Chemistry • Year 12 • Module 7 • Lesson 3
Hydrocarbons: Structure, Homologous Series & Physical Properties
Build your core vocabulary, master the three homologous series, and recognise the structural features that control boiling point and solubility in non-polar hydrocarbons.
1. Term–definition match
Twelve definitions are shuffled below. In the right-hand column write the matching term from this list: alkane, alkene, alkyne, sp³ hybridisation, homologous series, saturated hydrocarbon, unsaturated hydrocarbon, London dispersion forces, branching effect, general formula, sp² hybridisation, boiling point. 12 marks
| # | Definition (shuffled) | Matching term |
|---|---|---|
| 1.1 | A hydrocarbon containing only C–C single bonds; general formula CnH2n+2. | |
| 1.2 | Weak intermolecular forces between non-polar molecules arising from temporary instantaneous dipoles; the only forces between hydrocarbon molecules. | |
| 1.3 | A hydrocarbon with at least one C=C double bond; general formula CnH2n. | |
| 1.4 | The temperature at which the vapour pressure of a liquid equals atmospheric pressure; used to separate hydrocarbon fractions at the Ampol Lytton refinery. | |
| 1.5 | A family of organic compounds sharing the same functional group, differing by –CH2– units, with gradually changing physical properties. | |
| 1.6 | An expression (e.g. CnH2n+2) describing every member of a homologous series with the same structural pattern. | |
| 1.7 | A hydrocarbon with at least one C≡C triple bond; general formula CnH2n−2. | |
| 1.8 | The hybridisation state of carbon atoms in alkanes; each carbon forms four equivalent bonds arranged tetrahedrally with bond angles of 109.5°. | |
| 1.9 | A hydrocarbon containing only single bonds, with the maximum possible number of hydrogen atoms. | |
| 1.10 | The observation that isomers with more branching have lower boiling points than their straight-chain counterparts, due to reduced surface area for intermolecular contact. | |
| 1.11 | The hybridisation state of carbon atoms at the C=C double bond in alkenes; produces a trigonal planar arrangement with bond angles of approximately 120°. | |
| 1.12 | A hydrocarbon with one or more C=C or C≡C bonds; contains fewer hydrogen atoms than the maximum. |
2. True or false — with correction
For each statement, circle T or F. If the statement is false, write the corrected version on the line below. 10 marks (1 for T/F, 1 for correction where needed)
2.1 Alkanes are described as “saturated” because they contain only C–C single bonds and have the maximum number of hydrogen atoms for their carbon count. T / F
2.2 Non-polar hydrocarbons have no intermolecular forces, which is why they have very low boiling points. T / F
2.3 Pentane (C5H12) has a higher boiling point than methane (CH4) because pentane has a longer carbon chain, greater surface area, and stronger London dispersion forces. T / F
2.4 Branched-chain isomers always have higher boiling points than their straight-chain counterparts of the same molecular formula. T / F
2.5 The carbon atoms at a C=C double bond in an alkene are sp²-hybridised, producing trigonal planar geometry with bond angles of approximately 120°. T / F
3. Complete the paragraph
Fill each blank with the correct word or phrase from the word bank. Use each term once only. 10 marks
Word bank: London dispersion forces • surface area • chain length • non-polar • hydrogen bonds • boiling point • homologous series • –CH2– • branching • insoluble
Alkanes, alkenes, and alkynes each form a , a family of compounds differing by one unit between successive members. All hydrocarbons are molecules, so the only intermolecular forces between them are . These forces increase in strength as increases because longer chains have more electrons and a greater for intermolecular contact. As a result, rises steadily up the series. Increased reduces boiling point by making the molecule more compact and reducing contact area. Hydrocarbons are in water because dissolving would require breaking water’s network without adequate replacement.
4. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “increases with”, “depends on”, “determines”). Aim for at least 6 labelled arrows. 6 marks
Supplied terms: chain length • surface area • London dispersion forces • boiling point • branching • solubility in water
5. Function recall
Answer each in 1–2 sentences using precise terms from the lesson. 10 marks (2 each)
5.1 What is the role of surface area in determining the strength of London dispersion forces between hydrocarbon molecules?
