R3.3 Electron sharing reactions

Most molecules have all their electrons in pairs. A species with an unpaired electron, a radical, reacts quickly with almost anything it meets, because forming a bond pairs the electron and releases energy. Radical reactions turn methane into the chloromethane solvents of industry, drive the chemistry of the upper atmosphere, and, through chlorine atoms from chlorofluorocarbons, opened the ozone hole over Antarctica.

Radicals: species with an unpaired electron

In nearly all stable molecules every electron is paired: either shared in a covalent bond or held as a lone pair. A species with an odd number of valence electrons cannot pair them all, so at least one electron is left on its own. Such species are called radicals (older books call them "free radicals").

A radical is represented by a dot, $\mathord{\bullet}$, placed on the atom that carries the unpaired electron. A chlorine atom is written ${{\mathrm{Cl}}}\mathord{\bullet}$ and the methyl radical $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$, with the dot next to the carbon atom because carbon carries the unpaired electron. The dot stands for one electron only, so in a Lewis formula the other valence electrons are still drawn as bonding pairs and lone pairs. The Lewis formulas of four common radicals show this.

Counting valence electrons confirms each structure. A chlorine atom has seven: three lone pairs and one unpaired electron. In $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$, carbon has four valence electrons and uses three of them in ${{\mathrm{C}{-}\mathrm{H}}}$ bonds, leaving one unpaired. Nitrogen monoxide, ${{\mathrm{NO}}}$, has $5 + 6 = 11$ valence electrons; an odd total always leaves one unpaired.

Atoms, molecules and ions can all be radicals

The definition says "molecular entity" rather than "molecule" because radicals come in every form. Each kind in the table has an unpaired electron, whatever its size or charge.

The ions in the last two rows are radicals because removing one electron from, or adding one electron to, a molecule whose electrons were all paired must leave one unpaired. Nitrogen monoxide and nitrogen dioxide are unusual in being radicals that are stable enough to bottle. Most radicals, such as $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$, exist for only a tiny fraction of a second before they react.

Why radicals are so reactive

An unpaired electron becomes paired when the radical forms a covalent bond, either by taking an atom from a molecule or by combining with another radical. Bond formation releases energy, so these reactions are usually exothermic. They also need very little activation energy: a radical does not have to break a bond completely before the new bond starts to form, and when two radicals meet there is no bond to break at all. Radical reactions are therefore very fast, and radicals rarely survive long enough to build up to a high concentration.

Homolytic fission: how radicals are formed

A covalent bond is a pair of electrons shared between two atoms. When the bond breaks, the pair can be divided in two ways. If both electrons go to one atom, ions are formed; this is heterolytic fission, the subject of R3.4. If the pair is split evenly, each atom keeps one electron and two radicals are formed.

Homolytic fission is shown with single-barbed curved arrows, called fish-hook arrows. Each fish-hook has half an arrowhead and shows the movement of one electron, from where it starts (the bond) to where it ends up (an atom). Breaking one bond homolytically therefore needs two fish-hooks, one towards each atom. The full, double-barbed curly arrow shows the movement of a pair of electrons and is used for the reactions in R3.4.

The fission of chlorine can be written as a Lewis formula with arrows, as in the diagram, or simply as an equation. The condition, ultraviolet light, goes over the arrow; $h\nu$ (a photon) is sometimes written instead.

Heat can also break these bonds: chlorine and methane react in the dark at about $400$–$500\ ^\circ\text{C}$, a thermal route used industrially to make chloromethanes. In the laboratory, light is the usual way to start the reaction.

Why halogen molecules break apart in light

Light is absorbed as photons, and one photon can break one bond if it carries at least the bond's energy. Light of shorter wavelength carries more energy per photon. One mole of photons of red light ($700\ \text{nm}$) carries about $171\ \text{kJ}$, one mole of violet photons ($400\ \text{nm}$) about $299\ \text{kJ}$, and one mole of $250\ \text{nm}$ ultraviolet photons about $479\ \text{kJ}$. The chart compares these energies with bond enthalpies from the data booklet.

