Advanced A/AS Level Organic Chemistry: Crude Oil, fractions and useful products

1.2 Sources of alkanes, physical properties, boiling points of alkanes and the fractional distillation of crude oil into useful products


Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry

Crude petroleum oil is a mixture of hydrocarbons, mainly alkanes, that can be separated by fractional distillation into a variety of useful products. The physical properties of alkanes are important pieces of data and note the abbreviations used: mpt = melting point  and  bpt = boiling point.

Sub-index for this page

(1) Sources of alkanes and the fractional distillation of crude oil

(2) The boiling point trend of alkanes and intermolecular forces

(3) The trend in enthalpy of vaporisation

(4) The solubility of alkanes


(1) Sources of alkanes and the fractional distillation of crude oil

The principal source of alkanes is crude oil and natural gas.

Natural gas is mainly methane (CH4), but crude oil contains a wide range of hydrocarbons and other compounds too.

The products from crude oil are separated by fractional distillation.

This is described on another page fractional distillation of crude oil, theory & uses of fractions.

(no repeats here for Advanced Level organic chemistry students)

In reality it is a fractional condensation process as each fraction isn't simply distilled over, but each fraction is tapped off at a particular condensation point within the negative temperature gradient up the column.

You should appreciate how the uses of the fractions is often related to its molecular size and hence the intermolecular forces (intermolecular bonding) e.g. boiling point and volatility, ease of combustion - flammability, viscosity ('stickiness'!).

See other notes for Isomerism of alkanes and the effect on boiling point

Other sources of alkanes

Another possible source of methane is called 'methane hydrate' also known as 'methane clathrate' and is believed to be the world's largest reserve of natural gas..

It consists of methane gas trapped in an ice cage of hydrogen bonded water molecules, which is why it is sometimes incorrectly called 'methane ice'.

Their formation of CH4.5.75H2O (*) is favoured by low temperatures and high pressure.

(* It is sometimes describes as a 'clathrate compound, but the formula is just a ratio of 1 : 5.75 CH4 : H2O in the crystal structure (or 4 : 23). It has a clearly defined regular crystal structure, but it is not a real compound, because there are no chemical bonds between the methane and water. All the molecules are held in place by intermolecular forces: transient dipole .... induced dipole for CH4...H2O interactions and hydrogen bonding for H2O...H2O interactions - hope you find this little diversion interesting! There are other structures known with a different ratio of components and with other trapped-caged gases too.)

Huge amounts of methane hydrate have been found under Arctic permafrost, beneath Antarctic ice, and also in sedimentary deposits along continental margins. Some are close to high-population areas than any natural gas field may allow countries that currently import natural gas to become self-sufficient. Though it is some task to find safe, economical ways to develop methane hydrate.

The methane gas has been formed over thousands of years from the action of anaerobic acting organisms that metabolise organic detritus material.


1. We are trying to reduce our dependency on fossil fuels, to minimise carbon dioxide emissions, to reduce the greenhouse effect and minimise global warming, so why produce even more.

2. Methane is a very powerful greenhouse gas and is beginning to leak out as the oceans and arctic regions are warming up - exploitation of this reserve cannot help matters e.g. accidental leaks.

(There is also concern about methane leaking from disused oil and gas wells e.g. Canada as 100, 000 wells that may present a 'greenhouse' hazard situation.)

