Alkanes

1. Overview

Alkanes are saturated hydrocarbons: their molecules contain only C-C and C-H single bonds. They are the main compounds in natural gas and a large part of crude oil, and burning them supplies much of the world's energy. This section covers how alkanes are made, why they are so unreactive, how they burn and react with chlorine or bromine, and the pollution that burning them in engines causes.

Key Definitions

• Alkane: a saturated hydrocarbon, whose molecules contain only C-C and C-H single bonds; open-chain alkanes have the general formula CₙH₂ₙ₊₂.
• Hydrogenation: the addition of hydrogen across a C=C double bond, by heating an alkene with H₂(g) and a platinum or nickel catalyst; it turns the alkene into an alkane.
• Cracking: the breaking of long-chain alkane molecules into smaller molecules, including at least one alkene, by heating them, usually with a catalyst such as aluminium oxide, Al₂O₃.
• Complete combustion: burning in a plentiful supply of oxygen, so that all the carbon in the fuel forms carbon dioxide and all the hydrogen forms water.
• Incomplete combustion: burning in a limited supply of oxygen, so that some of the carbon forms carbon monoxide or carbon (soot) instead of carbon dioxide; water is still formed.
• Free radical: a species with one or more unpaired electrons.
• Homolytic fission: the breaking of a covalent bond in which each of the two atoms takes one electron from the shared pair, so two free radicals form.
• Free-radical substitution: a reaction in which an atom in a molecule, such as a hydrogen atom in an alkane, is replaced by another atom, such as Cl or Br, through a chain of steps carried by free radicals: initiation, propagation and termination.
• Catalytic converter: a device in a vehicle exhaust in which hot platinum, palladium and rhodium catalysts, spread over a large surface area, convert carbon monoxide, oxides of nitrogen and unburnt hydrocarbons into carbon dioxide, nitrogen and water.

Content

Making alkanes 1: hydrogenation of an alkene

Heating an alkene with hydrogen over a metal catalyst adds H₂ across the C=C double bond. The product is the alkane with the same carbon chain.

• Reagent: hydrogen gas, H₂(g).
• Conditions: a platinum (Pt) or nickel (Ni) catalyst, and heat.

CH₃CH=CH₂(g) + H₂(g) → CH₃CH₂CH₃(g)

Here propene, C₃H₆, becomes propane, C₃H₈. The same reaction appears in 14.2 as a reaction of alkenes.

Making alkanes 2: cracking

Cracking breaks long alkane molecules into shorter ones.

• Conditions: heat, with aluminium oxide, Al₂O₃, as a catalyst. Without a catalyst, a much higher temperature is needed.

The chain can break at almost any C-C bond, so one heavy alkane gives a mixture of products. For example, one molecule of undecane can give heptane and butene:

C₁₁H₂₄ → C₇H₁₆ + C₄H₈

At least one product is an alkene. Two alkanes, C₇H₁₆ and C₄H₁₀, would need 16 + 10 = 26 hydrogen atoms, but undecane has only 24, so one piece must form a C=C bond.

A useful rule: if the products are alkanes and alkenes with one C=C each, exactly one product is an alkane and the rest are alkenes. If hydrogen, H₂, also forms, every hydrocarbon product is an alkene. Always check that carbon and hydrogen both balance.

Why cracking is useful

Fractional distillation separates crude oil into fractions, each a mixture of hydrocarbons with boiling points in a narrow range. The amounts do not match demand:

• There is more of the heavy fractions, such as fuel oil, than can be sold. Their molecules are large (high $M_{\text{r}}$), so they are viscous and hard to ignite.
• There is less of the light fractions, such as petrol (roughly 5 to 10 carbon atoms per molecule), than is needed.
• Crude oil contains very little alkene, yet alkenes are needed to make polymers and other chemicals.

Cracking turns surplus heavy fractions into two useful kinds of product: shorter alkanes of lower $M_{\text{r}}$, which are volatile and ignite easily, so they suit petrol; and alkenes such as ethene and propene, the starting materials for polymers such as poly(ethene) (14.2 and 20.1).

Worked example 1: cracking tetradecane

Question: Tetradecane, C₁₄H₃₀, comes from a crude oil fraction heavier than petrol. A refinery cracks it to make octane, C₈H₁₈, for petrol and ethene, C₂H₄, for making poly(ethene).