The halogen–halogen bonds are among the weakest covalent bonds: ${{\mathrm{Cl}{-}\mathrm{Cl}}}$ is $242\ \text{kJ mol}^{-1}$ and ${{\mathrm{Br}{-}\mathrm{Br}}}$ $193\ \text{kJ mol}^{-1}$. Violet and near-ultraviolet light carries enough energy to break them, so a mixture of methane and chlorine reacts in sunlight but not in the dark. A bond can only be broken by light that the molecule absorbs. Chlorine is yellow-green because it absorbs violet and near-ultraviolet light, which splits its molecules. Alkanes are colourless and absorb no visible or near-ultraviolet light, and their ${{\mathrm{C}{-}\mathrm{H}}}$ bonds ($414\ \text{kJ mol}^{-1}$) would in any case need ultraviolet radiation of wavelength below about 290 nm, which hardly reaches the ground. The halogen molecule is therefore the species that splits to start the reaction.

The same comparison explains a key fact about chlorofluorocarbons (CFCs) in the upper atmosphere. The ${{\mathrm{C}{-}\mathrm{Cl}}}$ bond ($324\ \text{kJ mol}^{-1}$) is much weaker than the ${{\mathrm{C}{-}\mathrm{F}}}$ bond ($492\ \text{kJ mol}^{-1}$), so the UV radiation there breaks ${{\mathrm{C}{-}\mathrm{Cl}}}$ bonds and releases chlorine atoms, while the ${{\mathrm{C}{-}\mathrm{F}}}$ bonds survive and fluorine atoms are not released.

The reverse of homolytic fission

Running homolytic fission backwards joins two radicals: each supplies one electron and the two electrons form a new shared pair. This releases energy equal to the bond enthalpy of the bond formed, for example $242\ \text{kJ mol}^{-1}$ when two chlorine atoms combine:

Radical combination needs almost no activation energy, because no bond has to be broken. Radical combination is what ends radical chains in termination steps, and it is why the concentration of radicals in a reacting mixture always stays very low.

Why alkanes are unreactive

Alkanes are saturated hydrocarbons: they contain only carbon and hydrogen joined by single bonds. They burn readily once lit, but apart from combustion they react with very few substances. Acids, alkalis, oxidizing agents such as acidified potassium manganate(VII) and most other common reagents have no effect on them at room temperature. Two features of their bonding explain this.

• Strong bonds. ${{\mathrm{C}{-}\mathrm{H}}}$ ($414\ \text{kJ mol}^{-1}$) and ${{\mathrm{C}{-}\mathrm{C}}}$ ($346\ \text{kJ mol}^{-1}$) bonds are strong, so any reaction that must begin by breaking one has a high activation energy.

• Essentially non-polar bonds. Carbon (electronegativity 2.6) and hydrogen (2.2) differ by only 0.4, so the ${{\mathrm{C}{-}\mathrm{H}}}$ bond is almost non-polar, and ${{\mathrm{C}{-}\mathrm{C}}}$ bonds are completely non-polar. An alkane has no significant partial charges ($\delta+$ or $\delta-$) to attract ions or polar molecules, so reagents that react by donating or accepting electron pairs (R3.4) have nowhere to attack.

Being unreactive is not the same as being stable in energy terms. The combustion of methane is strongly exothermic, $\Delta H_{\text{c}}^{\ominus} = -891\ \text{kJ mol}^{-1}$ (data booklet), so carbon dioxide and water are far lower in energy than methane and oxygen: methane is thermodynamically unstable with respect to its combustion products. Yet natural gas and air can be mixed in a cooker without reacting until a spark or flame supplies the activation energy. Methane is kinetically stable: the reaction is energetically favourable, but its high activation energy makes it extremely slow at room temperature (R1.1, R2.2).

Radicals provide one of the few routes by which alkanes react at moderate temperatures. A halogen atom is reactive enough to remove a hydrogen atom from an alkane with only a small activation energy, so once the halogen molecule has been split, the kinetic stability of the alkane is overcome.

Free-radical substitution of methane

A mixture of methane and chlorine is stable in the dark at room temperature. In ultraviolet light, or bright sunlight, it reacts to form chloromethane and hydrogen chloride:

One hydrogen atom of methane has been replaced by a chlorine atom, so this is a substitution reaction. Because the reaction takes place through radicals it is called free-radical substitution, and because the halogen is introduced into the molecule it is also a halogenation.

