4.1.2 - Alkanes

Properties of Alkanes

An alkane is arguably of the most basic homologous series in all of organic chemistry. They are saturated hydrocarbons, which means they contain only $\sigma$-bonds between only carbon and hydrogen atoms. Since $\sigma$-bonds have free rotation, alkanes do not show geometric stereoisomerism, unlike alkenes (see 4.1.3 - Alkenes).

The carbon atoms in an alkane do not have double bonds, and so are surrounded by four bonding electron pairs. Using knowledge of the electron-pair repulsion theory, this means each carbon atom is at the centre of a tetrahedral structure, with bond angle 109.5$^\circ$.

As for the bonding between alkanes, since they do not contain disparately electronegative atoms, polar bonds do not exist. This means that permanent dipole-permanent dipole and permanent dipole-induced dipole interactions - as well as hydrogen bonds - do not occur. In fact, the only type of intermolecular interaction between alkane molecules in London Dispersion Forces (LDFs). This means that the melting and boiling points of alkanes are relatively low compared with other homologous series.

That being said, the range of melting and boiling point within the alkane series shows a trend, increasing with carbon-chain length. This is because as any molecule gets longer, there is more opportunity for instantaneous dipoles to form, and so more LDFs form. On top of this, the surface area of the molecule increases, and so there are more surface contact-points for LDFs to form between. This also applies to a more branched molecule, however a branched six-carbon alkane has less surface area than a straight-chained six-carbon alkane.

Reactions of Alkanes

As mentioned above, there is essentially no polarity on an alkane which, coupled with the C$-$C and C$-$H bonds having high bond enthalpies, means that alkanes are not very reactive. In order for an alkane to react, strong radicals or intense heat must be involved.


As with many organic compounds, alkanes can react with either sufficient or insufficient oxygen gas in a (respectively) complete or incomplete combustion reaction. The equations for the complete and incomplete combustions of butane are shown below:

$$ \text{C$_4$H$_10$} _\text{(l)} + 6 \text{$\frac{1}{2}$O$_2$} _\text{(g)} \rightarrow \text{4CO$_2$} _\text{(g)} + \text{5H$_2$O} _\text{(g)} $$

$$ \text{C$_4$H$_10$} _\text{(l)} + 4 \text{$\frac{1}{2}$O$_2$} _\text{(g)} \rightarrow \text{4CO} _\text{(g)} + \text{5H$_2$O} _\text{(g)} $$

You will notice that a product of incomplete combustion is carbon monoxide. This is an odourless, colourless and - most importantly - toxic gas. It is important that a fire has sufficient ventilation to avoid carbon monoxide poisoning.


As well as requiring heat to react with oxygen, alkanes require can react with halogen radicals. This is known as a free radical substitution, as one of the hydrogen atoms is replaced with a halogens.

The mechanism has three steps, the first of which involves the formation of two halogen radicals by homolytic fission of a diatomic molecule, under the influence of ultraviolet radiation. This is called initiation.

For this step, the number of radicals increases to two.

$$ \text{Cl$_2$} \xrightarrow{\text{UV light}} \text{2Cl$\large^\centerdot$} $$

The second step of a free radical substitution is the propagation. In this step, one of the radicals attacks the alkane, ‘stealing’ a hydrogen to form hydrochloric acid. The alkane becomes a radical.

The alkane radical then attacks a ‘non-radicalised’ chlorine molecule and ‘steals’ a chlorine to form a halogenoalkane or haloalkane. The other chlorine becomes a radical.

For this step, the number of radicals is maintained at two.

$$ \begin{aligned} \text{CH$_3$CH$_2$CH$_3$} + \text{Cl$\large^\centerdot$} &\rightarrow \text{CH$_3$CH$_2$C$\large^\centerdot$H$_2$} +\text{HCl} \\ \text{CH$_3$CH$_2$C$\large^\centerdot$H$_2$} + \text{Cl$_2$} &\rightarrow \text{CH$_3$CH$_2$CH$_2$Cl} + \text{Cl$\large^\centerdot$} \end{aligned} $$

The final step of a free radical substitution is the termination. In this step radicals come together to form a covalent bond between them. Since the propagation step produces two different radicals, three different products can be formed in termination.

If two chlorine radicals meet, chlorine gas is formed.

If two alkane radicals meet, they join to form a longer alkane.

If a chlorine radical meets an alkane radical, a halogenoalkane is formed.

In this step, the number of radicals decreases to zero.

$$ \begin{aligned} \text{Cl$\large^\centerdot$} + \text{Cl$\large^\centerdot$} &\rightarrow \text{Cl$_2$} \\ \text{CH$_3$CH$_2$C$\large^\centerdot$H$_2$} + \text{CH$_3$CH$_2$C$\large^\centerdot$H$_2$} &\rightarrow \text{CH$_3$CH$_2$CH$_2$CH$_2$CH$_2$CH$_3$} \\ \text{CH$_3$CH$_2$C$\large^\centerdot$H$_2$} + \text{Cl$\large^\centerdot$} &\rightarrow \text{CH$_3$CH$_2$CH$_2$Cl} \end{aligned} $$


The reaction mechanism shown above is actually a very inefficient way to synthesise halogenoalkanes. Just on a surface level, the atom economy is not 100%, but less because of the hydrochloric acid produced as waste.

And even looking deeper into the mechanism, the desired product is only a third of the possible products of the termination step. Therefore a mixture of organic and inorganic products is formed, which then has to be separated.

Still more inefficiency lies in the randomness of which a position of the carbon chain the halogen radicals will attack, meaning that a mixture of positional isomers will be created. Finally, more than one radical may attack, causing a polysubstitution where a only a monosubstitution is desired or vice versa.

To conclude, using free radical substitution to artificially synthesise haloalkanes is very inefficient, and other methods are preferred. This reaction mechanism will likely only occur naturally, where UV light from the sun initiates the reaction.