• (a) Write an equation for cracking one molecule of tetradecane into octane and ethene only.
• (b) Calculate the mass of each product made from 1.00 kg of tetradecane, if every molecule cracks this way.
• (c) Suggest a different set of products from one molecule of tetradecane.

Step 1: balance carbon, then check hydrogen. Octane uses 8 of the 14 carbon atoms. The other 6 make 3 molecules of ethene.

C₁₄H₃₀ → C₈H₁₈ + 3C₂H₄

Hydrogen: 18 + (3 × 4) = 30, which matches tetradecane.

Step 2: moles of tetradecane. $M_{\text{r}}$ = (14 × 12.0) + (30 × 1.0) = 198.0, and 1.00 kg = 1000 g. $$n = \frac{1000}{198.0} = 5.0505\ \text{mol}$$

Step 3: masses of the products. One mole of tetradecane gives 1 mol of octane ($M_{\text{r}}$ = 114.0) and 3 mol of ethene ($M_{\text{r}}$ = 28.0). $$m(\text{octane}) = 5.0505 \times 114.0 = 575.8\ \text{g}$$ $$n(\text{ethene}) = 3 \times 5.0505 = 15.1515\ \text{mol}$$ $$m(\text{ethene}) = 15.1515 \times 28.0 = 424.2\ \text{g}$$ Check: 575.8 + 424.2 = 1000.0 g, so mass is conserved.

Step 4: another split. Any split works if carbon and hydrogen balance, with one alkane and the rest alkenes.

Answer: (a) C₁₄H₃₀ → C₈H₁₈ + 3C₂H₄. (b) 576 g of octane and 424 g of ethene (3 s.f.). (c) For example, C₁₄H₃₀ → C₉H₂₀ + C₅H₁₀.

Why alkanes are unreactive

Under normal conditions, alkanes do not react with acids, alkalis or aqueous oxidising agents such as acidified potassium manganate(VII), which does oxidise alkenes. Two features of their bonds explain this.

1. The bonds are strong. C-H is 410 kJ mol⁻¹ and C-C is 350 kJ mol⁻¹, so breaking them takes a lot of energy.
2. The bonds have very little polarity. The electronegativities of carbon (2.5) and hydrogen (2.1) differ by only 0.4, and C-C is not polar at all. So there is no δ+ carbon atom to attract a nucleophile, such as OH⁻, and no region of high electron density, such as a C=C bond, to attract an electrophile.

Bond	Bond energy / kJ mol⁻¹	Electronegativity difference
C-H	410	0.4
C-C	350	0
O-H	460	1.4

Both factors matter: O-H is even stronger than C-H, yet alcohols react with sodium (16.1) because O-H is strongly polar.

Alkanes do react with oxygen once ignited, with chlorine or bromine in UV light (the attacking species are free radicals, not polar reagents), and on strong heating with a catalyst (cracking).

Complete and incomplete combustion

Alkanes burn in oxygen in very exothermic reactions, which is why they are used as fuels.

Complete combustion needs a plentiful supply of oxygen. All the carbon becomes CO₂ and all the hydrogen becomes H₂O:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

For any alkane $\text{C}_n\text{H}_{2n+2}$, one mole makes $n$ mol of CO₂ and $(n+1)$ mol of H₂O, and needs $\tfrac{3n+1}{2}$ mol of O₂.

Incomplete combustion happens when the oxygen supply is limited. Some carbon forms carbon monoxide, CO, or solid carbon (soot), which makes the flame yellow and smoky. Water still forms:

2C₃H₈(g) + 7O₂(g) → 6CO(g) + 8H₂O(l)

C₃H₈(g) + 2O₂(g) → 3C(s) + 4H₂O(l)

In an engine, fuel and air have very little time to mix, so some fuel burns incompletely, or not at all, even when there is enough air overall.

To balance a combustion equation, balance carbon, then hydrogen, then oxygen. Halves such as 3½O₂ are allowed.

Worked example 2: oxygen for complete and incomplete combustion

Question: Heptane, C₇H₁₆ ($M_{\text{r}}$ = 100.0), is a liquid alkane.