The equation for the reaction does not show what happens to individual particles. The reaction takes place in a series of simple steps, called its mechanism, which falls into three stages: initiation, propagation and termination.

Initiation

The initiation step creates radicals from a molecule that has none. Ultraviolet light breaks the weakest bond present, ${{\mathrm{Cl}{-}\mathrm{Cl}}}$, by homolytic fission:

Only a tiny fraction of the chlorine molecules is split at any moment, but each chlorine atom formed can start a long sequence of reactions.

Propagation

In a propagation step, a radical reacts with a molecule to form a new molecule and a new radical. The number of radicals does not change. Chlorination of methane has two propagation steps. First, a chlorine atom removes a hydrogen atom from methane, forming hydrogen chloride and a methyl radical. Then the methyl radical removes a chlorine atom from a chlorine molecule, forming chloromethane and a new chlorine atom:

The fish-hook arrows for these two steps show where each electron goes. In both steps one bond breaks homolytically and one new bond forms from one electron supplied by each partner.

Adding the two propagation steps and cancelling the radicals, which appear on both sides, gives the overall equation ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4}} {}+{} \mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{Cl} {}+{} \mathrm{HCl}}}$. The chlorine atom used in the first step is given back in the second, so it can start the cycle again.

Termination

A termination step removes radicals: two radicals combine to form a molecule, pairing their unpaired electrons in a new bond. Any two radicals present can combine:

Termination is rare compared with propagation, because radicals are present at very low concentration and so seldom meet each other. They are far more likely to collide with methane or chlorine molecules, which are present in vast excess.

A chain reaction

Because every propagation step regenerates a radical, the two propagation steps repeat over and over: each chlorine atom formed by initiation can convert a great many methane molecules before a termination step removes it. A reaction of this kind is a chain reaction. The cycle diagram shows how the three stages fit together.

The chain explains why a little light can produce a large amount of product, and why the reaction can be violent. A mixture of hydrogen and chlorine is stable in the dark but explodes when exposed to bright sunlight or a camera flash: chlorine atoms start a very fast chain (${{\mathrm{Cl}\mathord{\bullet} {}+{} \mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{HCl} {}+{} \mathrm{H}\mathord{\bullet}}}$, then ${{\mathrm{H}\mathord{\bullet} {}+{} \mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{HCl} {}+{} \mathrm{Cl}\mathord{\bullet}}}$), and the heat released speeds it up further.

Evidence of this kind supports the mechanism but does not prove it. Further support comes from the products themselves. Small amounts of ethane are always found in the product mixture, and the only reasonable source is the termination step ${{\mathord{\bullet}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}} {}+{} \mathord{\bullet}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\xrightarrow{}\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{6}}}}$. Adding a substance that reacts rapidly with radicals stops the reaction, as expected if radicals carry the chain.

Why a mixture of products forms

Free-radical substitution is hard to control. The reaction of methane with chlorine gives chloromethane, but never chloromethane alone. Three features of the mechanism lead to a mixture.

Further substitution

Chloromethane still has three ${{\mathrm{C}{-}\mathrm{H}}}$ bonds, and a chlorine atom can remove a hydrogen atom from it just as it can from methane. The radical formed, $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{Cl}}}$, then reacts with chlorine to give dichloromethane. The same steps can continue until all four hydrogen atoms have been replaced:

Each further substitution has the same general form, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4-n }}\mathrm{Cl}{\vphantom{X}}_{\smash[t]{n }} {}+{} \mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3-n }}\mathrm{Cl}{\vphantom{X}}_{\smash[t]{n +1}} {}+{} \mathrm{HCl}}}$, so the product is a mixture of all four chloromethanes and hydrogen chloride. The four compounds have quite different boiling points and are separated by fractional distillation.

Boiling points at 101 kPa, rounded to the nearest degree.

The proportions can be shifted by changing the ratio of reactants. A large excess of methane makes it much more likely that a chlorine atom collides with a methane molecule than with a chloromethane molecule, so chloromethane is the main product. An excess of chlorine favours the more substituted products, and enough chlorine converts methane completely into tetrachloromethane: ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4}} {}+{} 4\,\mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{CCl}{\vphantom{X}}_{\smash[t]{4}} {}+{} 4\,\mathrm{HCl}}}$.