(2) The boiling point trend of alkanes and intermolecular bonding forces

  • Alkanes are non–polar molecules where the only intermolecular force operating is the weakest possible, that is the instantaneous dipole – induced dipole intermolecular forces.
    • These are sometimes called London–dispersion forces and occur between ALL molecules, even single atoms of the noble gases.
    • Van der Waals forces include all types of intermolecular forces which are not due to an actual chemical bond BUT sometimes this name is used just to mean these instantaneous dipole – induced dipole dispersive forces and not permanent dipole - permanent dipole interactions (sorry but it can be confusing!).
    • The electronegativities are: C (2.5) and H (2.1) and produces a virtually non–polar bond and any very small effects will tend to cancel out e.g. H–C–H situations and so alkanes are the least polar organic molecules i.e. as near non–polar molecules you will get.
  • A transient δ+ in one alkane molecule induces a transient δ in a neighbouring alkane molecule, so causing a very weak and transient electrical attraction.
    • Note that these partial charges on the alkane molecule are shown as a delta + (δ+) or a delta – (δ) and they are tiny charges compared to a full single plus charge e.g. on an Na+ sodium ion or a full single minus charge  on a Cl chloride ion.
  • These electrical attractive forces act between ANY atoms or molecules and is primarily a function of the number of electrons in the molecule, though their spatial distribution can be significant.
  • The larger the alkane molecule, i.e. the greater the number of electrons in it, the more polarizable it is and the greater the chance of a random instantaneous dipole occurring to induce a dipole in a neighbouring molecule, so increasing the intermolecular attractive forces.
    • Hence the larger the alkane molecule the higher the boiling point (see diagram and graphs below).
  • The force arises from the instantaneous and random asymmetry of the electron fields in the atomic orbitals because of the random behaviour of electrons in the atomic or molecular orbitals and particle collisions and vibrations.
  • This polarisation can readily occur when particles collide with each other e.g. in liquids or vibrate against each other e.g. in a solid. In this situation electron clouds from neighbouring atoms/molecule will repel each other and the distortion of the charge distribution causes the polarization. Under these circumstances, contact between any two atoms/molecules can produce temporary or transient polarisation.
  • transient dipole - induced dipole attractions between alkane molecules
  • The graphs below, only consider the series of linear alkanes from CH4 to C14H30.

  • See other notes for Isomerism of alkanes and the effect on boiling point

Space filling diagrams to illustrate the different magnitudes of the intermolecular bonding forces between two alkanes of different molecular sizes (different number of carbon atoms, different numbers of electrons).

This gives rise to octane having a boiling point of 174oC and octadecane a boiling point of 317oC.


In the above diagram the alkane molecules have been drawn in a linear manner and they would be described as linear alkanes because there are no branches in the carbon chain. However, as illustrated in the 2nd diagram below, they are flexible from propane onwards and even the two methyl groups of ethane can freely rotate with respect to each other!

diagram of intermolecular bonding forces between linear alkanes doc brown's advanced A level organic chemistry notes


Graph 1 red line = alkanes

The red line graph shows the boiling point of alkanes from methane CH4 (boiling point -164oC/109 K)  to tetradecane C14H30 (boiling point 254oC/527 K). [Remember K = oC + 273]

Note: The red line represents linear alkanes in all the graphs 1-3.

A plot of number of electrons in an alkane molecule versus its boiling point (K) shows a steady rise with a gradually decreasing gradient.

Graph 1 is probably the best graph to look at the relative effects on intermolecular forces (intermolecular bonding) on boiling point because it is the distortion of the electron clouds (e.g. in non-polar alkanes, that gives rise to these, weak, but not insignificant, intermolecular forces.


Graph 2 red line = alkanes

A plot of the molecular mass of an alkane molecule versus its boiling point (K) shows a steady rise with a gradually decreasing gradient.


Graph 3 red line = alkanes

A plot of number of carbon atoms in an alkane molecule versus its boiling point (K) shows a steady rise with a gradually decreasing gradient.

graph of boiling point versus carbon number for linear alkanes doc brown's advanced A level organic chemistry notes

See other notes for Isomerism of alkanes and the effect on boiling point

(3) The trend in enthalpy of vaporisation

As well as melting points and boiling points, the enthalpy of vapourisation (ΔHvap/kJ/mol) and the enthalpy of fusion (melting) of organic compounds are also important pieces of data.

The values of these 'latent heats' are primarily determined by the generally weak (but not always!) intermolecular forces between the molecules, and, you must clearly distinguish this intermolecular bonding from the strong intramolecular bonding binding the atoms together in an organic molecule e.g. C-C, C-H, C=O, C-O etc.

As the length of the carbon chain increases for linear alkanes, the energy required for vaporisation steadily increases.

instantaneous dipole induced dipole intermolecular bonding forces between alkane molecules advanced organic chemistry revision notes doc brown

Expectedly, as the molecule increases in size, the instantaneous dipole – induced dipole intermolecular forces steadily increase too.

