Alkanes

AS · 15 min

Alkanes are the saturated hydrocarbons, $\ce{C_{n}H_{2n+2}}$ : methane in natural gas, propane and butane in bottled gas, octane in petrol. They are the most important fuels in the world and the starting materials for almost every other organic compound, yet chemically they are remarkably unreactive. This note explains that unreactivity, then covers the reactions alkanes do undergo (combustion and free-radical substitution with halogens), how they are made by cracking and by hydrogenating alkenes, and the pollutants that their combustion in car engines produces. The free-radical substitution mechanism is one of the most frequently examined mechanisms at AS.

Structure and physical properties

Every carbon atom in an alkane is sp³ hybridised, with four $\sigma$ bonds arranged tetrahedrally at $109.5^\circ$ . The molecules contain only C–C and C–H bonds.

Boiling points rise with chain length. Each additional $\ce{CH2}$ adds electrons, so the instantaneous dipole–induced dipole forces between molecules get stronger. Methane to butane are gases at room temperature; pentane to about $\ce{C16H34}$ are liquids; longer alkanes are waxy solids.
Branching lowers the boiling point. A branched molecule is more compact, with less surface area in contact with its neighbours, so the intermolecular forces are weaker. Pentane boils at $36\ ^\circ\text{C}$ , 2,2-dimethylpropane at $10\ ^\circ\text{C}$ .
Alkanes are insoluble in water. They cannot form hydrogen bonds with water, so mixing them would disrupt water's hydrogen bonding without any compensating attraction.

Why alkanes are unreactive

Key result

Alkanes are generally unreactive, particularly towards polar reagents (nucleophiles, electrophiles, acids, bases, aqueous oxidising agents), because:

C–C and C–H bonds are strong (C–H about $410\ \text{kJ mol}^{-1}$ ), so a lot of energy is needed to break them.
C–H bonds are almost non-polar. Carbon and hydrogen have similar electronegativities, so there is no $\delta+$ or $\delta-$ site to attract a nucleophile or electrophile.

The reactions alkanes do undergo, combustion and halogenation, need a large input of energy (a flame, a spark or ultraviolet light) and go by free radicals, which do not need a polar site to attack.

Combustion

Alkanes burn in a plentiful supply of oxygen to give carbon dioxide and water. This complete combustion is very exothermic, which is why alkanes are used as fuels.

\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}

\ce{2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(l)}

In a limited supply of oxygen, incomplete combustion produces carbon monoxide (a toxic gas) or carbon (soot):

\ce{2C8H18(l) + 17O2(g) -> 16CO(g) + 18H2O(l)}

\ce{CH4(g) + O2(g) -> C(s) + 2H2O(l)}

Method

Balancing a combustion equation for $\ce{C_{x}H_{y}}$

Balance carbon: $x$ $\ce{CO2}$ (or CO, or C).
Balance hydrogen: $\frac{y}{2}$ $\ce{H2O}$ .
Count the oxygen atoms on the right and halve to get the number of $\ce{O2}$ . For complete combustion this is $x + \frac{y}{4}$ .
If you have a fraction, double every coefficient.

Pollution from the internal combustion engine

Car engines burn petrol or diesel (mixtures of alkanes) in a confined space, very quickly, at high temperature. Three pollutants come out of the exhaust:

pollutant	how it forms	why it is harmful
carbon monoxide, CO	incomplete combustion when oxygen is limited	toxic: it binds to haemoglobin more strongly than oxygen does, reducing the blood's ability to carry oxygen
oxides of nitrogen, NO and $\ce{NO2}$	nitrogen and oxygen from the air combine at the high temperature of the engine: $\ce{N2 + O2 -> 2NO}$ , then $\ce{2NO + O2 -> 2NO2}$	cause acid rain ( $\ce{NO2}$ forms nitric acid); help form photochemical smog; $\ce{NO2}$ irritates the lungs
unburnt hydrocarbons	fuel that does not burn completely	react with oxides of nitrogen in sunlight to form photochemical smog; some are carcinogenic

Carbon dioxide, from complete combustion, is not toxic but is a greenhouse gas that contributes to climate change.