Substitution at different positions

In larger alkanes, the hydrogen atoms are not all equivalent. Propane has six hydrogen atoms on its end carbon atoms and two on its middle carbon atom. A chlorine atom can remove either kind, giving two different radicals, and these lead to two structural isomers.

Chlorination of propane therefore gives a mixture of 1-chloropropane and 2-chloropropane, before any further substitution is considered. With butane, pentane and larger alkanes the number of possible monosubstituted and disubstituted isomers increases rapidly, which makes free-radical substitution a poor way to make a single pure halogenoalkane from a large alkane.

Products of termination

Termination steps join whatever radicals meet. In the chlorination of methane, $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}} + \mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ gives ethane, which can itself be chlorinated to chloroethane. In the chlorination of ethane, two ethyl radicals give butane: ${{2\,\mathord{\bullet}\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}\xrightarrow{}\mathrm{C}{\vphantom{X}}_{\smash[t]{4}}\mathrm{H}{\vphantom{X}}_{\smash[t]{10}}}}$. These products contain more carbon atoms than the starting alkane and add further compounds to the mixture.

Halogenation of other alkanes

Every alkane reacts with chlorine and bromine in UV light by the same mechanism. Ethane, for example, reacts with chlorine as follows.

The ethyl radical can be written $\mathord{\bullet}{{\mathrm{C}{\vphantom{X}}_{\smash[t]{2}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}}}$ or ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}}}\mathord{\bullet}$; in either form the unpaired electron is on a carbon atom.

Chlorine and bromine compared

Bromine reacts with alkanes much more slowly than chlorine, even though the ${{\mathrm{Br}{-}\mathrm{Br}}}$ bond ($193\ \text{kJ mol}^{-1}$) is easier to break than ${{\mathrm{Cl}{-}\mathrm{Cl}}}$. The difference lies in the first propagation step. A chlorine atom removing a hydrogen atom from methane is slightly exothermic ($414 - 431 = -17\ \text{kJ mol}^{-1}$), but a bromine atom doing the same is endothermic, because ${{\mathrm{H}{-}\mathrm{Br}}}$ ($366\ \text{kJ mol}^{-1}$) is weaker than ${{\mathrm{C}{-}\mathrm{H}}}$:

This step has a higher activation energy, so the chain runs more slowly. Fluorine reacts explosively with alkanes even in the dark, while iodine hardly reacts at all.

Observing the reaction: bromine and hexane

The reaction of bromine with a liquid alkane can be followed by its colour. A few drops of a dilute solution of bromine are added to hexane in two test tubes. One is wrapped in foil or kept in a dark cupboard; the other is placed in bright sunlight or under a UV lamp. The tube kept in the dark stays orange. The tube in the light slowly becomes colourless as the bromine is used up, and misty fumes of hydrogen bromide appear at its mouth; they turn damp blue litmus paper red because hydrogen bromide dissolves in the water on the paper to form an acid.

The two tubes act as a controlled comparison: the only difference between them is the light, so light is shown to be needed. Alkenes behave differently: they decolourize bromine quickly even in the dark, by addition across the double bond (R3.4), and this difference is used to distinguish alkenes from alkanes. Bromine is toxic and corrosive, and its vapour irritates the lungs, so the experiment is done in a fume cupboard with gloves and eye protection, and halogenated waste is collected separately.

Alkyl side chains on a benzene ring

The same mechanism substitutes hydrogen atoms in an alkyl group attached to a benzene ring. Methylbenzene, ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{6}}\mathrm{H}{\vphantom{X}}_{\smash[t]{5}}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$, reacts with bromine in UV light at the methyl group, not at the ring:

The radical chain removes a hydrogen atom from the ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$ group, which behaves as a small alkane attached to the ring. Substitution on the ring itself needs different conditions and a different, electrophilic mechanism (R3.4, HL).

Radicals in the stratosphere: CFCs and the ozone layer

Radical chain reactions are not confined to the laboratory. The clearest example of their effect on the whole planet is the destruction of stratospheric ozone by chlorine atoms from chlorofluorocarbons.

