Chemistry • Year 12 • Module 7 • Lesson 9
Structure & Properties of Alcohols
Apply boiling point, solubility, and IMF reasoning to real data, a graph, a comparative table, and an Australian context case study.
1. Read and interpret the boiling point graph
The inline graph below plots the boiling points of the C1–C5 primary alcohols and their corresponding alkane chains. Use the graph to answer the questions. 8 marks
Boiling point data from NIST WebBook (standard pressure). Both series plotted against carbon chain length C1–C5.
1.1 Describe the trend in boiling point for the primary alcohol series as chain length increases from C1 to C5. 1 mark
1.2 At C2, the boiling point of ethanol is 78°C while ethane is −89°C — a difference of 167°C. Explain this difference in terms of intermolecular forces. 3 marks
1.3 The gap in boiling point between alcohols and alkanes appears to decrease slightly as chain length increases from C1 to C5 (gap at C1 = 226°C; gap at C5 = 102°C). Explain this trend. 2 marks
1.4 Using the graph, estimate the boiling point of hexan-1-ol (C6). Justify your estimate. 2 marks
2. Interpret C4H10O isomer data
Four C4H10O alcohol isomers are listed below with their properties. Use the data to answer the questions. 7 marks
| Compound | Formula | BP (°C) | Water solubility | Hexane solubility |
|---|---|---|---|---|
| Butan-1-ol | CH3CH2CH2CH2OH | 118 | Partially miscible | Miscible |
| 2-methylpropan-1-ol | (CH3)2CHCH2OH | 108 | Partially miscible | Miscible |
| Butan-2-ol | CH3CH(OH)CH2CH3 | 100 | Miscible | Partially miscible |
| 2-methylpropan-2-ol | (CH3)3COH | 83 | Fully miscible | Poorly miscible |
2.1 Classify each of the four compounds as primary (1°), secondary (2°), or tertiary (3°) and state the number of carbons bonded to C-OH for each. 2 marks
2.2 All four compounds have the same molecular formula C4H10O and the same -OH group. Use the data to identify the relationship between branching and boiling point. 2 marks
2.3 2-methylpropan-2-ol is fully miscible with water while butan-1-ol is only partially miscible, despite having the same carbon count. Using IMF reasoning, explain this difference. 3 marks
3. Compare and contrast: alcohol class properties
Complete the two-column table comparing primary and tertiary alcohols on five criteria. The first row is done as an example. 8 marks (2 each row × 4 remaining rows)
| Feature | Primary alcohol (e.g. butan-1-ol) | Tertiary alcohol (e.g. 2-methylpropan-2-ol) |
|---|---|---|
| Example (done) Number of C bonded to C-OH | 1 other carbon | 3 other carbons |
| Molecular shape / branching | ||
| Relative boiling point (constitutional isomers) | ||
| Relative water solubility (same carbon count) | ||
| H-bond donor/acceptor ability |
4. Australian case study — ethanol in beverages and glycerol in cosmetics
Read the stimulus below, then answer the question. 5 marks
Stimulus. Ethanol (CH3CH2OH) is the basis of the Australian alcohol beverage industry, regulated under the Alcohol Beverages Advertising Code (ABAC). The Australian Bureau of Statistics reported that Australians consumed approximately 9.5 litres of pure alcohol per person in 2022–23. Ethanol is a primary alcohol with a boiling point of 78°C, fully miscible with water, and carries an OH group that enables it to interact readily with biological molecules — leading to its CNS depressant effects at moderate doses. Glycerol (propane-1,2,3-triol) is a tri-alcohol (three -OH groups) used extensively in Australian pharmacy products including hand creams and wound dressings because of its high viscosity and strong water-attracting properties. It has a boiling point of 290°C and is fully miscible with water. During the COVID-19 pandemic (2020–21), Australia saw a surge in domestic ethanol production for hand sanitiser, with the Australian government fast-tracking approvals for food-grade ethanol to be redirected to sanitiser manufacture.
Q4. Using lesson content, compare ethanol and glycerol in terms of their boiling points and water miscibility. In your response, identify the structural feature(s) responsible for glycerol’s dramatically higher boiling point (290°C vs 78°C) and explain why both compounds are fully miscible with water. 5 marks
5. Predict and justify — methanol toxicity
A Safe Work Australia safety bulletin warns workers handling methanol that it is fatally toxic via ingestion, inhalation, or skin absorption, while ethanol is consumed socially in moderate amounts without the same acute lethality. 4 marks
Q5. A student claims: “Methanol must be more toxic than ethanol because it has a smaller molecule and therefore stronger IMF with biological molecules.” Evaluate this claim. Identify what is wrong with the reasoning and provide the correct biochemical explanation for methanol’s specific toxicity (as described in the lesson). 4 marks
Q1.1 — Trend description
Boiling point increases steadily as carbon chain length increases from C1 (methanol, 65°C) to C5 (pentan-1-ol, 138°C). The trend is approximately linear.
Q1.2 — Ethanol vs ethane BP difference (3 marks)
Ethanol contains a hydroxyl group (-OH). The highly polar O-H bond makes the hydrogen a hydrogen bond donor (δ+) and the oxygen’s lone pairs hydrogen bond acceptors, enabling hydrogen bonds (~20–25 kJ/mol) between ethanol molecules [1]. Ethane has only C-H and C-C bonds; only weak London dispersion forces act between ethane molecules [1]. Breaking the extensive H-bond network in ethanol requires far more energy than overcoming dispersion forces in ethane, resulting in a boiling point 167°C higher despite ethanol and ethane having similar molecular masses [1].
