Elimination from halogenoalkanes, and CFCs and the ozone layer

Hydroxide as a base: elimination

Hydroxide ions do not always substitute. Heated under reflux with potassium hydroxide dissolved in ethanol, 2-bromopropane gives off a gas that decolourises bromine water: propene. The halogenoalkane has lost a hydrogen atom from one carbon and the bromine from the carbon next to it, and a double bond has formed between them. That is elimination: a small molecule, here HBr, is removed from a larger one.

There is no state symbol for a solution in ethanol, so the equation is written without them: CH₃CHBrCH₃ + KOH → CH₂=CHCH₃ + KBr + H₂O.

The hydroxide ion is the same species as in substitution, but here it acts as a base, a proton acceptor. Instead of attacking the δ+ carbon, its lone pair takes an H⁺ from the carbon next to the C–Br carbon. The electrons of that C–H bond move in between the two carbons to form the second bond of the C=C, and the C–Br bond breaks with both electrons going to bromine.

Describe the drawing

2-bromopropane with its three carbons displayed: C1 carrying three hydrogens, C2 carrying one hydrogen and the bromine, C3 a methyl group, with δ+ on C2 and δ− on Br. A hydroxide ion at the upper left, lone pair and negative charge on O, acts as a base. Arrow 1 runs from the lone pair to an H on C1; arrow 2 from that C1–H bond to the C1–C2 bond; arrow 3 from the C2–Br bond to the bromine. After KOH in ethanol, heated under reflux: propene with the new C=C between C1 and C2, water and a bromide ion.

All three arrows happen in one step, and each needs to start and end in the right place. The first starts on a lone pair of the hydroxide oxygen and ends on the hydrogen, not the carbon. The second starts on the C–H bond and ends between the two carbons. The third starts on the C–Br bond and ends on the bromine. The products are the alkene, water and a bromide ion.

The word for the role matters. Call the hydroxide ion a base in elimination, because it accepts a proton, and a nucleophile in substitution, because it donates its pair to carbon. 'Hydroxide attacks the halogenoalkane' fits both and explains neither.

Elimination

A reaction in which a small molecule is removed from a larger one, leaving a double bond.

Base

A proton acceptor: a species that accepts an H⁺ ion using a lone pair.

Propene boils at −48 °C, so it leaves the reflux condenser as a gas and can be collected over water. Bubbled through bromine water, it turns the orange solution colourless, the test for C=C. The alcohol that substitution would have given does not decolourise bromine water, so the test tells the two routes apart.

Substitution or elimination

A halogenoalkane and hydroxide ions always have both reactions open to them, and both happen to some extent in any mixture. The conditions decide which is the main one. In water, the hydroxide ion is surrounded by water molecules and behaves mainly as a nucleophile; warm aqueous conditions favour substitution and give an alcohol. In ethanol, heated strongly, it behaves more as a base, and elimination gives an alkene.

The structure of the halogenoalkane matters too. At a primary carbon there is room for the hydroxide to reach the δ+ carbon, and substitution is the main reaction even in ethanol. At a tertiary carbon three alkyl groups crowd round the δ+ carbon, so the hydroxide ion finds the hydrogens on the neighbouring carbons easier to reach, and elimination wins.

Describe the drawing

A decision diagram starting from halogenoalkane plus hydroxide. The first question, the solvent, splits into water, warm, and ethanol, hot; under each the second question is primary or tertiary. Water with a primary halogenoalkane leads to a large box, substitution giving an alcohol with hydroxide as nucleophile. Ethanol with a tertiary one leads to a large box, elimination giving an alkene with hydroxide as base. The mixed cases lead to smaller boxes: both, mainly substitution; and both, substitution still main but more elimination. A dashed arrow joining the two large boxes says the other reaction always runs too, as the minor one.

Temperature pushes in the direction the solvent does. Elimination turns two reactant particles into three product particles, the alkene, water and a halide ion, and a higher temperature favours it over substitution, which turns two into two. A hot, concentrated solution of potassium hydroxide in ethanol is the condition for elimination; a warm, dilute aqueous solution is the condition for substitution.

Applied to particular compounds, the two sets of conditions give these main products. Secondary halogenoalkanes sit between the two extremes, so the conditions decide more for them than for a primary or a tertiary compound.