The ozone layer

The stratosphere is the layer of the atmosphere from about 10 to 50 km above the ground. It contains a small concentration of ozone, ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{3}}}}$, most of it between about 15 and 35 km: the ozone layer. Ozone and oxygen absorb almost all the high-energy ultraviolet radiation from the Sun before it reaches the ground, and ozone is the main absorber of the UV-B radiation that causes sunburn, skin cancer and cataracts. In the process, ozone is constantly broken down and re-formed:

Left alone, these reactions keep the amount of ozone roughly constant. The bonds in ozone are weaker than the ${{\mathrm{O}{=}\mathrm{O}}}$ double bond in oxygen ($498\ \text{kJ mol}^{-1}$), lying between that and the ${{\mathrm{O}{-}\mathrm{O}}}$ single bond ($144\ \text{kJ mol}^{-1}$), so lower-energy UV radiation can split ozone than oxygen (the HL treatment of resonance explains why the two bonds in ozone are identical and intermediate). This is also why chlorine atoms can break down ozone but not oxygen molecules in the stratosphere.

How CFCs destroy ozone

CFCs such as dichlorodifluoromethane, ${{\mathrm{CCl}{\vphantom{X}}_{\smash[t]{2}}\mathrm{F}{\vphantom{X}}_{\smash[t]{2}}}}$, were widely used from the 1930s as refrigerants, aerosol propellants and blowing agents for foams, because they are non-toxic, non-flammable and very unreactive. That lack of reactivity means nothing removes them in the lower atmosphere. Over several years they drift up into the stratosphere, where UV radiation of short wavelength breaks the weakest bond, ${{\mathrm{C}{-}\mathrm{Cl}}}$, by homolytic fission:

The chlorine atom then destroys ozone in two steps, each of which consumes one radical and produces another:

The chlorine atom is regenerated, so it acts as a catalyst. One chlorine atom can destroy up to about 100 000 ozone molecules before it is locked up in a less reactive compound, for example when ${{\mathrm{ClO}}}\mathord{\bullet}$ combines with nitrogen dioxide (another radical) in a termination step to form chlorine nitrate, ${{\mathrm{ClONO}{\vphantom{X}}_{\smash[t]{2}}}}$. Nitrogen monoxide from high-flying aircraft destroys ozone by a similar cycle: ${{\mathrm{NO} {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{3}}\xrightarrow{}\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$, then ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{O}\xrightarrow{}\mathrm{NO} {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$.

Fluorine atoms are not released, because the ${{\mathrm{C}{-}\mathrm{F}}}$ bond ($492\ \text{kJ mol}^{-1}$) is too strong to be broken by the UV radiation that splits ${{\mathrm{C}{-}\mathrm{Cl}}}$ ($324\ \text{kJ mol}^{-1}$). Any fluorine atoms that did form would also be removed quickly as stable hydrogen fluoride.

Prediction, discovery and the Montreal Protocol

In 1974 Mario Molina and F. Sherwood Rowland predicted that chlorine from CFCs would reach the stratosphere and destroy ozone by this catalytic cycle. In 1985 scientists of the British Antarctic Survey reported that springtime ozone over Antarctica had fallen sharply since the late 1970s, the first evidence of what became known as the ozone hole. Satellite measurements showed the scale of the loss.

The hole forms over Antarctica in spring because extremely cold clouds there convert chlorine compounds into forms that release chlorine atoms as soon as sunlight returns. In response, governments agreed the Montreal Protocol in 1987, which phased out the production of CFCs and other ozone-depleting substances; every member of the United Nations has since ratified it. Molina and Rowland shared the 1995 Nobel Prize in Chemistry with Paul Crutzen for their work on stratospheric ozone.

Because CFC molecules survive in the atmosphere for 50 to 100 years, the ozone hole stopped growing only slowly and has just begun to shrink. The 2022 scientific assessment by the World Meteorological Organization and the UN Environment Programme projected that ozone will return to its 1980 level around 2040 over most of the world, around 2045 over the Arctic and around 2066 over Antarctica.

