Chemistry • Year 12 • Module 7 • Lesson 6
Reactions of Alkenes
Apply the four alkene addition reactions to real data, Australian industrial contexts, and multi-step reasoning — including Markovnikov’s rule, the bromine water test, and hydration equilibrium.
1. Reaction conditions comparison table
The table below lists experimental observations for four reactions labelled P–S. Study the data and answer the questions that follow. 8 marks
| Reaction | Reagent added to alkene | Catalyst used | Temperature | Observation |
|---|---|---|---|---|
| P | H2(g) | None | 25°C (room temp.) | No detectable reaction |
| Q | H2(g) | Ni powder | 180°C | Rapid reaction; product is bromine-inactive |
| R | H2O (steam) | H3PO4 | 300°C, 65 atm | Alcohol produced (confirmed by Lucas test) |
| S | H2O (steam) | H3PO4 | 300°C, 1 atm (atmospheric) | Very little alcohol detected; mostly unreacted alkene |
1.1 Identify the reaction type for Reaction Q and explain why the same reagents at room temperature (Reaction P) produced no reaction. 2 marks
1.2 Identify the reaction type for Reaction R. Write a balanced equation for the reaction of ethene (CH2=CH2) under the conditions described, and name the product. 3 marks
1.3 Compare Reactions R and S. Using Le Chatelier’s principle, explain why changing only the pressure from 65 atm to 1 atm significantly reduces the alcohol yield. 3 marks
2. Interpret the graph — ethanol yield vs reaction pressure
The graph below shows the equilibrium yield of ethanol produced by the industrial hydration of ethene (CH2=CH2 + H2O ⇌ CH3CH2OH) as a function of reaction pressure at a fixed temperature of 300°C with H3PO4 catalyst. 7 marks
2.1 Describe the trend in equilibrium ethanol yield as pressure increases from 0 to 80 atm. 2 marks
2.2 Estimate the ethanol yield at the industrial operating pressure of 65 atm from the graph, and read off the approximate yield at 10 atm. Calculate the difference. 2 marks
2.3 Use Le Chatelier’s principle to explain the shape of the curve — why does increasing pressure increase ethanol yield in this reaction? Write the balanced equation as part of your response. 3 marks
3. Australian industrial context — case study
Read the information below and answer the questions. 5 marks
Context. The Manildra Group operates a fuel-grade and industrial-grade ethanol production facility at Nowra, NSW. Unlike the direct hydration route used in petrochemical plants overseas, Manildra’s ethanol is produced by fermentation of wheat starch — a renewable feedstock — then rectified (distilled) to high purity. By contrast, Goodman Fielder and similar manufacturers use the catalytic hydrogenation of vegetable oils (rich in unsaturated C=C bonds from oleic and linoleic acid chains) to produce the semi-solid fats used in margarines sold throughout Australia. The partial hydrogenation deliberately leaves some C=C bonds unreacted to control texture.
3.1 The passage describes two industrial processes involving alkene reactions. Identify which reaction type (from the four studied in this lesson) is used in each industry and write a representative balanced equation for each. 2 marks
3.2 Goodman Fielder’s process uses “partial hydrogenation” — deliberately leaving some C=C bonds unreacted. Predict what would happen to the melting point of the margarine if all C=C bonds were fully hydrogenated. Justify your prediction using lesson content. 2 marks
3.3 Manildra uses fermentation, not direct hydration of ethene. Identify one advantage and one disadvantage of the fermentation route compared to direct hydration for ethanol production. 1 mark
4. Sequence the steps — applying Markovnikov’s rule
The five steps below describe how to apply Markovnikov’s rule to find the major product of HBr addition to but-1-ene (CH3CH2CH=CH2). They are listed out of order. Write the correct order (1 to 5) in the “Order” column. 5 marks
| Order | Step |
|---|---|
| Write the balanced equation with the major product: CH3CH2CH=CH2 + HBr → CH3CH2CHBrCH3 and name the product: 2-bromobutane. | |
| Count H atoms on each carbon of the C=C: C1 (=CH2) has 2 H atoms; C2 (=CH-) has 1 H atom. | |
