The physical properties of every fuel, lubricant, and wax on Earth are controlled by the same simple idea: the strength of the intermolecular forces between hydrocarbon molecules, which depends directly on chain length, molecular shape, and available surface area.
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Methane boils at −162°C and escapes as a gas the moment it leaves a pipe. Octane boils at 126°C and remains a liquid in a petrol tank on a hot day. Paraffin wax, made of long-chain alkanes, is solid at room temperature and melts only when heated over a flame.
Before reading on, explain why increasing carbon chain length causes a large increase in boiling point. What changes between the molecules, and why does that mean more energy is needed to separate them?
📚 Core Content
Alkanes are the structural baseline for the entire module. Once you understand the geometry and physical properties of alkanes, every comparison involving alkenes, alkynes, branching, and functional groups becomes much easier to explain.
Alkanes contain only C–C and C–H single bonds, so they are described as saturated hydrocarbons. Every carbon in an alkane is sp3-hybridised, with tetrahedral geometry and bond angles of 109.5°. In three dimensions, a straight-chain alkane is not actually flat — it adopts a low-energy zigzag arrangement.
The first eight straight-chain alkanes are foundational reference points for physical-property questions:
| Name | Formula | State at 25°C | Boiling point (°C) |
|---|---|---|---|
| Methane | CH4 | Gas | −162 |
| Ethane | C2H6 | Gas | −89 |
| Propane | C3H8 | Gas | −42 |
| Butane | C4H10 | Gas | −1 |
| Pentane | C5H12 | Liquid | 36 |
| Hexane | C6H14 | Liquid | 69 |
| Heptane | C7H16 | Liquid | 98 |
| Octane | C8H18 | Liquid | 126 |
A strong answer does not stop at “larger molecules have higher boiling points.” It explains why larger hydrocarbon molecules create stronger dispersion forces at the electron-cloud level and why that means more energy is needed to separate them.
London dispersion forces arise because electron clouds fluctuate momentarily, producing instantaneous dipoles that induce dipoles in nearby molecules. Longer hydrocarbon chains have more electrons, more polarisable electron clouds, and larger surface area for contact between neighbouring molecules. That means more simultaneous temporary attractions between adjacent molecules.
As chain length increases, the total dispersion force between molecules increases. Because boiling requires molecules to separate into the gas phase, stronger intermolecular attraction means more energy is required, so the boiling point rises.
1. State the IMF present: both are non-polar alkanes, so the only intermolecular forces are London dispersion forces.
2. Identify the structural difference: hexane has a longer carbon chain and larger molecular surface area than pentane.
3. Link to force strength: greater surface area allows more simultaneous instantaneous dipole-induced dipole interactions.
4. Conclude with boiling point: stronger dispersion forces require more energy to overcome, so hexane has the higher boiling point.
Two molecules can have the same formula and the same molecular mass, yet different boiling points. The deciding factor is often shape: straight chains present more surface area for intermolecular contact than compact branched structures.
Branching makes a hydrocarbon more compact and less able to lie alongside neighbouring molecules across a long surface. Even when two molecules have the same number of atoms and identical molecular formula, the branched isomer typically has the lower boiling point because fewer simultaneous dispersion interactions can occur.
Hydrocarbon series differ not only in formula but also in local geometry. The bond type present at carbon directly determines hybridisation, bond angle, and the three-dimensional shape around that carbon.
| Series | Bond type | Key carbon hybridisation | Geometry | Bond angle |
|---|---|---|---|---|
| Alkane | C–C single bond | sp3 | Tetrahedral | 109.5° |
| Alkene | C=C double bond | sp2 at C=C | Trigonal planar | ~120° |
| Alkyne | C≡C triple bond | sp at C≡C | Linear | 180° |
When comparing compounds with the same carbon count, their boiling points are often quite similar because all remain largely non-polar and still rely mainly on dispersion forces. The major trend remains chain length. Differences between alkane, alkene, and alkyne of equal carbon number are usually smaller than the effect of adding several carbons to the chain.
“Like dissolves like” is only the shortcut. Full-mark answers explain solubility by naming the intermolecular forces broken and the new ones formed — and whether the energy trade-off is favourable.
Hydrocarbons are non-polar and experience only London dispersion forces. Water is strongly polar and held together by a large hydrogen-bond network. For a hydrocarbon to dissolve in water, water-water hydrogen bonds would need to be disrupted, but the hydrocarbon cannot replace them with interactions of similar strength. As a result, dissolution is not energetically favourable and the hydrocarbon remains insoluble.
Hydrocarbons dissolve readily in other non-polar solvents such as hexane or heptane because only dispersion forces are involved on both sides. Disrupting hydrocarbon-hydrocarbon interactions and replacing them with similar hydrocarbon-solvent interactions carries little energetic penalty.
✍️ Activities
Use the patterns from the lesson to predict physical properties before checking the answers. Focus on naming the IMF first, then linking structure to force strength.
“Non-polar means no intermolecular forces.” Wrong. Non-polar hydrocarbons still have London dispersion forces.
“Boiling point depends on bond strength inside the molecule.” Wrong. Boiling separates molecules from each other, so the key factor is intermolecular force strength.
“Like dissolves like” is enough by itself. Wrong. Use it as a rule, then explain the IMF logic underneath it.
🧠 Worked Examples
All four are straight-chain alkanes, so the only intermolecular forces present are London dispersion forces.