5.2 Why does branching in an alkane lower its boiling point compared with a straight-chain isomer?
5.3 Why are hydrocarbons insoluble in water but soluble in hexane?
5.4 Explain why two hydrocarbons with the same molecular formula (structural isomers) can have different boiling points. Name the structural feature responsible and explain how it affects London dispersion forces.
5.5 State the general formula and the hybridisation state of the key carbon atoms for each of: alkane, alkene, alkyne.
Q1 — Term–definition matches
1.1 alkane • 1.2 London dispersion forces • 1.3 alkene • 1.4 boiling point • 1.5 homologous series • 1.6 general formula • 1.7 alkyne • 1.8 sp³ hybridisation • 1.9 saturated hydrocarbon • 1.10 branching effect • 1.11 sp² hybridisation • 1.12 unsaturated hydrocarbon
Q2 — True / false with correction
2.1 True.
2.2 False. Correction: non-polar hydrocarbons do experience intermolecular forces — London dispersion forces (also called van der Waals forces). These forces are the reason hydrocarbons have measurable boiling points. A molecule with no intermolecular forces would be a gas at all temperatures and could not be condensed.
2.3 True.
2.4 False. Correction: branched-chain isomers have lower boiling points than their straight-chain counterparts, because the compact shape reduces surface area for intermolecular contact and therefore weakens London dispersion forces.
2.5 True. Carbon atoms forming a C=C double bond in an alkene are sp²-hybridised, resulting in three sigma bonds in a trigonal planar arrangement and one unhybridised p-orbital contributing to the pi bond. The bond angles between the sigma bonds are approximately 120°.
Q3 — Cloze paragraph
In order: homologous series • –CH2– • non-polar • London dispersion forces • chain length • surface area • boiling point • branching • insoluble • hydrogen bonds
Q4 — Sample concept map arrows
- chain length —increases→ surface area
- surface area —determines strength of→ London dispersion forces
- London dispersion forces —determine→ boiling point
- branching —decreases→ surface area
- branching —lowers→ boiling point
- chain length —has no effect on→ solubility in water (all non-polar hydrocarbons are insoluble regardless of length)
Award 1 mark per correctly labelled arrow. Any biologically valid phrasing accepted.
Q5.1 — Role of surface area in LDF
Greater surface area means more sites at which instantaneous dipole–induced dipole interactions can occur simultaneously between adjacent molecules. This increases the total London dispersion force between them, requiring more energy to overcome during boiling.
Q5.2 — Why branching lowers boiling point
Branching makes the hydrocarbon molecule more compact and rounded. Compact molecules cannot lie alongside neighbouring molecules across a long surface, so the contact area is smaller, fewer simultaneous dispersion interactions occur, and less energy is needed to separate the molecules, giving a lower boiling point.
Q5.3 — Solubility in water vs hexane
Hydrocarbons are non-polar and can only form dispersion forces. Water molecules are held together by a strong hydrogen-bond network; dissolving a hydrocarbon would require disrupting those bonds without replacing them with interactions of comparable strength — so dissolution is not favourable and the hydrocarbon is insoluble. In hexane (also non-polar), only dispersion forces need to be disrupted and reformed, making mixing energetically favourable.
Q5.4 — Why structural isomers can have different boiling points
Two structural isomers with the same molecular formula have identical molecular mass, but their shapes differ. The key structural feature is branching: a branched isomer has a more compact, rounded shape and therefore less surface area for intermolecular contact between neighbouring molecules. Fewer simultaneous London dispersion force interactions result, so less energy is required to separate the molecules during boiling — giving a lower boiling point for the branched isomer compared with its straight-chain counterpart.
Q5.5 — General formulas and hybridisation
Alkane: CnH2n+2; all carbons are sp³-hybridised (tetrahedral, 109.5°). Alkene: CnH2n; carbons at the C=C bond are sp²-hybridised (trigonal planar, ~120°). Alkyne: CnH2n−2; carbons at the C≡C bond are sp-hybridised (linear, 180°).