Catalytic converters

A catalytic converter in the exhaust contains a ceramic honeycomb coated with platinum, palladium and rhodium, giving a large surface area. These heterogeneous catalysts convert the three pollutants into less harmful gases:

\ce{2CO(g) + 2NO(g) -> 2CO2(g) + N2(g)}

\ce{2C8H18(g) + 50NO(g) -> 16CO2(g) + 18H2O(g) + 25N2(g)}

In the first reaction NO is reduced (N: $+2 \to 0$ ) and CO is oxidised (C: $+2 \to +4$ ). How the catalyst works (adsorption, weakening of bonds, desorption) is covered in Nitrogen oxides and acid rain.

Free-radical substitution

In ultraviolet light (or sunlight), alkanes react with chlorine or bromine. A hydrogen atom is replaced by a halogen atom: a substitution. The reaction goes by a free-radical mechanism.

\ce{CH4(g) + Cl2(g) ->[UV] CH3Cl(g) + HCl(g)}

Key result

Reagents and conditions: chlorine or bromine; ultraviolet light (sunlight); room temperature. No reaction occurs in the dark at room temperature.

Product: a halogenoalkane and a hydrogen halide. With bromine, the orange-brown colour of bromine fades slowly as it is used up.

The mechanism

The mechanism of the reaction between methane and chlorine has three stages.

Initiation. Ultraviolet light supplies the energy to break the Cl–Cl bond, the weakest bond present, by homolytic fission:

\ce{Cl2 ->[UV] 2Cl^.}

Each chlorine atom takes one electron from the shared pair. Two half-headed arrows can be drawn, one from the Cl–Cl bond to each Cl atom.
Two chlorine free radicals form, each with an unpaired electron.

Propagation. Each step uses up one radical and produces another, so the chain continues:

\ce{CH4 + Cl^. -> CH3^. + HCl}

\ce{CH3^. + Cl2 -> CH3Cl + Cl^.}

A chlorine radical removes a hydrogen atom from methane. The C–H bond breaks homolytically: one electron goes with the H to pair with the radical's electron, forming H–Cl; the other stays on carbon, leaving a methyl radical, $\ce{CH3^.}$ .
The methyl radical collides with a chlorine molecule and takes one chlorine atom, forming the C–Cl bond of chloromethane. The Cl–Cl bond breaks homolytically, releasing a new chlorine radical.
That chlorine radical can attack another methane molecule, so one initiation event can lead to thousands of product molecules.

Termination. Two radicals meet and pair their electrons, removing radicals from the system:

\ce{Cl^. + Cl^. -> Cl2}

\ce{CH3^. + Cl^. -> CH3Cl}

\ce{CH3^. + CH3^. -> C2H6}

The third termination step explains why small amounts of ethane are found in the product mixture, which is evidence for the radical mechanism.

Tip

Adding the two propagation steps gives the overall equation: $\ce{CH4 + Cl2 -> CH3Cl + HCl}$ . The chlorine radical is used in the first step and regenerated in the second, so it behaves like a catalyst for the chain. This is a good check that your propagation steps are right.

Further substitution: a mixture of products

Chloromethane still has C–H bonds, so a chlorine radical can attack it too:

\ce{CH3Cl + Cl^. -> CH2Cl^. + HCl}

\ce{CH2Cl^. + Cl2 -> CH2Cl2 + Cl^.}

Repeating this gives dichloromethane, trichloromethane ( $\ce{CHCl3}$ ) and tetrachloromethane ( $\ce{CCl4}$ ). The product is always a mixture, which must be separated by fractional distillation. This is the main disadvantage of free-radical substitution as a synthetic method.

To favour monosubstitution ( $\ce{CH3Cl}$ ), use an excess of methane: a chlorine radical is then much more likely to hit a methane molecule than a chloromethane molecule.
To favour complete substitution ( $\ce{CCl4}$ ), use an excess of chlorine.

With larger alkanes, substitution can happen at any carbon, giving positional isomers. Propane gives both 1-chloropropane and 2-chloropropane, plus multiply substituted products.

Making alkanes

Hydrogenation of alkenes

\ce{CH2=CHCH3 + H2 ->[Ni][heat] CH3CH2CH3}

Key result

Hydrogenation: hydrogen gas with a nickel catalyst at about $150\ ^\circ\text{C}$ (or a platinum catalyst at room temperature). This is an addition reaction: one mole of $\ce{H2}$ adds across each C=C.