Hydrofluorocarbons (HFCs), such as ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{2}}\mathrm{FCF}{\vphantom{X}}_{\smash[t]{3}}}}$, replaced CFCs in most refrigerators and air conditioners. They contain no chlorine and so do not destroy ozone, but many are powerful greenhouse gases, and the Kigali Amendment to the Montreal Protocol (2016) now limits their use as well. CFCs are an example of a chemical whose harmful effects appeared only after decades of global use, and the Montreal Protocol an example of international action taken on scientific evidence.

Applying the ideas in exam questions

Paper 1A: identifying a propagation step

Paper 1A: counting isomers

Paper 1B: bromination of hexane under a lamp

A student investigated the effect of light intensity on the reaction between bromine and hexane. $5.0\ \text{cm}^3$ of hexane and $1.0\ \text{cm}^3$ of a dilute solution of bromine in hexane were mixed in a test tube placed at different distances from a UV lamp, and the time taken for the orange colour to disappear was measured with a stopwatch. A tube kept in the dark was still orange after 30 minutes. The results are in the table.

Paper 2: chlorination of ethane

Paper 2: CFCs and ozone

Self-check

• To do:

  I can define a radical and explain why radicals are highly reactive.

• To do:

  I can represent radicals with a dot on the correct atom, such as ${{\mathrm{Cl}}}\mathord{\bullet}$ and $\mathord{\bullet}{{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}}}$, and decide from an electron count whether a species is a radical.

• To do:

  I can give examples of radicals that are atoms, molecules, cations and anions.

• To do:

  I can describe homolytic fission and draw it with two fish-hook arrows.

• To do:

  I can explain, using bond enthalpies, why halogen molecules split in UV light and why CFCs release chlorine but not fluorine atoms.

• To do:

  I can explain why alkanes are unreactive in terms of bond strength and bond polarity, and why they are kinetically stable but thermodynamically unstable.

• To do:

  I can write equations for the initiation, propagation and termination steps in the reactions of methane and ethane with chlorine or bromine, with the condition.

• To do:

  I can explain why free-radical substitution is a chain reaction.

• To do:

  I can explain why free-radical substitution gives a mixture of products, and deduce the isomers that can form.

• To do:

  I can describe how bromine and an alkane react in light and in the dark, and the observations.

• To do:

  I can explain how chlorine atoms from CFCs catalyse the destruction of ozone, with equations.

(b) $7 + 1 = 8$, all paired: three ${{\mathrm{C}{-}\mathrm{H}}}$ bonds and a lone pair on carbon. Not a radical (it is an anion).

(c) $7 + 6 = 13$, odd. ${{\mathrm{ClO}}}\mathord{\bullet}$ is a radical; it appears in the destruction of ozone.

(d) $7 + 1 = 8$, four lone pairs. Not a radical.

Propagation step 2, ${{\mathord{\bullet}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}} {}+{} \mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{Cl} {}+{} \mathrm{Cl}\mathord{\bullet}}}$: one ${{\mathrm{Cl}{-}\mathrm{Cl}}}$ bond broken, one ${{\mathrm{C}{-}\mathrm{Cl}}}$ bond formed.

The two steps add up to $-99\ \text{kJ mol}^{-1}$, which is $\Delta H$ for the overall reaction ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4}} {}+{} \mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{Cl} {}+{} \mathrm{HCl}}}$, as expected, since the radicals cancel.

(b) One ${{\mathrm{C}{-}\mathrm{H}}}$ bond broken, one ${{\mathrm{C}{-}\mathrm{Cl}}}$ bond formed:

Comment. Both real propagation steps are exothermic, so the chain runs easily once started. The alternative step is endothermic by about $90\ \text{kJ mol}^{-1}$ and would have a high activation energy, so it does not compete. This is evidence that the chlorine atom removes a hydrogen atom to form ${{\mathrm{HCl}}}$, as in the accepted mechanism.

Six structural isomers. Arrangements such as 3,4 or 2,4 are the same as 1,2 and 1,3 counted from the other end, and 3,3 is the same as 2,2.

(c)

• the chlorine atom is regenerated «in the second step», so each one can react with many ozone molecules/it acts as a catalyst/chain reaction ✓

(d)

• HFCs contain no chlorine/no C–Cl bond, so no chlorine atoms are released; C–F and C–H bonds are too strong to be broken ✓

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