A-level Chemistry/OCR/Chains and Rings/Alkanes

The Alkanes

Alkanes are saturated hydrocarbons. This means that they contain only carbon and hydrogen atoms and they contain no 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 the carbon can form another C-C bond so they have the general formula of $C_n{H}_{2n}$

Physical Properties

The alkanes are the simplest 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 $C{O}_2+H_{2}O$

It will under go substitution reactions (Say with $C{l}_2$) to create $C_2{H}_{5}Cl+HCl$

It will also under go 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 oxygens & thus are a major fuel.

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, this catalyst is either $Si{O}_2$ or $A{l}_2{O}_3$. These reactions also need high temperatures (773k or 500c is usually used).

The products from cracking can vary, for instance, 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: $C_8{H}_{18}\to 4C_2{H}_4+H_2$

Under conditions of 773k and with an $A{l}_2{O}_3$ catalyst present.

Reforming

Reforming reactions are reactions where one turns a straight chain alkane 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 aluminium oxide 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. But, 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:

$C{l}_2\to C{l}^{\bullet }+C{l}^{\bullet }$

Propagation:

$C{l}^{\bullet }+C{H}_4\to HCl+C{H}_3^{\bullet }$

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:

$C{H}_3^{\bullet }+C{l}_2\to C{H}_{3}Cl+C{l}^{\bullet }$

Or

$C{H}_3^{\bullet }+C{H}_{3}Cl\to C_2{H}_6+C{l}^{\bullet }$

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:

$C{H}_3^{\bullet }+C{l}^{\bullet }\to C{H}_{3}Cl$

$C{l}^{\bullet }+C{l}^{\bullet }\to C{l}_2$

$C{H}_3^{\bullet }+C{H}_3^{\bullet }\to C_2{H}_6$

Henceforth, the reaction ends.

Problems

As we know from above, alkanes burn readily in oxygen & give off carbon dioxide, which is a weak greenhouse gas and thus a negligible contributor to global warming. But this is not the only problem. Alkanes must burn in excess oxygen or they combust incompletely, depending on the amount of oxygen produced, they can combust incompletely to create a mixture of carbon dioxide & carbon monoxide, or, they can combust incompletely to create only carbon monoxide or, if there's very little oxygen around, they combust to make carbon (soot you see in yellow flames).

Carbon monoxide is particularly dangerous as it replaces the oxygen in haemoglobin which can cause people to suffocate.(Due to the fact that carbon monoxide and haemoglobin have 240 times more affinity than oxygen and haemoglobin)