Chemistry • Year 12 • Module 7 • Lesson 20
Organic Reactions Mastery
Apply your knowledge of Module 7 reactions to real test data, industrial contexts, and multi-step reasoning. Includes data table and graph interpretation.
1. Identify unknown compounds from test results
A chemist tests five unknown organic compounds (A–E) using a series of standard chemical tests. Each compound has the molecular formula C4HxOy (x and y vary). The results are recorded below. 9 marks
| Compound | Tollens’ reagent | Fehling’s solution | K2Cr2O7/H2SO4 | Bromine water | Na2CO3 solution |
|---|---|---|---|---|---|
| A | Silver mirror formed | Brick-red ppt | Orange → green | No change | No fizzing |
| B | No reaction | No reaction | Orange → green | No change | No fizzing |
| C | No reaction | No reaction | No change (stays orange) | No change | No fizzing |
| D | No reaction | No reaction | Orange → green | No change | Fizzing (CO2 released) |
| E | No reaction | No reaction | No change (stays orange) | Decolourises | No fizzing |
1.1 Identify each compound A–E by functional group class. Justify your identification for compounds A, C, and D. 5 marks
1.2 A student claims compound B could be “either a secondary alcohol or a primary alcohol.” Assess this claim. What additional test(s) could be used to distinguish between them? 2 marks
1.3 Compound E decolourises bromine water. Write a balanced equation for the addition of Br2 across a C=C bond using but-1-ene as the specific example. State the name of the organic product. 2 marks
2. Interpret graph — ester yield vs temperature for industrial esterification
The graph below shows the equilibrium yield of ethyl ethanoate (%) as a function of reaction temperature for two industrial conditions: with catalyst and without catalyst. Data adapted from industrial process engineering reports for Australian ethanol-based esterification (e.g. Manildra Group, NSW). 8 marks
2.1 Describe the trend in ester yield with increasing temperature for each condition (with and without catalyst). 2 marks
2.2 Estimate the maximum yield achieved with catalyst, and the temperature at which it occurs. Explain, using the role of H2SO4 in this reaction, why the catalyst increases yield at all temperatures. 3 marks
2.3 At temperatures above ~100°C, the yield with catalyst begins to fall. Propose a reason for this, using your understanding of equilibrium and the nature of esterification. 3 marks
3. Case study — Australian industrial organic chemistry
Read the passage and answer the questions using your Module 7 chemistry knowledge. 8 marks
Stimulus. Australia’s organic chemical industry operates on feedstocks produced by three main pathways: petrochemical cracking (Qenos, Altona, Victoria — produces ethylene/ethene from naphtha), grain-based fermentation (Manildra Group, Nowra, NSW — produces ~65 million litres of fuel-grade ethanol per year from wheat starch), and biopharmaceutical synthesis (CSL, Parkville, Victoria — uses esterification and amide formation in drug manufacturing). The AWRI (Australian Wine Research Institute) uses controlled fermentation at ~18–22°C (lower than industrial 35°C) to preserve ester aromas. Ampol’s Lytton refinery (Brisbane) uses catalytic cracking and halogenation steps in fuel production.
3.1 Manildra Group produces ethanol via fermentation. Write the balanced equation for this reaction and identify all the specific conditions required. Why is the reaction carried out in a sealed vessel? 3 marks
3.2 The AWRI uses fermentation at ~18–22°C rather than the industrial ~35°C. Using your knowledge of reaction rates and equilibrium, explain the trade-off involved in choosing a lower temperature for wine ester production. 3 marks
3.3 Qenos produces ethene (ethylene) from naphtha. Outline one Module 7 addition reaction that ethene could undergo, including reagents, conditions, and the product name. 2 marks
4. Sequence the synthesis steps
The steps below describe a three-step synthesis of propyl ethanoate from ethanol and 1-bromopropane. They are presented in the wrong order. Write the correct step number (1, 2, 3) in the Order column. 6 marks (2 each)
| Order | Step description | Equation (balanced) | Conditions |
|---|---|---|---|
| Step X: React ethanoic acid with propan-1-ol in the presence of an acid catalyst under reflux. Collect the ester layer by separatory funnel. | CH3COOH + CH3CH2CH2OH ⇌ CH3COOCH2CH2CH3 + H2O | ||
| Step Y: Oxidise ethanol to ethanoic acid using excess acidified potassium dichromate under reflux. | CH3CH2OH + 2[O] → CH3COOH + H2O | ||
| Step Z: React 1-bromopropane with aqueous NaOH under reflux to produce the alcohol needed for esterification. | CH3CH2CH2Br + NaOH(aq) → CH3CH2CH2OH + NaBr |
For each step in the correct order, also write the complete conditions (reagent, catalyst, temperature, equipment) in the Conditions column above.
