyl radicals combine — and its presence is strong evidence for the radical mechanism, since there is no other way to form a C–C bond from these reactants.

(d) Homolytic fission: the bonding pair splits evenly, one electron going to each atom, so two radicals are produced. It is represented with single-barbed arrows (fish hooks), each showing the movement of a single electron — in contrast with the double-barbed curly arrow used for the movement of an electron pair in heterolytic fission.

Check it. Two mechanical checks that catch nearly every error on this topic. First, count radicals in every step: initiation has 0 in and 2 out; propagation has 1 in and 1 out; termination has 2 in and 0 out. Label each step by that count rather than by position. Second, add the propagation steps and confirm that the radicals cancel to give the overall equation exactly. If they do not, one of your steps is wrong — a very common slip is writing \( \mathrm{Br\cdot} + \mathrm{C_2H_6} \rightarrow \mathrm{C_2H_5Br} + \mathrm{H\cdot} \), which is energetically wrong and does not sum correctly.

Breaking the C–H bond in the initiation step. Initiation breaks the halogen–halogen bond, not a bond in the alkane. The reason is energetic: the Br–Br bond enthalpy is about 193 kJ mol⁻¹ and Cl–Cl about 242, while C–H is about 414 — so ultraviolet photons of ordinary energy can break the halogen bond and cannot break C–H. Writing \( \mathrm{C_2H_6} \rightarrow \mathrm{\cdot C_2H_5} + \mathrm{H\cdot} \) as initiation loses the mark and shows the wrong picture of why light is needed at all.

📝Practise

Work through these on paper, then reveal the answer.

1. Define a radical and explain why radicals are highly reactive. Give one example each of a radical that is an atom and one that is a molecule.

A radical is a molecular entity possessing an unpaired electron. Radicals are highly reactive because an unpaired electron is a high-energy arrangement: pairing it by forming a new covalent bond releases energy, so a radical reacts readily with almost any species it collides with, and its lifetime is very short. Radicals are also, necessarily, exceptions to the octet rule — an odd number of valence electrons cannot all be paired. Atom: \( \mathrm{Cl\cdot} \), formed by homolytic fission of \( \mathrm{Cl_2} \); \( \mathrm{H\cdot} \) is another. Molecule: \( \mathrm{NO} \) or \( \mathrm{NO_2} \), each of which has an odd total number of valence electrons; \( \mathrm{O_2} \) also qualifies, having two unpaired electrons. Radicals may equally be cations or anions — charge is irrelevant to the definition.

2. Write the full mechanism for the reaction of methane with chlorine in UV light, giving one initiation step, two propagation steps and three termination steps, each labelled.

Initiation (0 radicals in, 2 out): \( \mathrm{Cl_2} \xrightarrow{\ \mathrm{UV}\ } 2\,\mathrm{Cl\cdot} \). Propagation 1 (1 in, 1 out): \( \mathrm{Cl\cdot} + \mathrm{CH_4} \rightarrow \mathrm{HCl} + \mathrm{\cdot CH_3} \). Propagation 2 (1 in, 1 out): \( \mathrm{\cdot CH_3} + \mathrm{Cl_2} \rightarrow \mathrm{CH_3Cl} + \mathrm{Cl\cdot} \). Termination (2 in, 0 out), any of: \( \mathrm{Cl\cdot} + \mathrm{Cl\cdot} \rightarrow \mathrm{Cl_2} \); \( \mathrm{\cdot CH_3} + \mathrm{Cl\cdot} \rightarrow \mathrm{CH_3Cl} \); \( \mathrm{\cdot CH_3} + \mathrm{\cdot CH_3} \rightarrow \mathrm{C_2H_6} \). Check: the two propagation steps sum to \( \mathrm{CH_4} + \mathrm{Cl_2} \rightarrow \mathrm{CH_3Cl} + \mathrm{HCl} \), the overall equation, with the radicals cancelling.

