Alkanes

Lessons

Senior 5 · Chemistry

Alkanes

4 lessons

Lesson 4 of 4

Reactions and Uses of Alkanes

25 min

Why "paraffins"?

An old name for the alkanes is paraffins, from the Latin parum affinis, "little affinity". Alkanes ignore acids, alkalis, and oxidising agents such as acidified potassium manganate(VII). Their C–C and C–H bonds are strong, non-polar and have no lone pairs or π electrons for a reagent to attack. Yet alkanes power most of the world's vehicles. They react in two situations: when they are heated with oxygen, and when they meet atoms with an unpaired electron, called free radicals.

Definition

Free radical: an atom or group of atoms with an unpaired electron, shown by a dot: $\mathrm{Cl\bullet}$ , $\mathrm{{\bullet}CH_3}$ .

Homolytic fission: breaking a covalent bond so that each atom keeps one of the two shared electrons, forming two radicals.

Free-radical substitution: a reaction in which an atom (usually H) is replaced, by a chain mechanism with initiation, propagation and termination steps.

Cracking: breaking long-chain alkanes into shorter alkanes and alkenes, using heat or a catalyst.

Combustion

With enough oxygen, an alkane burns completely to carbon dioxide and water, releasing a lot of energy:

\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)} \qquad \Delta_cH = -890\ \mathrm{kJ\,mol^{-1}}

\mathrm{C_8H_{18}(l) + \tfrac{25}{2}O_2(g) \rightarrow 8CO_2(g) + 9H_2O(l)} \qquad \Delta_cH = -5470\ \mathrm{kJ\,mol^{-1}}

Per gram, methane releases $\frac{890}{16.0} = 55.6\ \mathrm{kJ}$ and octane $\frac{5470}{114} = 48.0\ \mathrm{kJ}$ . With too little oxygen, combustion is incomplete, giving carbon monoxide or soot (carbon):

\mathrm{2CH_4(g) + 3O_2(g) \rightarrow 2CO(g) + 4H_2O(l)}

Carbon monoxide is colourless and odourless. It binds to haemoglobin far more strongly than oxygen does, so the blood carries less oxygen. The US Centers for Disease Control and Prevention (CDC) lists headache, dizziness, weakness, upset stomach, vomiting, chest pain and confusion as the common symptoms. Never burn charcoal (an imbabura) or run a generator or a gas heater in a closed room; anyone affected needs fresh air at once and medical help.

Halogenation: free-radical substitution

A mixture of methane and chlorine does nothing in the dark, but reacts in ultraviolet light or sunlight. The light provides the energy to break the weakest bond present, Cl–Cl, and starts a chain reaction.

Figure 4.1: Free-radical substitution of methane. A half-headed ("fish-hook") arrow shows the movement of ONE electron; a dot shows an unpaired electron.

Initiation makes radicals. Propagation uses one radical and makes another, so a single Cl• can convert thousands of methane molecules. Termination removes radicals by pairing them; the ethane formed this way is evidence that $\mathrm{{\bullet}CH_3}$ radicals exist.
Further substitution: Cl• can also take an H from $\mathrm{CH_3Cl}$ , so dichloromethane, trichloromethane and tetrachloromethane form too. An excess of methane favours $\mathrm{CH_3Cl}$ ; an excess of chlorine gives $\mathrm{CCl_4}$ .
Selectivity: in larger alkanes, a hydrogen on a tertiary carbon is removed fastest and one on a primary carbon slowest, because the radical formed is more stable (3° > 2° > 1°). Bromine is far more selective than chlorine.

Cracking and reforming

Crude oil contains more long-chain alkanes than the market needs and fewer short ones. Thermal cracking uses high temperature (about 450–750 °C) and pressure; catalytic cracking uses a zeolite catalyst at about 450 °C and low pressure. Both give a shorter alkane and at least one alkene. Reforming turns straight chains into branched, cyclic and aromatic hydrocarbons, which burn more smoothly in petrol engines.

Figure 4.2: Cracking turns long, less useful alkanes into shorter alkanes and alkenes.

\mathrm{C_{10}H_{22}(l) \rightarrow C_8H_{18}(l) + C_2H_4(g)}

Uses, and their cost to the air

Use	Alkanes
cooking and heating fuel	LPG (propane, butane), natural gas and biogas (methane), kerosene
transport fuel	petrol ( $\mathrm{C_5}$ – $\mathrm{C_{10}}$ ), diesel, jet kerosene
lubricants, waxes, jelly	long chains: engine oil, candles, petroleum jelly
roads and roofs	bitumen
solvents	hexane, for example to extract vegetable oil from seeds
raw materials	cracking gives ethene and propene for plastics; methane gives hydrogen for ammonia

Burning alkanes releases carbon dioxide, the main greenhouse gas behind climate change; methane that leaks unburnt is itself a strong greenhouse gas. Engines also emit carbon monoxide, soot particles that damage the lungs, unburnt hydrocarbons, and nitrogen oxides, which together form smog in sunlight. A catalytic converter removes much of this, for example $\mathrm{2CO(g) + 2NO(g) \rightarrow 2CO_2(g) + N_2(g)}$ .