| Identify the C=C double bond in but-1-ene: it is between carbon 1 (C1) and carbon 2 (C2) of the four-carbon chain. | |
| Apply Markovnikov’s rule: H adds to C1 (more H atoms); Br adds to C2 (fewer H atoms, more substituted). | |
| Confirm that the major product (2-bromobutane) has Br on the secondary carbon, making it more stable than the primary-carbon alternative (1-bromobutane). |
Q1.1 — Reaction type for Q; why P fails (2 marks)
Reaction Q is hydrogenation. [1] At room temperature (Reaction P), the activation energy for H–H bond breaking is too high and H2 cannot adsorb effectively to the alkene surface without the Ni catalyst. The catalyst lowers the activation energy by providing a surface for adsorption; without it and without elevated temperature, the rate is negligible. [1]
Q1.2 — Reaction R: type, equation, product name (3 marks)
Reaction type: hydration. [1] Balanced equation: CH2=CH2 + H2O → CH3CH2OH [1]. Product name: ethanol (ethan-1-ol). [1]
Q1.3 — Le Chatelier and pressure (3 marks)
The hydration equation is: CH2=CH2(g) + H2O(g) ⇌ CH3CH2OH(g). [1] There are 2 moles of gas on the left and 1 mole on the right. By Le Chatelier’s principle, increasing pressure favours the side with fewer moles of gas — i.e. the product side — shifting the equilibrium right and increasing ethanol yield. [1] At 1 atm (Reaction S), the pressure is insufficient to drive the equilibrium toward the product; the reverse reaction (dehydration) is competitive, so very little ethanol accumulates. [1]
Q2.1 — Trend description (2 marks)
As pressure increases from 0 to 80 atm, the equilibrium ethanol yield increases in a positive but decelerating (concave-down) curve [1] — the yield rises steeply at low pressures and then approaches a plateau at very high pressures, suggesting diminishing returns [1].
Q2.2 — Yield estimates and difference (2 marks)
At 65 atm: approximately 46% (accept 43–49%). At 10 atm: approximately 8% (accept 6–12%). Difference: approximately 38 percentage points (accept 32–43 pp). [1 mark for each reading; accept ±3 pp from the model curve.]
Q2.3 — Le Chatelier explanation (3 marks)
Balanced equation: CH2=CH2(g) + H2O(g) ⇌ CH3CH2OH(g). [1] There are 2 mol gas on the left and 1 mol on the right. Le Chatelier’s principle: when pressure is increased, the system responds by shifting the equilibrium to reduce the pressure — i.e. toward the side with fewer moles of gas [1] — which is the product (ethanol) side, so yield increases. The plateau effect at very high pressures reflects the system approaching complete conversion, limited also by practical constraints such as reactor cost and competing side reactions. [1]
Q3.1 — Reaction types and equations (2 marks)
Manildra — fermentation (not an alkene addition; note: the lesson contrasts this with hydration). If the question intends the alternative petrochemical route: hydration, e.g. CH2=CH2 + H2O → CH3CH2OH (H3PO4, 300°C, high pressure). [1] Goodman Fielder — hydrogenation: for a representative unsaturated fatty acid chain with one C=C: —CH=CH— + H2 → —CH2—CH2— (Ni catalyst, ~150–200°C). [1]
Q3.2 — Full hydrogenation and melting point (2 marks)
Full hydrogenation would remove all C=C bonds from the fatty acid chains, converting the unsaturated oil completely to saturated fat [1]. Saturated fatty acid chains pack more closely due to their linear shape, increasing intermolecular dispersion forces and raising the melting point significantly — the product would be a hard, waxy solid rather than a semi-solid spread [1].
Q3.3 — Fermentation vs hydration (1 mark)
Advantage of fermentation: uses renewable plant-based glucose/starch; does not depend on crude oil-derived ethene. Disadvantage: slow (batch process), lower purity (~15% — must distil to increase concentration), not readily scalable to the same throughput as continuous hydration processes. [1 mark for one valid advantage AND one valid disadvantage.]
Q4 — Markovnikov sequence (5 marks: 1 each correct position)
Correct order: Step 3 → Step 2 → Step 4 → Step 5 → Step 1 (in table row order: Identify C=C → Count H → Apply rule → Confirm stability → Write equation). [1 mark per correctly placed step, 5 marks total.]