Chemistry • Year 12 • Module 7 • Lesson 12
Reactions of Alcohols
Apply dehydration, substitution, and oxidation of alcohols to data tables, conditions summaries, and real-world Australian contexts.
1. Conditions summary table — complete the blanks
The table below summarises the three alcohol reactions from this lesson. Fill in every blank cell. Use precise chemical names and terms. 12 marks
| Reaction type | Starting material → product | Reagent / catalyst | Temperature / equipment | Observable change | Arrow type |
|---|---|---|---|---|---|
| Dehydration | Alcohol → (1.1) ___________ | (1.2) ___________ | (1.3) ___________; distillation | Colourless gas evolved; alkene decolourises Br2/CCl4 | (1.4) ___________ |
| Substitution with HX | Alcohol + HBr → (1.5) ___________ + H2O | (1.6) ___________ | Heat under reflux | Separate organic layer forms; haloalkane odour | → |
| Oxidation (1° → aldehyde) | Primary alcohol → (1.7) ___________ | K2Cr2O7/H2SO4 | (1.8) ___________ | (1.9) ___________ | → |
| Oxidation (1° → carboxylic acid) | Primary alcohol → (1.10) ___________ | Excess K2Cr2O7/H2SO4 | Reflux | Orange → green | → |
| Oxidation (2° → ketone) | Secondary alcohol → (1.11) ___________ | K2Cr2O7/H2SO4 | Reflux | Orange → green | → |
| Oxidation (3°) | Tertiary alcohol → (1.12) ___________ | K2Cr2O7/H2SO4 | Any | Stays orange | — |
2. Interpret oxidation test data — identifying unknown alcohols
A chemistry class tested five unknown alcohols (A–E) with acidified K2Cr2O7 solution under the conditions shown. All alcohols have molecular formula C4H10O. Results are recorded below. 10 marks
| Alcohol | Test 1: K2Cr2O7/H+, distillation | Test 2: K2Cr2O7/H+, reflux | Product (Test 2) | Alcohol class |
|---|---|---|---|---|
| A | Orange → green; fruity-smelling distillate | Orange → green | Carboxylic acid | (2.1) ___ |
| B | Stays orange | Stays orange | No reaction | (2.2) ___ |
| C | Orange → green; distillate collected | Orange → green | Carboxylic acid | (2.3) ___ |
| D | Orange → green; no separate distillate noted | Orange → green | Ketone | (2.4) ___ |
| E | Stays orange | Stays orange | No reaction | (2.5) ___ |
Source: simulated class experiment data.
2.6 Identify the alcohol class of A, B, C, D, and E. Explain how the data in Tests 1 and 2 together distinguish between a primary, secondary, and tertiary alcohol. 4 marks
2.7 Alcohols B and E both have molecular formula C4H10O and both gave no reaction with K2Cr2O7/H+. Explain why at least two different structural isomers would give these results, and name one example of each. 2 marks
3. Australian breathalyser chemistry — graph interpretation
Australian police use the acidified dichromate breathalyser reaction: ethanol in breath reduces orange Cr2O72− to green Cr3+. The device measures absorbance of orange light (430 nm) after the reaction — higher ethanol means more Cr2O72− is reduced, so less orange light is absorbed (lower absorbance at 430 nm). The graph below shows absorbance at 430 nm plotted against ethanol concentration in mg per 100 mL of breath for calibration samples. 8 marks
3.1 Describe the relationship between ethanol concentration in breath and absorbance at 430 nm shown in Figure 3.1. 2 marks
3.2 Use Figure 3.1 to estimate the absorbance reading that corresponds to the Australian legal breath-alcohol limit of 0.05 mg/100 mL. 1 mark
3.3 The breathalyser uses acidified K2Cr2O7 to oxidise ethanol. State the colour change observed when ethanol is present and explain, using your knowledge of alcohol oxidation, what happens to (a) the ethanol and (b) the dichromate species during the reaction. Identify which species is the oxidising agent and which is the reducing agent. 3 marks
3.4 A breath sample gives an absorbance of 0.20. Use Figure 3.1 to estimate whether the driver is above or below the Australian legal limit, and justify your answer quantitatively. 2 marks
4. Case study — Manildra Group ethylene production, Nowra NSW
Read the scenario below and answer the question. 5 marks
The Manildra Group facility at Nowra, New South Wales, produces ethylene from bio-ethanol derived from wheat starch. Ethanol vapour is passed over an acid catalyst at approximately 170°C. The ethylene produced is used to manufacture bio-based polyethylene and other industrial chemicals. In 2022 the facility had a production capacity of approximately 50 000 tonnes of ethylene per year. Because the feedstock is agricultural (wheat) rather than petroleum-derived, the process is considered a lower-carbon pathway to ethylene compared to steam-cracking of fossil fuels.
