Year 12 ChemistryModule 7 · Organic Chemistry⏱ ~45 min5 MC · 3 Short AnswerLesson 3 of 23CoreCore
Hydrocarbons, Structure, Homologous Series & Physical Properties
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 top-band marks.
Today's hook: Methane boils at −162°C and escapes as a gas, 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?
0/5TASKS
1
You’re here
Connect: retrieve the homologous-series pattern
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
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?
Learning Intentions
goals
Know
The structural features of alkanes, alkenes, and alkynes
The first eight alkane names, formulas, and typical states at 25°C
The boiling point and solubility trends within homologous series
The geometry and bond angle around alkane, alkene, and alkyne carbons
Understand
Why London dispersion forces increase with chain length
Why branching lowers boiling point even at constant molecular formula
Why hydrocarbons are insoluble in water but dissolve in non-polar solvents
Why chain length dominates physical properties more strongly than unsaturation type
Can Do
Predict and explain boiling point order within a hydrocarbon series
Compare structural isomers using surface area and IMF reasoning
Explain solubility using IMF disruption and replacement arguments
State geometry and bond angles around carbon in each hydrocarbon series
Scan these before reading
vocab
AlkaneA saturated hydrocarbon with only C–C single bonds; general formula CnH2n+2.
AlkeneAn unsaturated hydrocarbon with at least one C=C double bond; general formula CnH2n.
AlkyneAn unsaturated hydrocarbon with at least one C≡C triple bond; general formula CnH2n−2.
Homologous seriesA family of organic compounds with the same functional group, differing by –CH₂– units, with gradually changing properties.
Van der Waals forcesWeak intermolecular forces between non-polar molecules; increase with molecular size, explaining boiling point trends in alkane series.
Branching effectBranched isomers have lower boiling points than straight-chain isomers due to reduced surface area and weaker van der Waals forces.
Core Content
High-Frequency Misconceptions:
"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.
2
Explain: construct the alkane series
1
Alkane Structure and the First Eight Members
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
Must Know: Memorise the C1–C8 alkane names, formulas, and gas-to-liquid transition around C5. These values give you the reference points needed to predict unfamiliar trends confidently.
Common Error: “Alkanes have no intermolecular forces” is wrong. Alkanes are non-polar, but all molecules experience London dispersion forces. Those dispersion forces are the sole reason alkanes have measurable boiling points.
Which alkane is the first to be a liquid at room temperature (25°C)?
3
Represent: chain length and boiling point
2
Why Boiling Point Increases with Chain Length
London forces · surface area · top-band 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.
top-band Structure: How to Explain “Hexane Has a Higher Boiling Point Than Pentane”
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.
Key Insight: For simple hydrocarbons, chain length usually matters much more than whether the molecule is an alkane, alkene, or alkyne. A long hydrocarbon nearly always has a much higher boiling point than a short one because dispersion-force strength scales strongly with size.
True or False: The reason octane boils at a higher temperature than methane is that octane has stronger covalent bonds inside the molecule.
4
Investigate: measure the trend, do not just assert it
3
Investigation: Boiling Point Across the Alkane Series Core
The syllabus asks you to investigate the trend, not only to explain it
Module 7 requires you to conduct an investigation into the physical property trends within a homologous series. The dot point is about handling evidence: naming variables, processing data, describing a pattern and stating what limits your conclusion. Work through this one properly and you have met it.
Inquiry question. How does boiling point change as chain length increases across the straight-chain alkanes?
Why supplied data. The first four alkanes are gases at room temperature, so their boiling points cannot be measured on a school bench. Processing a reliable published dataset is a legitimate secondary-source investigation and is what the dot point allows.
Variables.
Independent: number of carbon atoms in the chain, from 1 to 8.
Dependent: boiling point, in degrees Celsius.
Controlled: pressure held at 100 kPa; only unbranched isomers used; all values taken from the same reference source and the same measurement method.