The same reaction is used to harden vegetable oils into margarine, by hydrogenating some of their C=C bonds.

Cracking

Crude oil is separated by fractional distillation into fractions. The heavier fractions (long-chain alkanes) are produced in larger amounts than they are needed, while there is more demand for petrol (about $\ce{C5}$ – $\ce{C10}$ ) and for alkenes, the raw material for polymers and many other chemicals.

Definition

Cracking is the breaking down of long-chain alkanes into shorter-chain alkanes and alkenes of lower relative molecular mass, using heat and a catalyst.

Key result

Catalytic cracking: heat (about $500\ ^\circ\text{C}$ ) with a catalyst of aluminium oxide, $\ce{Al2O3}$ , or a zeolite (aluminosilicate). Thermal cracking: high temperature (up to $900\ ^\circ\text{C}$ ) and high pressure, without a catalyst.

Cracking breaks C–C bonds at random, so a range of products forms. A typical equation:

\ce{C12H26 -> C8H18 + 2C2H4}

Each cracking equation must balance in C and H. Because an alkane has the maximum possible number of hydrogens, cracking one alkane always gives at least one alkene (there is not enough hydrogen for all the products to be alkanes). Hydrogen gas can also be a product.

Cracking is useful because it:

converts surplus heavy fractions into petrol (shorter alkanes, often branched, which burn more smoothly in engines);
produces alkenes such as ethene and propene for making polymers, ethanol and other chemicals;
can produce hydrogen for the Haber process.

Worked examples

Routine: combustion equations

Write equations for (a) the complete combustion of butane, (b) the incomplete combustion of butane to carbon monoxide and water.

Solution

(a) $\ce{C4H10}$ : 4 $\ce{CO2}$ , 5 $\ce{H2O}$ ; oxygen atoms on the right $= 8 + 5 = 13$ , so $6\tfrac{1}{2}$ $\ce{O2}$ . Doubling:

\ce{2C4H10 + 13O2 -> 8CO2 + 10H2O}

(b) 4 CO, 5 $\ce{H2O}$ ; oxygen atoms $= 4 + 5 = 9$ , so $4\tfrac{1}{2}$ $\ce{O2}$ :

\ce{2C4H10 + 9O2 -> 8CO + 10H2O}

Routine: cracking equations

(a) Tetradecane, $\ce{C14H30}$ , is cracked to give octane, ethene and one other hydrocarbon. Identify the other hydrocarbon. (b) Explain why cracking is carried out.

Solution

(a) $\ce{C14H30 -> C8H18 + C2H4 + C_{x}H_{y}}$ . Carbon: $x = 14 - 8 - 2 = 4$ . Hydrogen: $y = 30 - 18 - 4 = 8$ . The other product is $\ce{C4H8}$ , a butene (an alkene).

(b) Long-chain alkanes are in excess supply and have few uses, while there is high demand for shorter-chain alkanes for petrol and for alkenes as a feedstock for polymers. Cracking converts the low-value heavy fractions into these more useful products.

Standard: the mechanism with ethane and bromine

Ethane reacts with bromine in ultraviolet light to form bromoethane. (a) Name the mechanism. (b) Write equations for the initiation step, the two propagation steps and two termination steps. (c) Explain why butane is found in small amounts in the products.

Solution

(a) Free-radical substitution.

(b) Initiation: $\ce{Br2 ->[UV] 2Br^.}$ (homolytic fission of Br–Br).

Propagation: $\ce{C2H6 + Br^. -> C2H5^. + HBr}$ , then $\ce{C2H5^. + Br2 -> C2H5Br + Br^.}$ .

Termination (any two): $\ce{Br^. + Br^. -> Br2}$ ; $\ce{C2H5^. + Br^. -> C2H5Br}$ ; $\ce{C2H5^. + C2H5^. -> C4H10}$ .

Standard: finding a formula from gas volumes

$10\ \text{cm}^3$ of a gaseous alkane is mixed with $100\ \text{cm}^3$ of oxygen (an excess) and exploded. After cooling to room temperature, the volume of gas is $75\ \text{cm}^3$ . Passing this through aqueous sodium hydroxide reduces the volume to $35\ \text{cm}^3$ . All volumes are at the same temperature and pressure. Find the formula of the alkane.