Worked example

Problem. Write equations for the complete and incomplete (to CO) combustion of propane. What volume of oxygen burns 10 $\mathrm{dm^3}$ of propane completely, and what volume of $\mathrm{CO_2}$ forms? (Gas volumes at the same temperature and pressure.)

Solution. Complete: $\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)}$ . Incomplete: $\mathrm{2C_3H_8(g) + 7O_2(g) \rightarrow 6CO(g) + 8H_2O(l)}$ .

By Avogadro's law, gas volumes are in the ratio of moles: 1 : 5 : 3. So 10 $\mathrm{dm^3}$ of propane needs 50 $\mathrm{dm^3}$ of oxygen and gives 30 $\mathrm{dm^3}$ of carbon dioxide.

Worked example

Problem. 2-methylpropane, $\mathrm{(CH_3)_3CH}$ , reacts with chlorine in UV light. Name the two monochloro products, predict their ratio from the number of each kind of hydrogen, and explain why the real ratio (about 63% to 37%) is different.

Solution. Replacing one of the nine primary H atoms (on the three $\mathrm{CH_3}$ groups) gives 1-chloro-2-methylpropane; replacing the single tertiary H gives 2-chloro-2-methylpropane. If every H were equally reactive, the ratio would be 9 : 1 (90% to 10%). In fact the tertiary product makes up about 37%, because the tertiary radical $\mathrm{(CH_3)_3C\bullet}$ is more stable than the primary radical, so a tertiary H is removed several times faster.

Warning

Use fish-hook arrows for radicals. A full arrowhead means a PAIR of electrons moves. In homolytic fission each atom takes ONE electron, so draw two half-headed arrows. And in the first propagation step Cl• takes a hydrogen ATOM, giving HCl and $\mathrm{{\bullet}CH_3}$ , not $\mathrm{CH_3Cl}$ and H•.

Tip

Check a radical mechanism by counting: initiation 0 radicals in → 2 out; propagation 1 in → 1 out; termination 2 in → 0 out.

Paul Morgan, A Level Chemistry: How to Draw Free Radical Substitution Mechanisms (8 min). Watch where each half-headed arrow starts and ends in the initiation and propagation steps.

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Practice

Write a balanced equation, with state symbols, for the complete combustion of butane.
Explain why carbon monoxide is poisonous, and give one way to avoid it at home.
Write the two propagation steps for the reaction of ethane with bromine in UV light.
Write a termination step for the reaction in question 3 that forms butane.
How many monochloro isomers can form from (a) propane; (b) 2,2-dimethylpropane? Name them.
Complete: $\mathrm{C_{14}H_{30} \rightarrow C_{10}H_{22} + \_\_\_}$ . Name a possible product.
Calculate the energy released when 1.00 kg of methane burns completely ( $\Delta_cH = -890\ \mathrm{kJ\,mol^{-1}}$ ).

Show answersHide answers

$\mathrm{2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l)}$ .
CO binds to haemoglobin much more strongly than oxygen, so the blood carries less oxygen to the tissues. Never burn charcoal or run an engine in a closed room; keep rooms with stoves ventilated.
$\mathrm{Br\bullet + C_2H_6 \rightarrow {\bullet}C_2H_5 + HBr}$ ; $\mathrm{{\bullet}C_2H_5 + Br_2 \rightarrow C_2H_5Br + Br\bullet}$ .
$\mathrm{{\bullet}C_2H_5 + {\bullet}C_2H_5 \rightarrow C_4H_{10}}$ .
(a) Two: 1-chloropropane and 2-chloropropane. (b) One: 1-chloro-2,2-dimethylpropane (all twelve H atoms are equivalent).
$\mathrm{C_4H_8}$ , for example but-1-ene (or $\mathrm{2C_2H_4}$ , ethene).
$n = \frac{1000}{16.0} = 62.5$ mol; energy $= 62.5 \times 890 = 5.56 \times 10^4\ \mathrm{kJ}$ .

1Why "paraffins"?

2Combustion

3Halogenation: free-radical substitution

4Cracking and reforming

5Uses, and their cost to the air

Why "paraffins"?

Combustion

Halogenation: free-radical substitution

Cracking and reforming

Uses, and their cost to the air