Elimination matters in synthesis because it turns a saturated compound into an alkene, which can then be built on. Substitution by hydroxide gives an alcohol directly. The conditions are chosen to match the product wanted, and some of the other is expected.

The alkenes from 2-bromobutane

Bromoethane and 2-bromopropane each give only one alkene, because every hydrogen that can be removed sits in an equivalent position. 2-bromobutane, CH₃CHBrCH₂CH₃, is different. The bromine is on carbon 2, and carbon 2 has two neighbours with hydrogens on them: carbon 1, the CH₃ at the end, and carbon 3, the CH₂ in the chain.

Describe the drawing

2-bromobutane drawn at the top with its carbons numbered 1 to 4, the bromine on carbon 2, δ+ on C2 and δ− on Br. Below, two copies of the mechanism. On the left, a hydroxide ion's lone pair arrows onto an H on carbon 1, the C1–H bond arrows into the C1–C2 bond and the C–Br bond arrows onto bromine, giving but-1-ene. On the right the same three arrows take an H from carbon 3 and form the double bond between C2 and C3, giving but-2-ene, drawn twice: E with the two CH₃ groups on opposite sides of the double bond and Z with them on the same side.

The three do not form in equal amounts. More but-2-ene forms than but-1-ene, and more of the E isomer than the Z, because an alkene with more alkyl groups on its double bond, held further apart, is more stable. A question that asks which alkenes form wants all three named, with the E and Z isomers counted separately.

A tertiary halogenoalkane can be symmetrical too. 2-bromo-2-methylpropane has three identical CH₃ groups round the C–Br carbon, so whichever one loses a hydrogen the product is the same, 2-methylpropene.

Position isomers

Structural isomers with the same carbon skeleton and the same functional group at a different position on the chain.

E/Z isomers

Stereoisomers that exist because rotation about a C=C bond is restricted and each carbon of the double bond carries two different groups.

Ozone in the stratosphere

Between about 15 and 35 kilometres up, in the stratosphere, ozone, O₃, is formed and broken down continuously. Ultraviolet light of short wavelength splits oxygen molecules into atoms, and an oxygen atom joins an oxygen molecule to make ozone:

O₂(g) → 2O(g)ultraviolet light

O(g) + O₂(g) → O₃(g)

Ozone itself absorbs ultraviolet light of slightly longer wavelength, UV-B, and splits back into O₂ and O. The two processes balance, and the ozone they maintain absorbs most of the Sun's UV-B before it reaches the ground. UV-B damages DNA: more of it reaching the surface means more skin cancer and more cataracts, and harm to crops and plankton.

Ozone at ground level is a different matter: there it is a pollutant that irritates the lungs. The benefit belongs to the stratospheric layer, and an answer about the ozone layer says where it is and what it absorbs.

Ozone layer

The region of the stratosphere where ozone is concentrated, which absorbs most of the ultraviolet radiation from the Sun.

CFCs and the chlorine radical cycle

Chlorofluorocarbons, CFCs, are halogenoalkanes that contain only carbon, chlorine and fluorine, such as CCl₂F₂ and CCl₃F. They are non-toxic, non-flammable, volatile and very unreactive, which made them ideal refrigerants, aerosol propellants, foam-blowing agents and cleaning solvents through the twentieth century. Being unreactive, they are not broken down in the lower atmosphere, and they drift up into the stratosphere over years.

There, ultraviolet light breaks the C–Cl bond homolytically. The C–F bond, at 467 kJ mol⁻¹ against 346 kJ mol⁻¹ for C–Cl, is too strong for the ultraviolet light that reaches that height, and survives. This is the same order of bond enthalpies that makes fluoroalkanes the slowest to hydrolyse.

