A-level Chemistry/OCR/Chains, Energy and Resources/Basic Concepts and Hydrocarbons/Alkanes

The Alkanes

Alkanes are saturated hydrocarbons. This means that they contain only carbon and hydrogen atoms and they contain no carbon-carbon double bonds.

The alkane homologous series has the general formula of $C_{n}H_{2n+2}$ except for the cycloalkanes in which 2 hydrogens are lost so that the carbon can form another C-C bond; they have the general formula of $C_{n}H_{2n}$

Physical Properties

The alkanes are the most complex of the organic molecules. There are a few things you will need to know about their physical properties in an exam:

Boiling/Melting Point

As the relative molecular mass, or number of carbon atoms in an alkane increases, so does its boiling or melting point:

Alkane	Molecular Mass	Molecular Formula	Boiling Point (K)
Methane	16	CH₄	109
Ethane	30	C₂H₆	185
Propane	44	C₃H₈	231
Butane	58	C₄H₁₀	273
Pentane	72	C₅H₁₂	309

As you should be aware, alkanes are held together by Van der Waal's forces. The larger a molecule is, the more electrons it has. This means it can form larger dipoles and its Van der Waal's forces will be larger. It will therefore take more energy to break the bonds and so the boiling and melting point will be higher.

As you should know, however, some alkanes can be branched instead of straight chained. Boiling point decreases as a molecule becomes more branched. This is because branched molecules cannot pack so tightly together, so their Van der Waal's forces must act over larger distances (intensity of the Van der Waal's forces decreases) and thus require less energy to break. For example, hexane has five isomers:

Isomer	Structural Formula	Boiling Point (K)
Hexane	CH₃CH₂CH₂CH₂CH₂CH₃	342
3-Methylpentane	CH₃CH₂CH(CH₃)CH₂CH₃	337
2-Methylpentane	CH₃CH(CH₃)CH₂CH₂CH₃	333
2,3-Dimethylbutane	CH₃CH(CH₃)CH(CH₃)CH₃	331
2,2-Dimethylbutane	CH₃CH(CH₃)(CH₃)CH₂CH₃	323

Reactions

The alkanes are fairly unreactive, as the C-H bond is non-polar. However they undergo three reactions.

Reactions of alkanes

Take the alkane $C_{2}H_{6}$ .

It will under go combustion to form $CO_{2}+H_{2}O$

It will under go substitution reactions (Say with $Cl_{2}$ ) to create $C_{2}H_{5}Cl+HCl$

It will also undergo cracking reactions to create an alkene, in this case, $C_{2}H_{4}+H_{2}$

The most important of these being combustion, alkanes are very volatile & burn easily when in the presence of plentiful amounts of oxygen & thus are major fuels.

Cracking

As said earlier, long chain molecules have higher boiling points and are more difficult to ignite. These long chain molecules can be broken into shorter chain (more useful) molecules through catalytic cracking. Any alkane from $C_{4}$ to $C_{50}$ can be cracked. As it is catalytic cracking it requires a catalyst, previously this catalyst was either $SiO_{2}$ or $Al_{2}O_{3}$ , but now zeolite is preferred. These reactions also need high temperatures (773K or 450 °C is usually used).

The products from cracking can vary. For example you can create several moles of one alkene and hydrogen from cracking an alkane if done right. However you can also create a mixture of alkenes & alkanes or just a mixture of alkenes.

Example: $C_{8}H_{18}\rightarrow \;4C_{2}H_{4}+H_{2}$

Under conditions of 773k and with an $Al_{2}O_{3}$ catalyst present.

Reforming

Reforming reactions are reactions where a straight chain alkane is converted into a branched alkane or cycloalkane to increase the octane number in petrol. (Octane numbers represent how much pressure a fuel can be put under before it combusts & causes knocking which damages engines).

Again, reforming uses an platinum or rhodium catalyst & a high temperature, but this time it also uses a higher pressure (40 atm or so). The by-product of reforming is hydrogen molecules.

Free radical substitution

Alkanes are very unreactive due to the non-polar C-H bond, so nucleophilic & electrophilic attack reactions are not possible. However, free radical substitutions are possible due to their reactivity. Free radical substitution involves a halogen & some UV light to initiate homolytic fission.

An example with methane.

Initiation reaction:

$Cl_{2}\rightarrow \;Cl^{\bullet }+Cl^{\bullet }$

Propagation:

$Cl^{\bullet }+CH_{4}\rightarrow HCl+CH_{3}^{\bullet }$

This sets into motion the chain reaction, where one free radical reacts with a stable species to form another stable species & a free radical.

The next propagation steps:

$CH_{3}^{\bullet }+Cl_{2}\rightarrow CH_{3}Cl+Cl^{\bullet }$

Or

$CH_{3}^{\bullet }+CH_{3}Cl\rightarrow C_{2}H_{6}+Cl^{\bullet }$

This goes on until 2 radicals meet & combine to form a stable species & thus once all the radicals have done this, the reaction terminates, hence why it is the termination step.

In this reaction we have 3 possible termination steps:

$CH_{3}^{\bullet }+Cl^{\bullet }\rightarrow CH_{3}Cl$

$Cl^{\bullet }+Cl^{\bullet }\rightarrow Cl_{2}$

$CH_{3}^{\bullet }+CH_{3}^{\bullet }\rightarrow C_{2}H_{6}$

Henceforth, the reaction ends.

Note that a free-radical hydrogen is never produced.

Problems

As we know from above, alkanes burn readily in oxygen & give off carbon dioxide, which is a weak greenhouse gas and thus a contributor to global warming.

Alkanes must burn in excess oxygen for complete combustion and the production of carbon dioxide:

CH₄ + 2O₂ -> CO₂ + 2H₂O

Incomplete combustion comes from a reduced supply of oxygen and produces water and carbon monoxide:

CH₄ + 1½O₂ -> CO + 2H₂O

Carbon monoxide is poisonous when inhaled. It binds irreversibly to haemoglobin in red blood cells and makes carboxyhaemoglobin. The red blood cells can no longer carry oxygen and this can cause suffocation if enough cells are carrying carboxyhaemoglobin.

UK regulations ensure that all gas equipment must be annually serviced and that all gas installations require adequate ventilation.