4.1 Write a balanced equation for the industrial dehydration of ethanol to ethylene, including the conditions used. State the reaction type and explain why concentrated acid at 170°C is needed rather than dilute acid at lower temperature. 5 marks
Q1 — Conditions table
1.1 alkene • 1.2 conc. H2SO4 (or conc. H3PO4) • 1.3 ~170°C • 1.4 → (one-way arrow) • 1.5 bromoalkane / haloalkane • 1.6 HBr (concentrated aqueous or gas) • 1.7 aldehyde • 1.8 warm / distillation • 1.9 orange → green; aldehyde distillate collected • 1.10 carboxylic acid • 1.11 ketone • 1.12 no reaction
Q2 — Unknown alcohol identification
2.1 A: primary (1°) — gives aldehyde on distillation, then carboxylic acid on reflux; both tests orange→green.
2.2 B: tertiary (3°) — no colour change in either test; no H on C–OH carbon.
2.3 C: primary (1°) — same pattern as A; a different structural isomer with molecular formula C4H10O.
2.4 D: secondary (2°) — gives colour change in both tests; Test 2 product is ketone (not further oxidised), no separate distillate in Test 1 because ketone boiling point higher than aldehyde and not removed selectively.
2.5 E: tertiary (3°) — same as B.
2.6 (4 marks): Test 1 (distillation) distinguishes 1° from 2°: only the 1° aldehyde has sufficiently low boiling point to be collected as a separate distillate under these conditions (1 mark). A colour change in Test 1 or Test 2 confirms the alcohol is oxidisable, i.e. 1° or 2° (1 mark). If both tests show no colour change, the alcohol is tertiary (1 mark). Final product in Test 2 determines 1° (carboxylic acid) vs 2° (ketone), because ketones cannot be further oxidised (1 mark).
2.7: The two C4H10O tertiary alcohols that resist oxidation are: 2-methylpropan-2-ol (tert-butanol, the only tertiary alcohol with this formula) (1 mark). One mark only available — there is actually only one tertiary C4H10O alcohol. Accept: explanation that both B and E are structural isomers both happening to be tertiary; or that one might be 2-methylpropan-2-ol and the other is stated correctly as the same compound. Award mark for recognising both results indicate tertiary alcohol and naming 2-methylpropan-2-ol (1 mark) and acknowledging it is the same compound / only one tertiary isomer exists (1 mark).
Q3 — Breathalyser graph
3.1: As ethanol concentration in breath increases, absorbance at 430 nm decreases. The relationship is approximately linear (inverse proportion). This is because greater ethanol concentration reduces more Cr2O72− (orange) to Cr3+ (green), leaving less orange species to absorb 430 nm light. (2 marks: 1 for describing the inverse trend; 1 for quantitative support or chemical explanation.)
3.2: Reading from Figure 3.1 at x = 0.05, absorbance ≈ 0.30 (accept 0.28–0.32). (1 mark)
3.3: Colour change: orange → green [1 mark]. (a) Ethanol (a primary alcohol) is oxidised by the acidified dichromate: the –OH group loses hydrogen atoms to form the C=O group; ethanol is first oxidised to ethanal (aldehyde) and then to ethanoic acid (carboxylic acid) under excess oxidant (reflux conditions in the lab equivalent) [1 mark — accept “ethanol is oxidised to a carboxylic acid or ethanoic acid”]. (b) Cr2O72− (orange dichromate ion) is reduced to Cr3+ (green chromium(III) ion); the colour change from orange to green signals this reduction [1 mark — identifying the change and naming both species]. The oxidising agent is Cr2O72−; the reducing agent is ethanol. Total 3 marks.
3.4: From Figure 3.1, absorbance 0.20 corresponds to approximately x = 0.06 mg/100 mL. This is above the Australian legal limit of 0.05 mg/100 mL, so the driver is over the limit. (2 marks: 1 for reading off ~0.06; 1 for comparing to 0.05 and concluding over limit.)
Q4 — Manildra ethylene production
Balanced equation: CH3CH2OH → CH2=CH2 + H2O (1 mark for balanced equation including H2O as product).
Conditions: conc. H2SO4 (or acid catalyst), ~170°C (1 mark for both conditions stated).
Reaction type: dehydration (elimination) (1 mark).
Why concentrated acid at 170°C: Concentrated acid acts as both a catalyst (protonates the –OH group) and a dehydrating agent that absorbs the water produced, pushing equilibrium toward the alkene (1 mark). At lower temperatures or with dilute acid, the reverse reaction (hydration, addition of water across C=C) is thermodynamically favoured; high temperature shifts the equilibrium toward the elimination products (1 mark). Total 5 marks.