Results. Boiling points at 100 kPa, with the change from each member to the next:
Alkane
Formula
Carbon atoms
Boiling point (°C)
Change from previous (°C)
Methane
CH4
1
-161.5
—
Ethane
C2H6
2
-88.6
+72.9
Propane
C3H8
3
-42.1
+46.5
Butane
C4H10
4
-0.5
+41.6
Pentane
C5H12
5
36.1
+36.6
Hexane
C6H14
6
68.7
+32.6
Heptane
C7H16
7
98.4
+29.7
Octane
C8H18
8
125.7
+27.3
Plot boiling point (vertical axis) against number of carbon atoms (horizontal axis) in your book before reading on. Label both axes with units.
Analysis. The plot is a rising curve, not a straight line. Boiling point increases at every step, but the size of each step falls steadily, from +72.9 °C to +27.3 °C.
Boiling point rises across a homologous series because each additional CH2 group adds electrons and surface area, giving a larger and more polarisable electron cloud, so dispersion forces between molecules are stronger and more energy is needed to separate them.
Why the steps get smaller. Each added CH2 is the same absolute addition, but a progressively smaller fraction of the molecule that already exists. Going from one carbon to two doubles the chain; going from seven to eight lengthens it by about a seventh. The increase in polarisability per added carbon therefore diminishes, and so does the increase in boiling point.
Common error: "The covalent bonds get harder to break as the chain gets longer." Boiling breaks intermolecular forces, not covalent bonds. The C–C and C–H bonds are intact in both the liquid and the gas; if they were breaking, the substance would decompose rather than boil.
Evaluating the evidence. A conclusion is only as good as its limits:
Pressure dependence. Boiling point is defined at a stated pressure. At the top of a mountain every value here would be lower, so the pressure must be quoted for the numbers to mean anything.
Measurement uncertainty. A school thermometer typically reads to ±0.5 °C. That is small against steps of 27 °C or more, so it cannot explain the falling increment — the pattern is real, not an artefact of precision.
Purity. A soluble impurity raises the boiling point of a liquid, so an impure sample would shift individual points and weaken the comparison.
Scope of the conclusion. Only unbranched isomers were used. The trend describes the straight-chain series and does not transfer to branched isomers, which boil lower — the next card shows exactly that.
Secondary data. These are published values, so the reliability rests on the source rather than on repeated trials of your own. Naming the source is part of the evaluation.
Across the alkane series the boiling point rises, but the increase from one member to the next gets smaller. What best explains the shrinking increment?
In this investigation, which is the controlled variable?
5
Respond: branching changes intermolecular contact
4
Branching Lowers 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.
Pentane
Formula: C5H12 Shape: Long straight chain Boiling point: 36°C
Critical Error: Do not explain branching differences using “stronger covalent bonds.” Boiling point depends on intermolecular forces, not the covalent bonds inside each molecule.
Which of the following does NOT correctly explain why a branched isomer has a lower boiling point than its straight-chain counterpart?
6
Feedback: compare hydrocarbon structures
5
Alkenes, Alkynes, Geometry and Physical Comparisons
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.
Method: For any geometry question, identify the bond type around the carbon first. Single bond only → tetrahedral, 109.5°; part of C=C → trigonal planar, ~120°; part of C≡C → linear, 180°.
Which row correctly matches bond type and geometry?
7
Apply: predict solubility and trends
6
Solubility of Hydrocarbons
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.
Hydrocarbon solubility
In water: Insoluble
In non-polar solvent: Soluble
In alcohols: Limited / partial
Reason
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
Common Error: Do not write “hydrocarbons are insoluble in everything.” They are insoluble in water, but highly soluble in non-polar organic solvents.
Hexane is insoluble in water because dissolving would require disrupting water’s _______ network, and hexane can only offer _______ forces in return, which are much weaker.
Worked Examples
Worked Example 1, Arrange hydrocarbons by increasing boiling point
Order methane, propane, butane, and heptane by increasing boiling point and explain the trend.