• (a) Write equations for its complete combustion and for its incomplete combustion to carbon monoxide.
• (b) Calculate the volume of oxygen, measured at room conditions, needed to burn 3.00 g of heptane in each way.

Step 1: equations. Seven carbon atoms give 7CO₂ (or 7CO), and 16 hydrogen atoms give 8H₂O. Then count the oxygen atoms on the right: 14 + 8 = 22 (11O₂) for complete combustion, and 7 + 8 = 15 (7½O₂) for incomplete combustion.

C₇H₁₆(l) + 11O₂(g) → 7CO₂(g) + 8H₂O(l)

C₇H₁₆(l) + 7½O₂(g) → 7CO(g) + 8H₂O(l)

Step 2: moles of heptane. $$n = \frac{3.00}{100.0} = 0.0300\ \text{mol}$$

Step 3: oxygen. Multiply by the mole ratio, then by $V_{\text{m}}$ = 24.0 dm³ mol⁻¹ (room conditions).

• Complete: $0.0300 \times 11 = 0.330$ mol of O₂, and $0.330 \times 24.0 = 7.92$ dm³.
• Incomplete: $0.0300 \times 7.5 = 0.225$ mol of O₂, and $0.225 \times 24.0 = 5.40$ dm³.

Answer: 7.92 dm³ of oxygen for complete combustion and 5.40 dm³ for incomplete combustion to CO (3 s.f.). Burning to CO needs only about 68% of the oxygen (7.5 ÷ 11), so CO forms when the air supply is short.

Pollutants from the internal combustion engine

Besides CO₂, water and unreacted air, a car's exhaust contains these pollutants:

• Carbon monoxide, CO, from incomplete combustion. It is toxic: it binds to the haemoglobin in red blood cells more strongly than oxygen does, so the blood carries less oxygen.
• Oxides of nitrogen, NO and NO₂, often written NOₓ.
• Unburnt hydrocarbons. In sunlight they react with oxides of nitrogen to form photochemical smog, which contains PAN (peroxyacetyl nitrate) and irritates the eyes and lungs.
• Carbon (soot) particles from incomplete combustion, which damage the lungs.

The oxides of nitrogen do not come from the fuel: alkanes contain no nitrogen. N₂ is very unreactive (N≡N is 944 kJ mol⁻¹), but the very high temperature in the engine lets nitrogen and oxygen from the air react:

N₂(g) + O₂(g) → 2NO(g)

In the air, NO is oxidised to NO₂, which forms nitric acid with water and oxygen and so adds to acid rain. Section 12.1 covers these reactions.

Catalytic removal of pollutants

A catalytic converter in the exhaust pipe contains a ceramic honeycomb coated with a thin layer of platinum, palladium and rhodium, which gives the hot gases a very large surface of catalyst. The metals are heterogeneous catalysts: solids, in a different phase from the reacting gases (8.3).

2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)

2CO(g) + O₂(g) → 2CO₂(g)

C₇H₁₆(g) + 11O₂(g) → 7CO₂(g) + 8H₂O(g)

These are redox reactions. CO and hydrocarbons are oxidised to CO₂ and H₂O. NO is reduced to N₂: the oxidation number of nitrogen falls from +2 to 0.

Consequences. Much less CO, NOₓ and unburnt hydrocarbon reaches the air, so there is less carbon monoxide poisoning, acid rain and smog. But a converter does not remove CO₂: it turns CO and hydrocarbons into more CO₂, a greenhouse gas.

Free-radical substitution of ethane

In the dark at room temperature, ethane and chlorine do not react. In ultraviolet (UV) light they react by free-radical substitution: chlorine atoms replace hydrogen atoms, one at a time.

C₂H₆(g) + Cl₂(g) → C₂H₅Cl(g) + HCl(g)

The organic product is chloroethane. Bromine reacts in the same way, giving bromoethane and hydrogen bromide, and its red-brown colour fades as it is used up.

The mechanism: initiation, propagation and termination

The terms used here are defined in 13.2. Write every radical with a dot for its unpaired electron, on the atom that carries it.

Initiation. UV light breaks the Cl-Cl bond by homolytic fission: each atom keeps one electron of the shared pair.

