Organic chemistry · Lesson 2

The forces between molecules

Boiling separates molecules without breaking a single covalent bond. Dispersion forces, dipole–dipole forces and hydrogen bonds explain boiling points and solubility across every organic family.

Natural gas, bottled camping gas, petrol, diesel and candle wax are all mixtures of alkanes. They contain the same two elements, joined by the same kinds of bonds. The only real difference between them is the length of the molecules: one carbon in natural gas, three or four in camping gas, around eight in petrol, sixteen or so in diesel, and twenty or more in wax.

That single difference turns a gas into a liquid and a liquid into a solid. And add one oxygen and one hydrogen to a small molecule and the change is even more dramatic: propane (C₃H₈) boils at −42 °C, but propan-1-ol (C₃H₈O) boils at 97 °C.

Boiling doesn’t break any covalent bonds. It only pulls molecules apart from each other. This lesson is about the forces between molecules, and how they explain boiling points, melting points and solubility.

Three kinds of intermolecular force

Dispersion forces act between all molecules. Electrons are always moving, so at any instant a molecule has a temporary dipole, which induces dipoles in its neighbours. Dispersion forces are stronger for molecules with more electrons (larger molecules) and more surface contact (straight chains rather than compact, branched ones).
Dipole–dipole forces act between polar molecules, which have a permanent dipole because of a polar bond such as C=O or C–Cl in an asymmetric molecule. The δ+ end of one molecule attracts the δ− end of the next.
Hydrogen bonds form when a hydrogen atom bonded to N, O or F is attracted to a lone pair on an N, O or F atom of a neighbouring molecule. They are the strongest of the three. Alcohols, carboxylic acids, amines and amides hydrogen bond; aldehydes, ketones and esters cannot hydrogen bond to each other (no O–H or N–H), although they can accept hydrogen bonds from water.

Boiling points

To compare boiling points, identify the strongest intermolecular force in each substance, then compare molecule size and shape. Comparing molecules of similar size isolates the effect of the functional group:

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Experiment: which liquid cools faster?

Choose two liquids, dip the probes, then lift them out. The faster a liquid evaporates, the more it cools the probe, and the weaker its intermolecular forces.

Within a homologous series, boiling point rises steadily with chain length: more electrons and more surface area mean stronger dispersion forces.
Branching lowers the boiling point: 2,2-dimethylpropane (10 °C) against pentane (36 °C).
At similar size: alkanes (dispersion only) < aldehydes, ketones, esters (dipole–dipole) < alcohols (hydrogen bonds) < carboxylic acids (two hydrogen bonds per pair of molecules).

Boiling points of three homologous series. The gap between alkanes and alcohols is hydrogen bonding; it matters less as the chains get longer and dispersion forces dominate.

Solubility in water

“Like dissolves like.” A substance dissolves in water if it can form attractions with water molecules comparable to those it breaks. Small alcohols, carboxylic acids, amines and amides hydrogen bond with water, so methanol, ethanol, propan-1-ol and ethanoic acid mix with water in all proportions. As the hydrocarbon chain gets longer, more of the molecule is non-polar and cannot interact strongly with water: butan-1-ol dissolves only moderately, and hexan-1-ol barely. Alkanes do not dissolve in water at all, but dissolve in non-polar solvents such as hexane.

Comparing boiling points for full marks

Name the intermolecular forces present in each substance, including dispersion forces.
Identify the strongest force and explain why it is stronger (for example, hydrogen bonding because of the O–H group).
Conclude: stronger intermolecular forces need more energy to overcome, so a higher boiling point. Never say covalent bonds break on boiling.

Common traps

Boiling separates molecules; it does not break C–H or C–C bonds.
Dispersion forces are present in polar molecules too, and in large molecules they can outweigh a small dipole.
Hydrogen bonding needs H bonded directly to N, O or F. The H atoms in CH₃ groups do not hydrogen bond.

Now put it to workFive worked examples, twenty practice questions and ten harder problems on this lesson.

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