CCl₂F₂(g) → •CClF₂(g) + Cl•(g)ultraviolet light

The chlorine radical then catalyses the decomposition of ozone in two propagation steps:

Cl•(g) + O₃(g) → ClO•(g) + O₂(g)

ClO•(g) + O₃(g) → 2O₂(g) + Cl•(g)

2O₃(g) → 3O₂(g)overall: add the two steps

Describe the drawing

A circular catalytic cycle. A box at the top left shows a CFC, CCl₂F₂, splitting in ultraviolet light to give a chlorine radical, which is fed into the cycle at the top. Down the right-hand arc Cl• reacts with ozone, which enters, to give ClO•, at the bottom, and oxygen, which leaves. Up the left-hand arc ClO• reacts with a second ozone to give two oxygen molecules and Cl• again. The centre reads overall 2O₃ to 3O₂, and a box below says Cl• is regenerated, so it is a catalyst.

The chlorine radical is used in the first step and regenerated in the second, so it is a catalyst: it speeds up the breakdown of ozone and is not used up. ClO• appears in the first step and is used in the second, so it is an intermediate and does not appear in the overall equation. One radical can go round the cycle many thousands of times before termination removes it, so a small amount of CFC breaks down a very large amount of ozone.

Most of the chlorine in the stratosphere is not free radicals at any moment. It is held in two reservoir compounds: hydrogen chloride, made when Cl• takes a hydrogen from methane, and chlorine nitrate, made when two radicals meet:

ClO•(g) + NO₂(g) → ClONO₂(g)a termination step

The thinning is worst over Antarctica in the southern spring. Through the polar winter, ice clouds form in the stratosphere, and on their surfaces HCl and ClONO₂ react to give Cl₂ and nitric acid, which stays in the ice. When sunlight returns in September it splits the Cl₂ into radicals all at once, and the ozone over the continent can fall by more than half within weeks.

The case against CFCs rests on three kinds of evidence from different research groups, each of which would have been weak on its own. Laboratory measurements of how unreactive CFCs are, and of how fast chlorine radicals react with ozone, predicted that CFCs would survive to reach the stratosphere and that their chlorine would break down ozone there. Measurements from the ground then found the loss: springtime ozone over the British Antarctic Survey's station at Halley Bay had fallen by about a third, and satellite records showed a hole over the whole continent.

Aircraft flown into the hole found ClO• at high concentration exactly where the ozone was missing. A mechanism, a measured loss and the radical caught where the damage was: together they made the case that CFCs were to blame.

The Montreal Protocol, which committed governments to phasing out CFCs, then served as a test of that case. If CFCs were the cause, stopping them should make stratospheric chlorine fall and the ozone recover, slowly, because CFC molecules already up there survive for decades. Both have been measured: chlorine in the stratosphere has been falling since the late 1990s, and the Antarctic ozone layer is on course to recover to its 1980 level around the 2060s.

The main replacements are hydrofluorocarbons, HFCs, such as CH₂FCF₃ in car air conditioning. They contain no chlorine, and the C–F bond is too strong for ultraviolet light in the stratosphere to break, so they release no radicals that attack ozone. They are, however, powerful greenhouse gases, and the 2016 Kigali Amendment to the Montreal Protocol now phases them down too, in favour of hydrocarbons such as 2-methylpropane and newer fluorinated alkenes.

CFC

A chlorofluorocarbon: a halogenoalkane containing only carbon, chlorine and fluorine.

HFC

A hydrofluorocarbon: a compound of hydrogen, fluorine and carbon with no chlorine, used to replace CFCs.

Catalyst

A substance that increases the rate of a reaction without being used up, by providing an alternative route with a lower activation energy.

Other boards: reagents, the oxygen-atom cycle and NO radicals

Cambridge gives the reagent for elimination as sodium hydroxide in ethanol, heated, and the chemistry is the same as with potassium hydroxide. OCR examines CFCs and the ozone layer but not elimination from haloalkanes; its elimination reaction is the dehydration of alcohols with an acid catalyst.

OCR writes the second propagation step with an oxygen atom in place of a second ozone molecule, because oxygen atoms are present in the stratosphere from the photolysis of O₂ and O₃:

ClO•(g) + O(g) → Cl•(g) + O₂(g)

O₃(g) + O(g) → 2O₂(g)overall, OCR's version

The chlorine radical is regenerated either way, so it is a catalyst in both versions. OCR also asks how nitrogen monoxide radicals, made in aircraft engines and by lightning, catalyse the breakdown of ozone by the same pattern: NO• + O₃ → NO₂• + O₂, then NO₂• + O → NO• + O₂, with NO• regenerated.