Chemistry • Year 12 • Module 7 • Lesson 1

Introduction to Organic Chemistry & IUPAC Nomenclature I

Apply IUPAC naming rules, interpret boiling point data for alkane homologous series, and reason about molecular structure from a real Australian industry context.

Apply · Data & Reasoning

1. Sequence the IUPAC naming steps

The seven steps below for naming the compound CH3CH2CH(CH3)CH2CH3 are shuffled. Write the correct order (1–7) in the “Order” column. 6 marks (1 per correct position)

Step (shuffled)Order (1–7)
Write the complete IUPAC name: 3-methylpentane.
Identify all substituents; here, one –CH3 (methyl) group is present.
Confirm the molecular formula using CnH2n+2: n = 6, so formula = C6H14.
Identify the functional group class: only C–C and C–H single bonds → alkane; suffix = –ane.
Number the chain from the end nearer the branch; methyl branch falls on C3 from either end.
Trace the longest continuous carbon chain: CH3–CH2–CH–CH2–CH3 = 5 carbons; parent chain = pentane.
Name the substituent with its locant: 3-methyl.
Stuck? Revisit lesson Card 4 — the three IUPAC steps (longest chain → lowest locants → alphabetical substituents) map onto these seven sub-steps.

2. Cause-and-effect chain — why alkane boiling points increase with chain length

Each cause box is filled. Complete the effect boxes to trace the mechanism from molecular structure to macroscopic property. 5 marks

Cause 1: Longer alkane chain = more electrons and larger molecular surface area.
↓
Effect 1:
Cause 2: London (dispersion) forces between molecules increase.
↓
Effect 2:
Cause 3: More energy must be supplied to separate molecules from the liquid phase.
↓
Overall outcome (so…):
Stuck? Think about what ‘stronger intermolecular force’ means for how hard it is to pull molecules apart when liquid turns to vapour.

3. Interpret boiling point data — straight-chain alkanes C1–C8

The graph below shows the normal boiling points (°C) of the first eight straight-chain alkanes. Use it to answer the questions. 8 marks

−160 −120 −80 −40 0 40 80 120 Normal boiling point (°C) methane ethane propane butane pentane hexane heptane octane Alkane (straight-chain, C1to C8) −162 −89 −42 −1 36 69 98 126 0°C

Boiling point data: NIST WebBook (webbook.nist.gov). Straight-chain alkanes at 1 atm.

3.1 Describe the overall trend in boiling point from methane to octane. 2 marks

3.2 At room temperature (25°C), which of the eight alkanes are gases and which are liquids? Use the graph to justify your answer. 2 marks

3.3 Estimate the boiling point of nonane (C9H20) by extrapolating the trend. Show your reasoning. 2 marks

3.4 Methane and ethane are the primary components of natural gas from the Karratha LNG (liquefied natural gas) facility in Western Australia. Using the graph, explain why LNG must be stored at approximately −162°C to keep these gases in liquid form. 2 marks

Stuck? To identify gases at 25°C, find which alkanes have boiling points below 25°C on the graph. For the LNG question, recall what boiling point physically means.

4. Compare and contrast — alkane vs alkene for the same chain length

Complete the comparison table for butane (C4H10) versus but-1-ene (C4H8). 7 marks (1 per row)

FeatureButaneBut-1-ene
IUPAC suffix
General formula used
Functional group present
Hybridisation of ALL carbons
Homologous series
Relative reactivity with Br2(aq)
One real-world application or source
Stuck? Row 6 — recall that alkenes are unsaturated (have a pi bond available for addition) while alkanes are saturated.