1
All four are straight-chain alkanes, so the only intermolecular forces present are London dispersion forces.
2
Boiling point increases with chain length because longer chains have larger electron clouds and greater surface area for contact.
3
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.
Worked Example 2, Explain boiling-point differences between isomers
Pentane and 2,2-dimethylpropane both have formula C5H12. Why does pentane have the higher boiling point?
1
The two molecules have the same molecular formula and molecular mass, so mass alone cannot explain the difference.
2
Pentane is elongated, while 2,2-dimethylpropane is compact and nearly spherical.
3
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.
Worked Example 3, Explain solubility of a hydrocarbon in two solvents
Predict the solubility of hexane in water and in heptane.
1
Hexane is non-polar and has only dispersion forces.
2
Water has a strong hydrogen-bond network. Hexane cannot form interactions strong enough to replace those H-bonds.
3
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.
Key Relationships, Consolidation
Alkane: CnH2n+2
Saturated, single bonds only All carbons: tetrahedral, 109.5°
Alkene: CnH2n
One C=C double bond C=C carbons: trigonal planar, ~120°
The Organic Structure tool shows that carbon forms 4 covalent bonds because it has how many valence electrons?
🔬Predict, Then Reveal+8 XP
Two alkanes: methane (CH₄) and octane (C₈H₁₈). Predict which has the higher boiling point, and give TWO structural reasons why.
Your predictionExpert answerCompare
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.
8
Practise independently
Activities
Activity 1, Predict, Rank, Explain
Use the patterns from the lesson to predict physical properties. Focus on naming the IMF first, then linking structure to force strength.
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
MC
Question bank
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
Comprehensive Answers
Activity 1, Guided Practice
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.
Multiple Choice Explanations
1. Correct response: 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. Correct response: 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. Correct response: Carbon in a triple bond is linear with a bond angle of 180°.
4. Correct response: 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. Correct response: “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.
Short Answer Model Answers
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].
9
Retrieve and reflect
Check what actually stuck
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.
Alkane: CnH2n+2, single bonds only, tetrahedral, 109.5°. Alkene: CnH2n, C=C double bond, trigonal planar at C=C carbons, ~120°. Alkyne: CnH2n−2, C≡C triple bond, linear at C≡C carbons, 180°.
Give the four-step top-band explanation for why hexane has a higher boiling point than pentane.
(1) Both are non-polar alkanes → only London dispersion forces present. (2) Hexane has a longer carbon chain and larger molecular surface area than pentane. (3) Greater surface area allows more simultaneous instantaneous dipole-induced dipole interactions. (4) Stronger dispersion forces require more energy to overcome → hexane has the higher boiling point.
List the three C5H12 isomers and rank them in order of decreasing boiling point. Explain the trend.
Pentane (36°C) > 2-methylbutane (28°C) > 2,2-dimethylpropane (10°C). All have the same molecular formula and molecular mass. As branching increases, the molecular shape becomes more compact, reducing effective surface area for intermolecular contact. Less contact area means fewer simultaneous dispersion interactions, so less energy is needed to separate the molecules and the boiling point falls.
Explain why hexane is insoluble in water but miscible with heptane. Use IMF reasoning.
Hexane is insoluble in water because dissolving would require breaking water’s strong hydrogen-bond network. Hexane (non-polar, dispersion forces only) cannot replace those H-bonds with interactions of comparable strength, so dissolution is not energetically favourable. Hexane mixes readily with heptane because both are non-polar hydrocarbons, the same type of dispersion forces that hold the pure liquids together also operate between hexane and heptane molecules, so mixing carries little energetic cost.
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?
But-1-yne has the higher boiling point. Both are C4 hydrocarbons dominated by dispersion forces, but the alkyne’s more polarisable electron density and less compact shape produce stronger intermolecular attractions than in butane. The C≡C carbons in but-1-yne are linear with a bond angle of 180°.
Revisit Your Initial Thinking
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.
Mark lesson as complete
Tick when you’ve finished all activities and checked your answers.