Stuck? Identify what you need for the final esterification step, then work backwards: you need both an acid AND an alcohol. Which starting materials give you these?Q1 — Identifying unknowns from test results
A = aldehyde (Tollens’ silver mirror + Fehling’s brick-red ppt = aldehyde; K2Cr2O7 orange→green confirms oxidation; no C=C as Br2 water unchanged).
B = primary or secondary alcohol (K2Cr2O7 orange→green = oxidised; no Tollens’/Fehling’s = not an aldehyde at the start; no CO2 = not a carboxylic acid; no C=C).
C = ketone OR tertiary alcohol (K2Cr2O7 stays orange = no oxidation; no positive test for aldehyde; no CO2; no C=C).
D = carboxylic acid (CO2 fizzing with Na2CO3 confirms acidic — carboxylic acid; K2Cr2O7 could stay or go green but the CO2 test is definitive).
E = alkene (Br2 water decolourises = C=C; K2Cr2O7 stays orange = no oxidisable functional group under these conditions).
1.2 The claim is partially correct but imprecise. B could be primary or secondary alcohol; both give K2Cr2O7 colour change. To distinguish: add Tollens’ reagent or Fehling’s solution after oxidation — if primary, oxidation first gives an aldehyde (positive Tollens’); if secondary, oxidation gives a ketone (no Tollens’/Fehling’s response). Alternatively, identify the carbon skeleton by looking at the oxidation product: primary → aldehyde (can be further oxidised); secondary → ketone (dead end).
1.3 CH3CH2CH=CH2 + Br2 → CH3CH2CHBrCH2Br. Product: 1,2-dibromobutane (room temperature, no catalyst, fume cupboard).
Q2 — Graph interpretation
2.1 With catalyst: yield rises from ~52% at 20°C to a peak of ~67% at ~80°C, then decreases to ~50% at 170°C. Without catalyst: yield rises from ~28% at 20°C to a peak of ~44% at ~80°C, then decreases similarly. Both curves show the same overall shape but the catalysed reaction achieves significantly higher yields at all temperatures.
2.2 Maximum yield ~67% at ~80°C. Conc. H2SO4 acts as both an acid catalyst (H+ activates the carbonyl group of the carboxylic acid, speeding both forward and reverse rates) and a dehydrating agent (absorbs the H2O produced), which shifts the equilibrium to the right (Le Chatelier’s principle) — increasing yield at all temperatures.
2.3 Above ~100°C, the equilibrium begins to shift back toward reactants because the reverse hydrolysis reaction is endothermic (absorbs energy) — increasing temperature provides energy that favours the reverse reaction. Additionally, very high temperatures can dehydrate the alcohol or cause other side reactions, reducing ester yield. The net effect is a decrease in equilibrium yield at higher temperatures for this exothermic esterification reaction.
Q3 — Case study answers
3.1 C6H12O6 → 2C2H5OH + 2CO2. Conditions: yeast (zymase enzyme) as catalyst, ~35°C, anaerobic (no O2). Sealed vessel: CO2 is produced and O2 must be excluded — oxygen would allow yeast to respire aerobically, converting ethanol to CO2 and H2O rather than producing it. The sealed vessel also prevents oxidation of ethanol to acetic acid by Acetobacter bacteria in the presence of air.
3.2 Lower temperature (~18–22°C): slower rate of fermentation (longer production time) but the yeast enzymes work more selectively, producing more of the desired ester and flavour compounds rather than off-flavours. At 35°C (industrial), the rate is faster but selectivity for specific aroma esters is lower and the risk of yeast stress and off-flavour production is higher. This is a trade-off between rate (speed of production) and selectivity (quality of product). For winemaking, quality outweighs speed.
3.3 Example — hydration: CH2=CH2 + H2O ⇌ CH3CH2OH. Conditions: H3PO4 catalyst, ~300°C, high pressure (~65 atm), high-pressure reactor. Product: ethanol. Alternatively: hydrogenation (H2, Ni, ~200°C → ethane); or halogenation (Br2, r.t. → 1,2-dibromoethane).
Q4 — Correct sequence and conditions
Correct order: Z (Step 1) → Y (Step 2) → X (Step 3)
Step 1 (Z) conditions: Reagent = NaOH(aq) (aqueous NaOH); catalyst = none; conditions = heat under reflux; equipment = round-bottom flask, reflux condenser.
Step 2 (Y) conditions: Reagent = K2Cr2O7/H2SO4 (excess, acidified); catalyst = H2SO4; conditions = heat under reflux; equipment = round-bottom flask, reflux condenser. (Observable: orange → green.)
Step 3 (X) conditions: Catalyst = conc. H2SO4 (a few drops); conditions = heat under reflux; arrow type = ⇌ (reversible); yield <100%.