5. Apply to a real scenario — Ampol Lytton refinery, Brisbane

The Ampol Lytton refinery in Brisbane processes crude petroleum oil to separate hydrocarbon fractions. Crude petroleum is a complex mixture of hundreds of organic compounds, predominantly straight-chain and branched alkanes ranging from C1 to C40+. Different fractions are separated based on their differing boiling points. 5 marks

5.1 Explain, using your understanding of homologous series and intermolecular forces, why crude petroleum separates into distinct fractions when heated in a fractionating column. 3 marks

5.2 One petroleum fraction (LPG — liquefied petroleum gas) contains mainly propane (C3H8) and butane (C4H10). Identify which homologous series these belong to, write their IUPAC names, and state one structural feature they share that places them in the same series. 2 marks

Stuck? Connect Section 3 (boiling point trend) with the idea that a fractionating column exploits that trend.
Answers — Do not peek before attempting

Q1 — Sequence

Correct order: 4 (identify alkane class) → 6 (find longest chain) → 2 (identify substituents) → 5 (number chain) → 7 (name substituent with locant) → 1 (write full name) → 3 (verify molecular formula). Numerical answer in order: Step 1 = 4, Step 2 = 6, Step 3 = 2, Step 4 = 5, Step 5 = 7, Step 6 = 1, Step 7 = 3.

Q2 — Cause-and-effect

Effect 1: London dispersion forces between molecules increase (more electrons and larger surface area means stronger temporary dipoles).

Effect 2: More energy must be input to overcome (break) the stronger intermolecular forces, raising the boiling point.

Overall outcome: Boiling point increases progressively with chain length within the alkane homologous series (as seen in the graph: methane −162°C → octane +126°C).

Q3 — Graph interpretation

3.1 Boiling point increases steadily from methane (−162°C) to octane (+126°C) as chain length increases. The rate of increase is approximately 20–40°C per additional CH2 unit (slightly concave curve — the increments decrease slightly for larger molecules). [1 for trend direction, 1 for quantitative reference or mention of the decreasing increment]

3.2 Gases at 25°C (boiling point below 25°C): methane, ethane, propane, butane (boiling points −162, −89, −42, −1°C respectively). Liquids at 25°C: pentane, hexane, heptane, octane (boiling points 36, 69, 98, 126°C). [1 for all four gases correct, 1 for liquids correct or for correct reasoning]

3.3 The increment from heptane (C7) to octane (C8) is 126 − 98 = 28°C. Extrapolating: nonane (C9) ≈ 126 + 27–28 ≈ 153–154°C. (Actual value: 151°C.) [1 for method, 1 for answer within ±10°C]

3.4 The boiling point of methane is −162°C and of ethane is −89°C — both well below room temperature. At temperatures above their boiling points these substances exist as gases, not liquids. To keep them in the liquid phase for storage and transport as LNG, the temperature must be maintained at or below the boiling point of the most volatile component (methane at −162°C). [1 for linking boiling point to gas/liquid state, 1 for specifying why −162°C is required]

Q4 — Compare and contrast

FeatureButaneBut-1-ene
IUPAC suffix–ane–ene
General formulaCnH2n+2CnH2n
Functional groupNone (C–C single bonds only)C=C double bond
HybridisationAll sp3C1 and C2: sp2; C3 and C4: sp3
Homologous seriesAlkanesAlkenes
Reactivity with Br2Very slow / no reaction at RTRapid decolourisation (addition)
Application / sourceLPG fuel, lighter fuelPetrochemical feedstock, polymer precursor

Q5 — Ampol Lytton refinery

5.1 Crude petroleum is a mixture of alkanes of different chain lengths. Because London dispersion forces increase with chain length (more electrons, larger surface area), shorter-chain alkanes have lower boiling points and vaporise at lower temperatures, while longer-chain alkanes vaporise at higher temperatures. When petroleum is heated in a fractionating column, compounds with similar boiling points condense at similar heights, separating into distinct fractions. [3 marks: 1 for dispersion forces increasing with chain length, 1 for linking to different boiling points, 1 for the separation mechanism in the column]

5.2 Both propane and butane belong to the alkane homologous series (all C–C and C–H single bonds, general formula CnH2n+2). IUPAC names: propane (3C) and butane (4C). They share the same functional group class (no reactive functional group; only C–C and C–H bonds) and the general formula CnH2n+2. [1 for naming both correctly as alkanes with IUPAC names, 1 for structural feature]

HSCScience • Chemistry • Year 12 • Module 7 • Lesson 1 • Worksheet 2 (Apply)