Solution

$\ce{C_{x}H_{y} + (x + \frac{y}{4})O2 -> xCO2 + \frac{y}{2}H2O}$

After cooling, water is liquid, so the $75\ \text{cm}^3$ is $\ce{CO2}$ + unreacted $\ce{O2}$ . Sodium hydroxide absorbs $\ce{CO2}$ , so $\ce{CO2} = 75 - 35 = 40\ \text{cm}^3$ and unreacted $\ce{O2} = 35\ \text{cm}^3$ .

By Avogadro's law, volume ratios equal mole ratios. $10\ \text{cm}^3$ of alkane gives $40\ \text{cm}^3$ $\ce{CO2}$ , so $x = 4$ .

Oxygen used $= 100 - 35 = 65\ \text{cm}^3$ , so $x + \frac{y}{4} = 6.5$ , giving $\frac{y}{4} = 2.5$ and $y = 10$ .

The alkane is $\ce{C4H10}$ , butane (or methylpropane). It fits $\ce{C_{n}H_{2n+2}}$ .

Exam-hard: monochlorination of 2-methylbutane

2-methylbutane, $\ce{(CH3)2CHCH2CH3}$ , reacts with chlorine in UV light.

(a) How many structural isomers of formula $\ce{C5H11Cl}$ can form? Name them. (b) Which of these are chiral? (c) Explain why this reaction is a poor way to make any one of them.

Solution

(a) Number the main chain $\ce{CH3-CH(CH3)-CH2-CH3}$ as C1 to C4 from the left. The methyl branch on C2 is equivalent to C1 (both are $\ce{CH3}$ groups on C2), so there are four different kinds of hydrogen: on C1 (or the branch), C2, C3 and C4. Four structural isomers:

Cl on C1: $\ce{ClCH2CH(CH3)CH2CH3}$ , 1-chloro-2-methylbutane
Cl on C2: $\ce{(CH3)2CClCH2CH3}$ , 2-chloro-2-methylbutane
Cl on C3: $\ce{(CH3)2CHCHClCH3}$ , 2-chloro-3-methylbutane
Cl on C4: $\ce{(CH3)2CHCH2CH2Cl}$ , 1-chloro-3-methylbutane

(b) 1-chloro-2-methylbutane: C2 carries H, $\ce{CH3}$ , $\ce{CH2Cl}$ , $\ce{C2H5}$ : chiral. 2-chloro-3-methylbutane: C2 carries H, Cl, $\ce{CH3}$ , $\ce{CH(CH3)2}$ : chiral. The other two have no carbon with four different groups. So there are six isomers including optical isomers.

(c) Free-radical substitution is not selective: chlorine radicals remove hydrogen atoms from every position, so all four isomers (plus their enantiomers) form, along with di- and polysubstituted products. The yield of any one is low and the mixture must be separated.

Exam-hard: energetics of the propagation steps

Use these bond energies ( $\text{kJ mol}^{-1}$ ): C–H 410, Cl–Cl 242, H–Cl 431, C–Cl 340, Br–Br 193, H–Br 366, C–Br 280.

(a) Calculate the enthalpy change of each propagation step for the chlorination of methane. (b) Do the same for bromination. (c) Suggest why bromination of methane is much slower than chlorination.

Solution

(a) $\ce{CH4 + Cl^. -> CH3^. + HCl}$ : break C–H, make H–Cl. $\Delta H = 410 - 431 = -21\ \text{kJ mol}^{-1}$ .

$\ce{CH3^. + Cl2 -> CH3Cl + Cl^.}$ : break Cl–Cl, make C–Cl. $\Delta H = 242 - 340 = -98\ \text{kJ mol}^{-1}$ .

(b) $\ce{CH4 + Br^. -> CH3^. + HBr}$ : $\Delta H = 410 - 366 = +44\ \text{kJ mol}^{-1}$ .

$\ce{CH3^. + Br2 -> CH3Br + Br^.}$ : $\Delta H = 193 - 280 = -87\ \text{kJ mol}^{-1}$ .