3. Explain why one initiation event can lead to the formation of many thousands of product molecules.

Because the propagation steps form a chain. In propagation step 1 a chlorine radical is consumed, but in step 2 a chlorine radical is regenerated. The regenerated radical then carries out step 1 again on another methane molecule, and the cycle repeats. Each turn of the cycle produces one molecule of product, and no radical is used up overall, so a single radical produced by one initiation event can go round the cycle thousands of times before it is finally removed. The cycle ends only when the radical happens to meet another radical in a termination step, and since radicals are present at extremely low concentration, such meetings are rare. The practical consequence is that only a very small amount of light is needed to sustain a large-scale reaction — and it also explains why the reaction can be explosive: nothing limits its rate once started.

4. Explain why the reaction of chlorine with propane produces both 1-chloropropane and 2-chloropropane, and why 1,2-dichloropropane and other polychlorinated products also form.

Two monochlorinated products. In propagation step 1 the chlorine radical abstracts a hydrogen atom, and propane has hydrogens in two different environments: six on the two end carbons and two on the middle carbon. Abstraction from an end carbon gives the propyl radical \( \mathrm{\cdot CH_2CH_2CH_3} \), which goes on to give 1-chloropropane; abstraction from the middle carbon gives the 2-propyl radical \( \mathrm{CH_3\cdot CHCH_3} \), which gives 2-chloropropane. The radical is not selective enough to attack only one position, so both form. Polychlorinated products. Once formed, 1-chloropropane and 2-chloropropane still contain C–H bonds, and a chlorine radical attacks them in exactly the same way, giving dichloro products, then trichloro, and so on. Using a large excess of propane makes it statistically more likely that a radical meets an unsubstituted molecule, which reduces polysubstitution but cannot eliminate it. This lack of selectivity is why radical substitution is a poor laboratory route to a single pure halogenoalkane.

5. Compare homolytic and heterolytic fission in terms of how the bonding pair is divided, the species produced, the arrows used, and the type of bond each typically occurs in.

Division of the pair: in homolytic fission the bonding pair splits evenly, one electron going to each fragment; in heterolytic fission both electrons go to one atom. Species produced: homolytic gives two radicals, each electrically neutral if the parent was; heterolytic gives two ions, one positive and one negative. Arrows: homolytic fission uses single-barbed arrows (fish hooks), each representing one electron, and two are needed; heterolytic fission uses a double-barbed curly arrow representing the movement of an electron pair. Typical bonds: homolytic fission occurs in non-polar bonds such as Cl–Cl or Br–Br, where neither atom has a stronger claim on the pair, and requires energy in the form of UV light or heat; heterolytic fission occurs in polar bonds such as C–Br, where the more electronegative atom takes both electrons, and is the basis of nucleophilic and electrophilic mechanisms in R3.4.

6. Chlorofluorocarbons (CFCs) in the stratosphere break down to release chlorine radicals but typically not fluorine radicals, and those chlorine radicals destroy ozone but not oxygen. Explain both observations in terms of bond enthalpies.

Chlorine, not fluorine. The C–Cl bond is weaker than the C–F bond (about 324 against 492 kJ mol⁻¹), because chlorine is a larger atom, so the shared pair is further from the nuclei and the overlap is poorer. Ultraviolet photons in the stratosphere carry enough energy to break C–Cl homolytically but not enough to break C–F, so chlorine radicals are released and fluorine radicals essentially are not. Ozone, not oxygen. The bond in \( \mathrm{O_2} \) is a full double bond with a bond enthalpy of about 498 kJ mol⁻¹. In ozone, \( \mathrm{O_3} \), the bonding is delocalised over three atoms, so each O–O bond has a bond order of about 1.5 and is correspondingly weaker and longer (about 364 kJ mol⁻¹). A chlorine radical can therefore break an ozone bond but not an oxygen double bond. The wider point, and the reason CFCs were so damaging, is that the chlorine radical acts catalytically: it is regenerated in a propagation cycle, so one radical destroys many thousands of ozone molecules before termination removes it.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The NASA Ozone Watch site — current stratospheric ozone data and the chemistry behind it, which turns the last practice question into something measurable.
Chemistry LibreTexts — radical halogenation, with the relative selectivity of chlorine and bromine explained more fully than the syllabus requires.
Your data booklet — the average bond enthalpy table. Every explanation on this page, including which bond breaks in initiation, comes back to numbers you are given.