Boiling point increases with chain length because longer chains have larger electron clouds and greater surface area for contact.
Order by chain length: methane (C1) < propane (C3) < butane (C4) < heptane (C7).
Answer: methane < propane < butane < heptane. Longer carbon chains experience stronger dispersion forces, so more energy is required to separate molecules into the gas phase.
The two molecules have the same molecular formula and molecular mass, so mass alone cannot explain the difference.
Pentane is elongated, while 2,2-dimethylpropane is compact and nearly spherical.
The longer straight-chain shape allows more surface contact between neighbouring molecules, producing stronger dispersion forces.
Answer: Pentane has the higher boiling point because its straight-chain structure provides greater surface area for intermolecular contact, leading to stronger dispersion forces than the compact branched isomer.
Hexane is non-polar and has only dispersion forces.
Water has a strong hydrogen-bond network. Hexane cannot form interactions strong enough to replace those H-bonds.
Heptane is also a non-polar hydrocarbon, so hexane-heptane interactions are of the same type and similar strength.
Answer: Hexane is insoluble in water but miscible with heptane. In water, hydrogen bonds would be disrupted without adequate replacement; in heptane, similar dispersion interactions form readily.
🧠 Check Your Understanding
1. Which statement best explains why butane has a higher boiling point than propane?
2. An alkyne with formula C4H6 is compared with an alkane C4H10. Which statement about water solubility is correct?
3. Which row correctly matches bond type, hybridisation, and geometry?
4. Three structural isomers all have formula C5H12. Which property would you expect to be lowest for the most branched isomer?
5. Which statement about the solubility of hydrocarbons is correct?
✍️ Short Answer
4. Explain why octane has a much higher boiling point than methane even though both are non-polar hydrocarbons. 3 MARKS
5. Compare pentane and 2,2-dimethylpropane. State which has the higher boiling point and explain why the difference exists even though both have molecular formula C5H12. 4 MARKS
6. A student compares hexane, hex-1-ene, hex-1-yne, and decane. Predict which compound has the highest boiling point and justify your answer using intermolecular-force reasoning. Then predict the solubility of hexane in water and heptane and explain both. 5 MARKS
1. Increasing boiling point: methane < pentane < octane. All are non-polar alkanes with dispersion forces only. Chain length increases from C1 to C5 to C8, so electron cloud size, surface area, and dispersion-force strength all increase.
2. Pentane has the higher boiling point because its straight-chain structure gives greater surface area for intermolecular contact. 2,2-Dimethylpropane is compact and branched, so fewer simultaneous dispersion interactions can occur.
3. In ethene, both carbons in the C=C are sp2-hybridised, trigonal planar, with bond angles of about 120°. In ethyne, both carbons in the C≡C are sp-hybridised, linear, with bond angles of 180°.
4. Hexane is insoluble in water because water’s hydrogen-bond network would be disrupted without enough compensating interaction. In ethanol, solubility is limited/partial rather than completely absent because ethanol contains both a polar OH group and a non-polar carbon chain, so it can interact somewhat with hydrocarbons.
1. B — Both molecules are non-polar alkanes, so the only intermolecular forces are London dispersion forces. Butane has greater chain length and surface area than propane, so dispersion forces are stronger and more energy is needed to boil it.
2. B — Both molecules are non-polar hydrocarbons and cannot form hydrogen bonds with water or provide interactions strong enough to compensate for breaking water’s H-bond network.
3. D — Carbon in a triple bond is sp-hybridised and linear, giving a bond angle of 180°.
4. C — All C5H12 isomers share the same molecular formula, molecular mass, and number of C–H bonds. The only property that differs is boiling point, which decreases as branching increases because the compact shape reduces surface area available for dispersion-force contact.
5. A — "Like dissolves like" at the IMF level: a non-polar hydrocarbon in a non-polar solvent replaces dispersion forces with similar dispersion forces, making dissolution energetically favourable. Water is incorrect because its strong H-bond network cannot be adequately replaced by hydrocarbon–water dispersion interactions.
Q4 (3 marks): Methane and octane are both non-polar hydrocarbons, so the only intermolecular forces present are London dispersion forces [1]. Octane has a much longer carbon chain, more electrons, and greater surface area than methane [1]. This produces stronger dispersion forces between octane molecules, so more energy is required to separate them into the gas phase, giving octane a much higher boiling point [1].
Q5 (4 marks): Pentane has the higher boiling point [1]. The two compounds have the same molecular formula and molecular mass, so the difference is not due to mass alone [1]. Pentane is a straight-chain molecule, while 2,2-dimethylpropane is much more compact and branched [1]. The straight-chain shape gives pentane greater surface area for intermolecular contact, so it experiences stronger dispersion forces and therefore a higher boiling point [1].
Q6 (5 marks): Decane has the highest boiling point [1] because it has the longest chain, largest surface area, and strongest dispersion forces of the four compounds [1]. Hexane, hex-1-ene, and hex-1-yne are all C6 hydrocarbons and remain largely non-polar, so they all rely mainly on dispersion forces; their boiling points are relatively similar compared with the much larger jump to decane [1]. Hexane is insoluble in water because water’s hydrogen-bond network would need to be disrupted, and hexane cannot replace those interactions with equally strong ones [1]. Hexane is soluble in heptane because both are non-polar hydrocarbons with similar dispersion-force interactions, so mixing is energetically favourable [1].
Return to your Think First response. You should now be able to sharpen it into a proper HSC-style explanation:
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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