Discover why a molecule's chain length and shape determine whether it's a gas, liquid, or solid — and master the intermolecular-force reasoning that earns Band 6 marks.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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
Saturated hydrocarbons · single bonds only · C1–C8 reference
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 has only single bonds — tetrahedral geometry, 109.5° bond angles, in a three-dimensional 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 |
London forces · surface area · Band 6 explanation structure
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.
Structural isomers · compact shape · reduced contact area
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.
Bond type → shape → bond angle · comparing series
Hydrocarbon series differ not only in formula but also in local geometry. The bond type present at carbon directly determines the shape and bond angle around that carbon — you can read the geometry straight from the structural formula.
| Series | Bond type | Geometry | Bond angle |
|---|---|---|---|
| Alkane | C–C single bond only | Tetrahedral | 109.5° |
| Alkene | C=C double bond | Trigonal planar | ~120° |
| Alkyne | C≡C triple bond | 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. In Lesson 4, you will see how the geometry difference between alkanes and alkenes explains why unsaturated hydrocarbons are far more reactive.
Water vs non-polar solvents · IMF disruption and replacement
“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.
In water: Insoluble
In non-polar solvent: Soluble
In alcohols: Limited / partial
Would require breaking H-bonding without forming comparable new interactions
Same IMF type: dispersion forces only
Contains both a polar OH region and a non-polar carbon chain
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.
All comparisons in this lesson are qualitative — explain using structure and intermolecular forces, not calculations.
Octane has a far higher boiling point (126°C vs −162°C). Two reasons: (1) Octane has a much larger molecular surface area, allowing more London dispersion force contact between molecules — larger electron clouds create stronger temporary dipoles. (2) Octane has greater molecular mass (114 g/mol vs 16 g/mol), meaning more electrons available for instantaneous polarisation. Both effects increase the energy needed to separate molecules from liquid to gas.
Activities
1. Rank methane, pentane, and octane in increasing order of boiling point. Justify each step using intermolecular-force reasoning.
2. Pentane (C5H12) and 2,2-dimethylpropane (C5H12) have the same molecular formula. Which has the higher boiling point, and why?
3. Describe the geometry (shape, bond angle) around the carbon atoms in: (a) ethane, (b) ethene, (c) ethyne.
4. Predict the solubility of hexane in water and in ethanol (which contains a polar OH group). Explain both.
Check Your Understanding
Click an option to check your answer. Aim for all 5 before looking at the answers accordion.
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 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. (a) Ethane: single bonds only → tetrahedral → 109.5°. (b) Ethene: C=C double bond → trigonal planar → ~120°. (c) Ethyne: C≡C triple bond → linear → 180°.
4. Hexane is insoluble in water because water’s hydrogen-bond network would be disrupted without enough compensating interaction from the non-polar hexane. 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 linear with 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.
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].
Review Drills
Name the three general formulas for alkanes, alkenes, and alkynes, and state the geometry and bond angle at the key carbons for each.
Give the four-step Band 6 explanation for why hexane has a higher boiling point than pentane.
List the three C5H12 isomers and rank them in order of decreasing boiling point. Explain the trend.
Explain why hexane is insoluble in water but miscible with heptane. Use IMF reasoning.
A student compares butane (C4H10) and but-1-yne (C4H6). Which has the higher boiling point, and why? What geometry and bond angle apply to the triple-bond carbons in but-1-yne?
Back at the start you were asked: methane escapes as a gas at −162°C, yet paraffin wax — made of the same type of molecule, just much longer — is solid at room temperature. What changes when you add more carbons?
Now you know the complete causal chain: adding more carbons increases chain length → larger electron clouds and greater surface area → stronger London dispersion forces between molecules → more energy required to separate molecules into the gas phase → higher boiling point. Methane (C1) experiences only the weakest dispersion forces, so it boils far below room temperature. Paraffin wax (C20+) has such strong cumulative dispersion forces that it remains solid at room temperature and only melts above a flame. The “type” of molecule is the same (C–H and C–C single bonds throughout) — only the length, and therefore the intermolecular force strength, changes.
Tick when you’ve finished all activities and checked your answers.
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