(c) The first propagation step is endothermic for bromine ( $+44\ \text{kJ mol}^{-1}$ ) but exothermic for chlorine. An endothermic step has a higher activation energy, so fewer collisions between $\ce{Br^.}$ and $\ce{CH4}$ succeed and the chain propagates more slowly.

Watch out

Hydrogen radicals. The first propagation step forms $\ce{CH3^.}$ and HCl, not $\ce{CH3Cl}$ and $\ce{H^.}$ . A step producing $\ce{H^.}$ is wrong.
Initiation by heat alone. At AS the condition is UV light. "Heat" or "sunlight and a catalyst" lose the mark.
Missing dots. Every radical in every step must carry its dot, $\ce{Cl^.}$ , $\ce{CH3^.}$ .
Cracking as "breaking down crude oil". Cracking breaks long-chain alkanes into shorter alkanes and alkenes. Fractional distillation separates crude oil; it does not break bonds.
Unbalanced cracking equations. Check C and H, and remember at least one product must be an alkene.
NOx from the fuel. Petrol contains almost no nitrogen. Nitrogen oxides form from nitrogen in the air at the high temperature in the engine.

Exam tip

"Describe the mechanism" for free-radical substitution: name each stage (initiation, propagation, termination), give the equations with radical dots, and state UV light. Two propagation steps are required; one termination step usually suffices unless more are asked for.
"Explain why alkanes are unreactive": strong C–H and C–C bonds and low polarity (no $\delta+$ / $\delta-$ sites). Both points are usually needed.
For pollution questions, link each pollutant to its source and effect, and give a balanced equation for its removal.
Gas-volume combustion questions are routine in Paper 1: water is a liquid at room temperature, alkali absorbs $\ce{CO2}$ , and volumes are proportional to moles.

Summary

Alkanes $\ce{C_{n}H_{2n+2}}$ : saturated, sp³ carbons, tetrahedral. Boiling point rises with chain length, falls with branching.
Unreactive because C–H and C–C bonds are strong and almost non-polar.
Complete combustion gives $\ce{CO2}$ and $\ce{H2O}$ ; incomplete gives CO or C.
Engine pollutants: CO (toxic), NO/ $\ce{NO2}$ (from $\ce{N2}$ and $\ce{O2}$ in air at high temperature; acid rain, smog), unburnt hydrocarbons (smog). Catalytic converters (Pt, Pd, Rh) convert them to $\ce{CO2}$ , $\ce{N2}$ and $\ce{H2O}$ .
Free-radical substitution with $\ce{Cl2}$ or $\ce{Br2}$ in UV light: initiation (homolytic fission), propagation (two steps, radical in and radical out), termination (two radicals combine).
Further substitution gives a mixture; excess alkane favours monosubstitution.
Alkanes are made by hydrogenation of alkenes ( $\ce{H2}$ , Ni catalyst, heat; or Pt) and by cracking (heat, $\ce{Al2O3}$ or zeolite catalyst), which turns long alkanes into shorter alkanes and alkenes.

Practice

Question

Give two reasons why alkanes do not react with aqueous acids, alkalis or oxidising agents.
Write balanced equations for (a) the complete combustion of propane, (b) the incomplete combustion of propane to carbon monoxide.
Complete the cracking equations: (a) $\ce{C10H22 -> C8H18 + ?}$ (b) $\ce{C16H34 -> C10H22 + 2 ?}$
Give the reagent, catalyst and product for the hydrogenation of but-2-ene.
Methane reacts with bromine in sunlight. Write equations for the initiation step and the two propagation steps, and name the type of fission in the initiation step.
Explain how nitrogen monoxide is formed in a car engine and write an equation for its removal, together with carbon monoxide, in a catalytic converter.
When methane reacts with chlorine, the product contains $\ce{CH3Cl}$ , $\ce{CH2Cl2}$ , $\ce{CHCl3}$ and $\ce{CCl4}$ . Explain why, and state how the proportion of chloromethane could be increased.
How many different monochloro products (structural isomers) are formed from (a) propane, (b) butane, (c) 2-methylpropane? Which one of all these products is chiral?
$20\ \text{cm}^3$ of a gaseous hydrocarbon is burnt completely in $200\ \text{cm}^3$ of oxygen. After cooling to room temperature, $160\ \text{cm}^3$ of gas remains; this falls to $100\ \text{cm}^3$ after passing through aqueous potassium hydroxide. Deduce the molecular formula of the hydrocarbon.
(a) Hexane reacts with chlorine in UV light. Write the two propagation steps leading to 2-chlorohexane. (b) Explain why the product mixture contains dodecane isomers, $\ce{C12H26}$ . (c) A 10.0 g sample of hexane gives 6.25 g of a mixture of monochlorohexanes ( $M_r$ 120.5). Calculate the percentage yield of monochlorohexanes. ( $M_r$ hexane 86.0)

