Chemistry • Year 12 • Module 7 • Lesson 3
Hydrocarbons: Structure, Homologous Series & Physical Properties
Apply your understanding of London dispersion forces, chain length, and branching to interpret real data, compare structural isomers, and reason about solubility in context.
1. Interpret the boiling point data
The graph below shows the boiling points of the straight-chain alkanes from methane (C1) to decane (C10) at standard atmospheric pressure. 10 marks
(a) Describe the trend shown by the data as the number of carbon atoms increases from C1 to C10. Quote at least two data points in your response. 2 marks
(b) Use the graph to estimate the boiling point of undecane (C11H24). State one assumption you are making in your estimate. 2 marks
(c) Explain, using London dispersion forces, why the boiling point increases with chain length. Your answer must include: (i) what LDF are; (ii) how chain length affects them; (iii) why stronger LDF require more energy. 3 marks
(d) The Ampol Lytton refinery in Brisbane uses fractional distillation to separate crude oil. Using the graph as evidence, explain why LPG (mainly propane, C3, and butane, C4) is collected near the top of the distillation tower, while diesel fractions (C12–C20) are collected near the bottom. 3 marks
2. Cause-and-effect chain — branching reduces boiling point
Complete the cause-and-effect chain below. Each box should be filled with the next consequence. The first and last steps are provided. 5 marks
Use: surface area • contact area • London dispersion forces • energy required to separate • boiling point3. Data table — comparing isomers
The table below shows boiling point data for four C6H14 isomers. Use the data to answer the questions. 6 marks
| Compound | Molecular formula | Structure type | Boiling point (°C) |
|---|---|---|---|
| Hexane | C6H14 | Straight chain | 69 |
| 2-methylpentane | C6H14 | One branch | 60 |
| 3-methylpentane | C6H14 | One branch (central) | 63 |
| 2,2-dimethylbutane | C6H14 | Two branches | 50 |
(a) Identify the trend shown in the table between degree of branching and boiling point. 1 mark
(b) Despite having the same molecular formula and molecular mass, these isomers have different boiling points. Explain why molecular mass alone is not sufficient to predict boiling point. 2 marks
(c) 2-methylpentane and 3-methylpentane have the same type and number of branches, yet their boiling points differ slightly (60°C vs 63°C). Suggest a reason for this small difference. 1 mark
(d) Predict whether 2,3-dimethylbutane (also C6H14) would have a boiling point above or below 50°C. Justify your prediction. 2 marks
4. Predict and justify
A student claims: “LNG (liquefied natural gas, mainly methane, CH4) and LPG (liquefied petroleum gas, mainly propane C3H8 and butane C4H10) are both used as fuels in Australian transport, but LNG must be stored at a much colder temperature than LPG.” 4 marks
Predict whether the student’s claim is consistent with London dispersion force theory and justify your prediction using structure, IMF strength, and boiling point.
Q1(a) — Describe the trend
As the number of carbon atoms increases from C1 to C10, the boiling point rises steadily from −162°C (methane, C1) to +174°C (decane, C10). The increase is large overall but the gaps between successive members decrease slightly at higher chain lengths, giving a slightly curved (concave) trend. Marking: 1 mark for direction of trend; 1 mark for quoting at least two correct data values.
Q1(b) — Estimate BP of C11
Reading the trend, each step from C9 to C10 adds roughly 23°C (174 − 151 = 23). A reasonable estimate for C11 is approximately 195–200°C. Assumption: the trend continues to increase by a similar increment at C11 (i.e. the relationship remains approximately linear in this region). 1 mark for a defensible estimate within ~20°C of the actual value (196°C); 1 mark for identifying an assumption.
Q1(c) — Explain using LDF
(i) London dispersion forces are weak intermolecular attractions that arise when temporary instantaneous dipoles in one molecule induce a dipole in a neighbouring molecule. (ii) Longer carbon chains have more electrons, more polarisable electron clouds, and greater surface area for contact, producing more simultaneous instantaneous dipole interactions. (iii) To boil a liquid, molecules must be fully separated from each other; stronger LDF require more energy (higher temperature) to overcome, so the boiling point is higher. 1 mark each for (i), (ii), (iii).
Q1(d) — Fractional distillation context
The graph shows that propane (C3, BP −42°C) and butane (C4, BP −1°C) have very low boiling points; at refinery operating temperatures these fractions are already gaseous or easily vaporised. They therefore rise to the top of the tower. Diesel fractions (C12–C20) have much longer chains, much stronger LDF, and boiling points well above 200°C; they remain liquid at higher temperatures and are drawn off near the bottom where temperatures are higher. The different boiling points caused by differences in LDF strength enable the physical separation of fractions. 1 mark for LPG/LDF explanation; 1 mark for diesel/LDF explanation; 1 mark for explicitly linking to tower position.
Q2 — Cause-and-effect chain blanks
In order: compact • contact area (surface area) • London dispersion force (instantaneous dipole–induced dipole) • energy required
1 mark per correct blank (4 blanks); 1 mark for overall outcome correctly stated (provided — no mark deducted).
Q3(a) — Trend in branching vs BP
As the degree of branching increases (0 branches → 1 branch → 2 branches), the boiling point decreases (69 → 60–63 → 50°C).
Q3(b) — Why mass alone is insufficient
All four isomers share the same molecular formula C6H14 and therefore the same molecular mass (86 g mol−1). Yet their boiling points span nearly 20°C. Boiling point depends on the strength of intermolecular forces, which is controlled by molecular shape and surface area, not molecular mass. Branching reduces surface area even when mass is constant, so mass cannot predict boiling point when structural isomers are being compared. 1 mark for identifying identical mass; 1 mark for explaining that shape/surface area, not mass, controls LDF strength.
Q3(c) — 2-methylpentane vs 3-methylpentane
Both have one branch but positioned differently. The branch at C3 (central) in 3-methylpentane makes the molecule slightly more symmetrical and possibly allows marginally better stacking/contact along part of the chain than the terminal branch in 2-methylpentane, explaining the slightly higher boiling point. Accept any reasonable structural argument.
Q3(d) — Predict BP of 2,3-dimethylbutane
2,3-dimethylbutane has two branches (same as 2,2-dimethylbutane), so its boiling point should be close to 50°C or slightly above/below. A prediction of approximately 50–60°C is reasonable. (Actual BP is 58°C.) 1 mark for prediction near 50–60°C; 1 mark for justification using number of branches → surface area → LDF strength.
Q4 — LNG vs LPG storage temperature
The student’s claim is consistent with LDF theory. Methane (CH4) has only 1 carbon atom, a very small electron cloud, and minimal surface area, so its LDF are extremely weak. Its boiling point is −162°C, meaning it must be stored at cryogenic temperatures to remain liquid as LNG. Propane (C3H8) and butane (C4H10) have longer chains, greater surface area, and stronger LDF; their boiling points (−42°C and −1°C) are much higher, so LPG can be stored as a liquid at moderate pressure without the extreme cooling required for LNG. 1 mark for correctly predicting consistent; 1 mark for structure/LDF reasoning for methane; 1 mark for structure/LDF reasoning for propane/butane; 1 mark for linking to storage temperatures.