Cl₂ → 2Cl•

Chlorine molecules absorb the UV light. The Cl-Cl bond (242 kJ mol⁻¹) is much weaker than C-H (410) or C-C (350), so it is the bond that breaks.

Propagation. A radical reacts with a molecule to give a new molecule and a new radical.

Cl• + CH₃CH₃ → CH₃CH₂• + HCl

CH₃CH₂• + Cl₂ → CH₃CH₂Cl + Cl•

The second step makes a new Cl•, which starts the first step again, so one initiation can lead to many cycles: a chain reaction. Adding the two steps and cancelling the radicals gives the overall equation.

Termination. Two radicals join, so radicals are used up and the chain stops.

Cl• + Cl• → Cl₂

CH₃CH₂• + Cl• → CH₃CH₂Cl

CH₃CH₂• + CH₃CH₂• → CH₃CH₂CH₂CH₃

The last step makes a small amount of butane. Finding a four-carbon product is evidence that ethyl radicals form.

Step	Radicals used	Radicals made
Initiation	0	2
Propagation	1	1
Termination	2	0

The mechanism with bromine is the same, with Br in place of Cl.

Why a mixture of products forms

A chlorine radical can also remove a hydrogen atom from chloroethane:

Cl• + CH₃CH₂Cl → CH₂ClCH₂• + HCl

CH₂ClCH₂• + Cl₂ → CH₂ClCH₂Cl + Cl•

Further substitution gives dichloroethanes and so on, up to hexachloroethane, C₂Cl₆. With the termination products, the result is a mixture that must be separated, for example by fractional distillation. Excess ethane favours chloroethane, because a Cl• is then far more likely to collide with ethane than with chloroethane; excess chlorine favours multiple substitution.

Worked example 3: bond energies and the propagation steps

Question: Use bond energies to calculate ΔH for each propagation step in the chlorination of ethane, and for the overall reaction. Then explain why the first propagation step forms HCl and not a hydrogen radical, H•. Treat all species as gases.

Bond energies / kJ mol⁻¹: C-H 410, Cl-Cl 242, C-Cl 340, H-Cl 431. As in 5.1, $\Delta H = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$.

Step 1: first propagation step. Cl• + CH₃CH₃ → CH₃CH₂• + HCl breaks C-H and forms H-Cl. $$\Delta H = 410 - 431 = -21\ \text{kJ mol}^{-1}$$

Step 2: second propagation step. CH₃CH₂• + Cl₂ → CH₃CH₂Cl + Cl• breaks Cl-Cl and forms C-Cl. $$\Delta H = 242 - 340 = -98\ \text{kJ mol}^{-1}$$

Step 3: overall reaction. The two steps add up to C₂H₆ + Cl₂ → C₂H₅Cl + HCl. $$\Delta H = (-21) + (-98) = -119\ \text{kJ mol}^{-1}$$ Check from the overall equation: (410 + 242) − (340 + 431) = −119 kJ mol⁻¹.

Step 4: the step that does not happen. Cl• + CH₃CH₃ → CH₃CH₂Cl + H• would break C-H and form C-Cl. $$\Delta H = 410 - 340 = +70\ \text{kJ mol}^{-1}$$

Answer: The propagation steps are −21 and −98 kJ mol⁻¹, and the overall reaction is −119 kJ mol⁻¹ (estimates, as C-H and C-Cl are average values). Making H• would be endothermic (+70 kJ mol⁻¹) while making HCl is exothermic (−21 kJ mol⁻¹), because H-Cl (431) is stronger than C-Cl (340). The energy favours forming HCl, and that is the step that happens.

With bromine (H-Br 366), the first propagation step is endothermic: 410 − 366 = +44 kJ mol⁻¹. Propagation steps are not always exothermic.

