Chemistry • Year 12 • Module 7 • Lesson 2
IUPAC Nomenclature II: Functional Group Classes & Isomers
Apply IUPAC naming to novel compounds, interpret boiling-point data for structural isomers, and use reasoning to distinguish isomer types in real contexts.
1. Interpret boiling-point data for structural isomers
The table below shows boiling points and selected properties for five C4 structural isomers. Study the data carefully, then answer the questions that follow. 9 marks
| Compound | Molecular formula | Functional group | Boiling point (°C) | Water solubility |
|---|---|---|---|---|
| Butane | C4H10 | Alkane | –1 | Insoluble |
| 2-methylpropane | C4H10 | Alkane | –12 | Insoluble |
| Butan-1-ol | C4H10O | Alcohol (–OH) | 118 | Miscible |
| Butan-2-ol | C4H10O | Alcohol (–OH) | 99 | Miscible |
| Butanal | C4H8O | Aldehyde (–CHO) | 75 | Slightly soluble |
Data: NIST WebBook / standard reference values.
1.1 Butane and 2-methylpropane share the molecular formula C4H10. Classify the type of structural isomers they represent and justify your answer with reference to their structural formulas (CH3CH2CH2CH3 vs CH3CH(CH3)CH3). 2 marks
1.2 2-methylpropane boils at –12°C while butane boils at –1°C, even though they have the same molecular formula and mass. Using your knowledge of intermolecular forces, explain why 2-methylpropane has a lower boiling point. 3 marks
1.3 Butan-1-ol (bp 118°C) and butan-2-ol (bp 99°C) are both C4H10O alcohols. (a) Classify the type of structural isomers they represent and justify. (b) Suggest why butan-1-ol has the higher boiling point, referring to the 1°/2° classification. 4 marks
2. Interpret a bar chart — boiling points of straight-chain alcohols
The bar chart below shows the boiling points of the first five straight-chain primary alcohols (methanol to pentan-1-ol). 8 marks
2.1 Describe the trend in boiling point as chain length increases from C1 to C5. 2 marks
2.2 Using your knowledge of intermolecular forces, explain why boiling point increases with chain length in this homologous series. 3 marks
2.3 Ethanol (C2, bp 78°C) is used in Australian E10 fuel blends at 10% volume. Using the graph, predict and justify what the boiling point of hexan-1-ol (C6) would be, then explain why E10 combustion behaviour is influenced by the relatively low boiling point of ethanol compared with petrol (typical bp 35–200°C). 3 marks
3. Cause-and-effect chain — amine vs amide reactivity with HCl
Complete the cause-and-effect chain. Fill in the empty boxes to show why ethanamine reacts with HCl while ethanamide does not. 5 marks
CH3CH2NH2
Lone pair on N is ________
Forms CH3CH2NH3+Cl–
CH3CONH2
Lone pair on N is ________
Ethanamide is neutral
4. Predict and justify
Q4. A TGA pharmacologist is reviewing two drugs with the molecular formula C3H7NO. Drug X reacts with aqueous hydrochloric acid at room temperature to form a white precipitate (salt). Drug Y does not react with HCl. 5 marks
(a) Identify the functional group most likely present in Drug X, and explain the structural feature responsible for its reactivity with HCl. 2 marks
(b) Identify the functional group most likely present in Drug Y, and explain why it does NOT react with HCl. 2 marks
(c) State the type of structural isomers that Drug X and Drug Y represent. Justify your answer. 1 mark
Q1.1 — Butane vs 2-methylpropane isomer type
Chain isomers. Both have the molecular formula C4H10 and the same functional group class (alkane, no functional group). They differ only in the shape of the carbon skeleton: butane has a straight, unbranched four-carbon chain (CH3CH2CH2CH3), while 2-methylpropane has a branched three-carbon chain with a methyl group at C2 (CH3CH(CH3)CH3). [1] Correct isomer type named + structural justification. [1]
Q1.2 — Why 2-methylpropane has a lower bp than butane
Both molecules experience the same type of intermolecular forces (London/dispersion forces only, as there is no polar functional group). Boiling point is determined by the strength of these forces, which depends on the surface area available for contact between molecules. Butane’s straight chain has greater surface area and so stronger London forces. [1] 2-methylpropane’s branched, more spherical shape has less surface area for intermolecular contact. [1] Therefore fewer/weaker London dispersion forces between 2-methylpropane molecules → less energy needed to separate them → lower boiling point. [1]
Q1.3 — Butan-1-ol vs butan-2-ol
(a) Position isomers. Both have the formula C4H10O [1], same functional group (–OH, alcohol), same 4-carbon skeleton, but –OH is at C1 in butan-1-ol (primary) and at C2 in butan-2-ol (secondary). Only the position of the group changes → position isomers. [1]
(b) Butan-1-ol (primary, –OH at C1 = terminal carbon) has a less hindered –OH group that can form stronger or more extensive hydrogen bonds with surrounding molecules compared with butan-2-ol (secondary, –OH at an internal, more branched position). The increased effective branching in butan-2-ol also reduces the molecular surface area available for London dispersion forces. Combined effect: butan-1-ol has a higher boiling point. [2]
Q2.1 — Trend in bp with chain length
Boiling point increases steadily as chain length increases from C1 (methanol, 65°C) to C5 (pentan-1-ol, 138°C). The increase is approximately 17–21°C per additional carbon unit. [2: 1 for direction of trend, 1 for quantitative description or comparative reference to at least two data points]
Q2.2 — Why bp increases with chain length
All primary alcohols can form hydrogen bonds via their –OH group (these are roughly constant per molecule). [1] As chain length increases, the total number of electrons and the molecular surface area both increase. [1] Larger surface area → stronger London (dispersion) intermolecular forces between chains → more energy required to separate molecules → higher boiling point. [1]
Q2.3 — Hexan-1-ol prediction + E10 context
The bp increase per additional carbon is roughly 17–20°C. Pentan-1-ol = 138°C; adding one more C→ hexan-1-ol predicted bp approximately 155–158°C. [1] Ethanol (bp 78°C) is more volatile than most petrol components (bp up to 200°C). Its lower boiling point means it vaporises more readily in the engine intake at lower temperatures. [1] In Australian conditions, this can improve cold-start combustion and reduce emissions, but the higher volatility also increases evaporative losses in fuel storage and affects the fuel’s Reid Vapour Pressure rating. [1]
Q3 — Cause-and-effect chain answers
Ethanamine row: Lone pair on N is free / not delocalised → the lone pair can accept H+ from HCl → reacts with HCl to form salt.
Ethanamide row: Lone pair on N is delocalised into the adjacent C=O by resonance → the lone pair is unavailable / cannot accept H+ → does NOT react with HCl.
[1 mark per correctly filled box, up to 5 marks total across both rows]
Q4 — TGA pharmacologist scenario
(a) Drug X contains an amine (–NH2). [1] The lone pair on the nitrogen atom is not delocalised; it acts as a Lewis base and accepts H+ from HCl to form an ammonium salt (R–NH3+Cl–). [1]
(b) Drug Y contains an amide (–CONH2). [1] The nitrogen lone pair is delocalised by resonance into the adjacent C=O; it is not available to accept H+, so ethanamide-type compounds are neutral and do not react with HCl. [1]
(c) Functional group isomers. [1] Both have the molecular formula C3H7NO, so they are structural isomers. The functional group is entirely different (amine vs amide), not just in a different position → functional group isomers by definition.