How do intermolecular forces explain the boiling points and solubilities of different organic families?
Relate the physical properties of organic compounds to their intermolecular forces and functional groups
A focused answer to the WACE Year 12 Chemistry dot point on the physical properties of organic compounds, explaining boiling points and water solubility in terms of dispersion forces, dipole-dipole forces and hydrogen bonding across the organic families, with a worked example and common exam mistakes.
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What this dot point is asking
The physical properties of organic compounds, especially boiling point and water solubility, are governed by the intermolecular forces between molecules. The stronger these forces, the more energy is needed to separate molecules, so the higher the boiling point.
The three intermolecular forces
- Dispersion forces exist between all molecules. They arise from temporary fluctuating dipoles and strengthen as molecular size and surface area increase.
- Dipole-dipole forces occur between polar molecules (those with a permanent dipole, such as ketones and haloalkanes) and are stronger than dispersion forces for molecules of similar size.
- Hydrogen bonding is the strongest, occurring when hydrogen is bonded to nitrogen, oxygen or fluorine. Alcohols, carboxylic acids and amines can hydrogen bond.
Boiling point trends across families
For molecules of comparable carbon number, the boiling point order reflects the strongest force present. Carboxylic acids and alcohols (hydrogen bonding) boil highest; aldehydes and ketones (dipole-dipole) are intermediate; alkanes (dispersion only) boil lowest. Carboxylic acids boil even higher than alcohols of similar size because they can form two hydrogen bonds, dimerising in the pure liquid.
Water solubility
A molecule is water-soluble if it can form favourable interactions (especially hydrogen bonds) with water. Small alcohols, carboxylic acids and amines are very soluble because their polar group hydrogen bonds with water. As the non-polar carbon chain grows, it dominates and solubility falls: methanol and ethanol mix freely with water, but longer-chain alcohols become increasingly insoluble. Alkanes are essentially insoluble because they are non-polar.
Carboxylic acid dimers and the volatility ladder
A detail that earns top marks is the dimerisation of carboxylic acids. Each group has both a hydrogen-bond donor (the ) and an acceptor (the C=O oxygen), so two molecules pair head-to-head through two hydrogen bonds. The effective particle is therefore roughly double the size and doubly bonded, which is why carboxylic acids boil markedly higher than alcohols of similar molar mass. Reading across a fixed carbon number, the volatility ladder runs from alkanes (most volatile, dispersion only), up through haloalkanes and aldehydes and ketones (dipole-dipole), then alcohols and amines (single hydrogen bonds), to carboxylic acids (double hydrogen bonds, least volatile). Being able to place an unknown family on this ladder, and justify it by the strongest force present, is the recurring physical-properties skill in Unit 4.
Why this matters
These ideas explain why alcohols are good solvents, why carboxylic acids are liquids while similar alkanes are gases, and why long-chain molecules are greasy and water-repellent. They also link to separation and purification techniques (such as distillation) used in synthesis, and underpin the property questions that recur in the examination.
Exam-style practice questions
Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
WACE 20226 marksThe boiling points of three compounds are: propane , ethanal , ethanol , which have similar molar masses. (a) Identify the strongest intermolecular force in each. (b) Explain the order of boiling points. (c) Explain why ethanoic acid (molar mass 60) boils at , higher than ethanol despite the similar size.Show worked answer →
A 6 mark question rewards the force identification, the trend explanation, and the dimer reasoning.
(a) Propane: dispersion forces only (non-polar). Ethanal: dipole-dipole (polar C=O) plus dispersion. Ethanol: hydrogen bonding (via ) plus dispersion.
(b) All have similar molar mass, so the order reflects the strongest force present. Propane has only weak dispersion forces, so it boils lowest. Ethanal adds dipole-dipole attraction, raising its boiling point above propane. Ethanol can form hydrogen bonds, the strongest of the three forces, so most energy is needed to separate its molecules and it boils highest.
(c) Ethanoic acid molecules form two hydrogen bonds each, pairing up as dimers in the liquid. Effectively the particles that must be separated are twice the size and held by double hydrogen bonding, so even more energy is needed than for ethanol, giving the higher boiling point.
Markers reward the three forces, the dispersion < dipole-dipole < hydrogen-bonding order, and the carboxylic-acid dimer explanation.
WACE 20205 marksExplain the trend in water solubility of the alcohols from methanol to octan-1-ol, and explain why ethanal is more soluble in water than butane.Show worked answer →
A 5 mark answer needs the chain-length trend and the polarity comparison.
Small alcohols (methanol, ethanol, propan-1-ol) are fully miscible with water because their group forms hydrogen bonds with water molecules, and the short carbon chain is a small part of the molecule. As the chain lengthens towards octan-1-ol, the large non-polar hydrocarbon part dominates: it cannot hydrogen bond with water and disrupts water's own hydrogen bonding, so solubility falls steadily and the longer alcohols are essentially insoluble.
Ethanal is more soluble in water than butane because ethanal is polar (its C=O group) and can form dipole interactions and hydrogen bonds (accepting through the oxygen) with water, whereas butane is non-polar and can only interact with water by weak dispersion forces, which cannot compete with water's hydrogen bonding.
Markers reward the hydrogen-bonding-then-chain-dominates trend for the alcohols, and the polar-versus-non-polar reasoning for ethanal versus butane.