Answers

The C–H and C–C bonds are strong, so a lot of energy is needed to break them; and the bonds are almost non-polar (C and H have similar electronegativities), so there are no $\delta+$ or $\delta-$ sites to attract polar reagents.
(a) $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$ (b) $\ce{2C3H8 + 7O2 -> 6CO + 8H2O}$
(a) $\ce{C2H4}$ (ethene). (b) $\ce{C3H6}$ (propene): carbon $16 - 10 = 6 = 2 \times 3$ ; hydrogen $34 - 22 = 12 = 2 \times 6$ .
Hydrogen gas, $\ce{H2}$ ; nickel catalyst with heat (or platinum at room temperature); product butane: $\ce{CH3CH=CHCH3 + H2 -> CH3CH2CH2CH3}$ .
Initiation: $\ce{Br2 -> 2Br^.}$ , homolytic fission. Propagation: $\ce{CH4 + Br^. -> CH3^. + HBr}$ ; $\ce{CH3^. + Br2 -> CH3Br + Br^.}$ .
In the engine the temperature is very high, so nitrogen and oxygen from the air react: $\ce{N2 + O2 -> 2NO}$ . In the catalytic converter (Pt/Pd/Rh): $\ce{2CO + 2NO -> 2CO2 + N2}$ .
Chloromethane also contains C–H bonds, so chlorine radicals can remove a hydrogen from it, and the resulting radical reacts with $\ce{Cl2}$ to give $\ce{CH2Cl2}$ ; the same process repeats to give $\ce{CHCl3}$ and $\ce{CCl4}$ . Using an excess of methane makes it much more likely that a chlorine radical collides with methane rather than with a chlorinated product, increasing the proportion of $\ce{CH3Cl}$ .
(a) Two: 1-chloropropane, 2-chloropropane. (b) Two: 1-chlorobutane, 2-chlorobutane. (c) Two: 1-chloro-2-methylpropane, 2-chloro-2-methylpropane. Only 2-chlorobutane is chiral (C2 carries H, Cl, $\ce{CH3}$ , $\ce{C2H5}$ ).
$\ce{CO2} = 160 - 100 = 60\ \text{cm}^3$ , so $x = 60 / 20 = 3$ . Unreacted $\ce{O2} = 100\ \text{cm}^3$ , so $\ce{O2}$ used $= 200 - 100 = 100\ \text{cm}^3$ , and $x + \frac{y}{4} = 100 / 20 = 5$ , giving $y = 8$ . The hydrocarbon is $\ce{C3H8}$ (propane).
(a) $\ce{CH3CH2CH2CH2CH2CH3 + Cl^. -> CH3CH^.CH2CH2CH2CH3 + HCl}$ (the radical is on C2); then $\ce{CH3CH^.CH2CH2CH2CH3 + Cl2 -> CH3CHClCH2CH2CH2CH3 + Cl^.}$ . (b) Two hexyl radicals can combine in a termination step, $\ce{C6H13^. + C6H13^. -> C12H26}$ ; because the radicals can have the unpaired electron on different carbons, several isomers of $\ce{C12H26}$ form. (c) $n(\text{hexane}) = 10.0 / 86.0 = 0.1163\ \text{mol}$ ; theoretical mass of $\ce{C6H13Cl} = 0.1163 \times 120.5 = 14.01\ \text{g}$ . Yield $= 6.25 / 14.01 \times 100 = 44.6\%$ .

Structure and physical properties

Why alkanes are unreactive

Combustion

Pollution from the internal combustion engine

Catalytic converters

Free-radical substitution

The mechanism

Further substitution: a mixture of products

Making alkanes

Hydrogenation of alkenes

Cracking

Worked examples

Practice

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