In other words the enthalpy of vaporisation (ΔHvap/kJ/mol) should steadily increase for every added -CH2- to the molecule.

Therefore to vaporise a molecule, the larger it is, the greater the volume of electron clouds to distort, the greater the kinetic energy required to overcome the intermolecular bonding forces of the instantaneous dipole – induced dipole attractions - see the graph below showing the expected trend and pictorially in the diagram above.

graph trend enthalpy of vaporisation for alkane molecules vapourisation vaporization advanced organic chemistry revision notes doc brown

The graph was obtained from the research paper https://www.umsl.edu/~chickosj/JSCPUBS/nC21C30.pdf

This begs the question, at what point does the enthalpy of vapourisation exceed the C-C bond enthalpy?

i.e. a point where the alkane molecule theoretically thermally decompose rather than vapourise on boiling?

The single (σ) C-C bond enthalpy is 346 kJ/mol.

The ΔHvap for C30H62 is ~152 kJ/mol, so, if we assume the linearity of the graph continues,

the ΔHvap for C60H122 is ~304 kJ/mol.

For each -CH2- the ΔHvap increases by ~5 kJ/mol, since 304 + (5 x 8) = 344 (~346)

Therefore for a ΔHvap of ~346 kJ/mol, the formula will be ~C65H132. and you might expect this to decompose at its boiling point, on a purely average bond enthalpy argument.

The highest boiling point I could find on the internet was for C43H88 bpt is 542oC.

In fact, these long-chain alkanes will probably start decomposing above 400oC in an inert atmosphere, because above this temperatures, hydrocarbon polymers like poly(ethene), thermally decompose giving off low molecular mass hydrocarbons.

This sort of data is important when considering the high temperature thermal degradation and combustion (incineration) of large hydrocarbon alkane molecules and poly(alkene) polymers e.g. trying to recycle the original monomer.

See also The thermal stability of poly(alkene) plastics

The scenario described above suggests that it might be impossible to distil high boiling compounds without simultaneous thermal decomposition at atmospheric pressure.

Not so, there are two ways of distilling high boiling compounds to avoid thermal decomposition.

(1) Distilling a homogeneous mixture under reduced pressure

The vapour pressure exerted by a liquid rises, exponentially, with increase in temperature.

A liquid boils when the vapour pressure it exerts, equals the ambient pressure e.g. water boils at 100oC at the 'normal' atmospheric pressure of 1 atm/101 kPa.

The vapour pressure of water at 25oC is ~0.03 atm, therefore if you pump out most of the air above water, it will boil at 25oC, room temperature!

Therefore, whenever you reduce the pressure above a liquid, it will boil at a lower temperature and this can be low enough to avoid thermal decomposition.

For more details see Vapour pressure and boiling point of a liquid

(2) Steam distillation

The substance required must not be soluble in water.

You pass steam through the mixture containing the substance you want to extract.

Even if the substance's vapour pressure is small at ~100oC, small amounts of it will distil over with the steam and collect in the condensed distillate with the water.

Therefore the initial separation/extraction occurs at a much lower temperature than the high boiling point

However, you do have to separate the desired product from the condensed steam distillate.

For more details see Steam distillation – theory, practice and uses

(4) Solubility of alkanes

Alkanes, like all hydrocarbons, are virtually insoluble in water.

The non-polar alkane molecules cannot hydrogen bond with water.

Neither are the weak hydrocarbon - water interactions strong enough to disrupt the strong hydrogen bonding between water molecules.

Alkanes, gases, liquids or solids, will dissolve in non-polar solvents like hexane or tetrachloromethane, where the solute-solute, solute-solvent and solvent-solvent intermolecular forces are of a similar magnitude and the dissolving process does not involve breaking strong intermolecular boding..

Doc Brown's Advanced Level Chemistry Revision Notes

Alkanes and Petrochemical Industry INDEX

 All Advanced Organic Chemistry Notes

 Index of GCSE/IGCSE Oil - Useful Products Chemistry Revision Notes

A basic introduction to the chemistry of alkanes



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