Q1.3 — Decreasing gap trend (2 marks)
As chain length increases, both series gain progressively more molecular surface area, which increases London dispersion forces in both [1]. The alkane boiling points rise faster per carbon added (gaining purely from dispersion) than the alcohol series (which already has a large H-bonding baseline). The H-bond contribution from the single -OH is roughly constant, while the relative contribution of chain-length dispersion forces grows in both series, narrowing the gap [1]. Accept: the non-polar chain in the alcohol increasingly resembles the alkane, so the incremental benefit of each added CH2 unit is similar for both.
Q1.4 — Estimated BP of hexan-1-ol (2 marks)
Extrapolating the alcohol trend: from C4 to C5 the increase is ~20°C (118 to 138). Applying a similar increment to C6 gives approximately 155–158°C. Accept any value in the range 155–162°C [1]. Justification: the trend is approximately linear with each CH2 group adding roughly 20°C due to increasing dispersion forces [1]. (Actual value: 157°C.)
Q2.1 — Classification (2 marks)
Butan-1-ol: Primary (1°), C-OH bonded to 1 other carbon. 2-methylpropan-1-ol: Primary (1°), C-OH (terminal CH2OH) bonded to 1 other carbon. Butan-2-ol: Secondary (2°), C-OH bonded to 2 other carbons (C1 and C3). 2-methylpropan-2-ol: Tertiary (3°), C-OH bonded to 3 other carbons (three CH3 groups).
Q2.2 — Branching and BP (2 marks)
As branching increases, boiling point decreases: butan-1-ol (straight, 1°, 118°C) > 2-methylpropan-1-ol (branched, 1°, 108°C) > butan-2-ol (2°, 100°C) > 2-methylpropan-2-ol (3°, most branched, 83°C) [1]. This is because all four compounds have the same -OH group and equivalent H-bonding; the difference arises from dispersion forces, which decrease as branching produces a more compact, spherical molecular shape with less surface area [1].
Q2.3 — Water solubility difference (3 marks)
Both compounds have one -OH group that forms H-bonds with water (identical H-bond interaction) [1]. Butan-1-ol’s straight, extended chain must be accommodated within water’s H-bond network, disrupting a larger portion of water-water H-bonds than the -OH alone can compensate for — only partial miscibility [1]. 2-methylpropan-2-ol’s compact, branched spherical structure exposes less non-polar surface area to water, causing less disruption to water’s H-bond network. The single -OH can compensate for this smaller disruption — fully miscible [1]. Note: both have the same carbon count; shape (not just carbon number) determines the difference.
Q3 — Compare/contrast table
| Feature | Primary (butan-1-ol) | Tertiary (2-methylpropan-2-ol) |
|---|---|---|
| Molecular shape/branching | Extended straight chain; maximum molecular surface area | Compact, spherical; highly branched; minimum surface area |
| Relative BP (isomers) | Highest BP of C4H10O isomers (118°C); strongest dispersion forces | Lowest BP (83°C); weakest dispersion forces due to compact shape |
| Relative water solubility (same C count) | Partially miscible; extended chain disrupts more water H-bonds | Fully miscible; compact shape minimises disruption to water’s H-bond network |
| H-bond donor/acceptor ability | Same as all alcohols: O-H = H-bond donor (δ+ H); lone pairs on O = acceptor. Equivalent to tertiary. | Identical to primary: O-H donor, lone pairs acceptor. Classification does NOT affect H-bond ability. |
Q4 — Ethanol vs glycerol (5 marks)
Both ethanol and glycerol are fully miscible with water because each -OH group can form H-bonds with water (O-H donor; lone-pair acceptor) and their non-polar regions are small enough that -OH interactions compensate for disruption of water’s H-bond network [1]. Ethanol has one -OH group; glycerol has three -OH groups [1]. Glycerol’s dramatically higher boiling point (290°C vs 78°C) is due to its three -OH groups, each able to act as both a H-bond donor and acceptor [1]. This creates a much more extensive H-bond network between glycerol molecules — approximately three times as many H-bonds per molecule as ethanol — requiring far more energy to overcome [1]. Glycerol also has a short three-carbon chain with all three carbons bearing -OH groups, so dispersion forces are modest; the H-bonding from three -OH groups is the dominant factor [1].
Q5 — Methanol toxicity claim evaluation (4 marks)
The student’s claim is wrong [1]. The reason methanol is more toxic than ethanol is not related to IMF strength or molecular size — it is a metabolic pathway difference [1]. Both methanol and ethanol are metabolised by the same enzyme, alcohol dehydrogenase. Ethanol → ethanal (acetaldehyde) → ethanoic acid — products that are tolerable in small amounts. Methanol → methanal (formaldehyde) → methanoic acid (formic acid) [1]. Formaldehyde specifically destroys proteins in the optic nerve causing permanent blindness, and formic acid causes severe metabolic acidosis. It is the difference in metabolic products, not IMF, that explains the dramatically different biological effects of these two structurally similar molecules [1].