Key Equations

Hydrogenation (H₂(g), Pt or Ni catalyst, heat; an open-chain alkene with one C=C and $n$ carbon atoms): $$\text{C}_n\text{H}_{2n} + \text{H}_2 \rightarrow \text{C}_n\text{H}_{2n+2}$$

Cracking (heat with Al₂O₃; $a + b = n$; more alkenes, or H₂, can also form): $$\text{C}_n\text{H}_{2n+2} \rightarrow \text{C}_a\text{H}_{2a+2} + \text{C}_b\text{H}_{2b}$$

Complete combustion (per mole of alkane): $$\begin{aligned}\text{C}_n\text{H}_{2n+2} &+ \tfrac{3n+1}{2}\,\text{O}_2 \\ &\rightarrow n\,\text{CO}_2 + (n+1)\,\text{H}_2\text{O}\end{aligned}$$

Incomplete combustion: one mole of alkane needs $\tfrac{2n+1}{2}$ mol of O₂ to form CO, or $\tfrac{n+1}{2}$ mol of O₂ to form carbon. Water forms in both.

Free-radical substitution of ethane (Cl₂ or Br₂, UV light; X = Cl or Br):

• Overall: C₂H₆ + X₂ → C₂H₅X + HX
• Initiation: X₂ → 2X•
• Propagation: X• + C₂H₆ → C₂H₅• + HX, then C₂H₅• + X₂ → C₂H₅X + X•
• Termination: X• + X• → X₂; C₂H₅• + X• → C₂H₅X; C₂H₅• + C₂H₅• → C₄H₁₀ (butane)

Engine and converter (all gases): N₂ + O₂ → 2NO in the engine; 2CO + 2NO → 2CO₂ + N₂ in the converter.

Enthalpy change from bond energies ($E$ in kJ mol⁻¹, for gases): $$\Delta H = \sum E(\text{broken}) - \sum E(\text{formed})$$

Common Mistakes to Avoid

• Mistake: making H• in a propagation step, or giving an initiation or termination step instead. Fix: in propagation, one radical goes in and one comes out: Cl• + C₂H₆ → C₂H₅• + HCl, not C₂H₅Cl + H•.
• Mistake: leaving out the dots, or writing ions such as Cl⁺ and Cl⁻. Fix: homolytic fission gives radicals, Cl₂ → 2Cl•. Put the dot on the atom with the unpaired electron: in CH₃CH₂•, it is on carbon.
• Mistake: defining a free radical as a species with a lone pair, or with a "free" or "unshared" electron. Fix: say "one or more unpaired electrons". A chlorine atom has seven outer electrons, one of them unpaired.
• Mistake: vague conditions such as "light", "heat" or "a catalyst". Fix: UV light for halogenation; H₂(g) with a Pt or Ni catalyst and heat for hydrogenation; heat with Al₂O₃ for cracking.
• Mistake: a cracking equation with two alkanes. Fix: count hydrogen. C₁₁H₂₄ → C₇H₁₆ + C₄H₁₀ has 26 hydrogen atoms on the right but 24 on the left, so the second product must be C₄H₈, an alkene.
• Mistake: naming only CO for incomplete combustion, or leaving out the water. Fix: soot can also form, and water still forms: C₃H₈ + 2O₂ → 3C + 4H₂O.
• Mistake: saying oxides of nitrogen come from the fuel or the catalytic converter, or that the converter turns NO into NO₂. Fix: N₂ and O₂ from the air react in the hot engine, N₂ + O₂ → 2NO, and the converter reduces NO to N₂: 2CO + 2NO → 2CO₂ + N₂.
• Mistake: saying the converter oxidises hydrocarbons to CO₂ and hydrogen, or that it removes CO₂. Fix: they form CO₂ and water, C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O, so the converter increases CO₂ rather than removing it.

Exam Tips

• Name the type of reaction in full when asked: "free-radical substitution" for an alkane with Cl₂ or Br₂ in UV light, not just "substitution", and "addition" for the hydrogenation of an alkene.
• Describe the mechanism by naming each step and writing it as a balanced equation, with a dot on every radical. Check that the two propagation steps add up to the overall equation. For "a termination step", any two radicals from the mechanism joining will do.
• Explain the unreactivity of alkanes with two points and data: C-H is strong (410 kJ mol⁻¹) and almost non-polar (electronegativities 2.5 and 2.1), so there is no δ+ site or electron-rich region to attract a polar reagent.
• Suggest cracking products by balancing carbon, then hydrogen, and say why each is useful: shorter alkanes for petrol, alkenes for polymers.
• For each pollutant, say where it comes from, what harm it does and how the converter removes it, with a balanced equation.