Chemistry • Year 12 • Module 7 • Lesson 19
Organic Reaction Pathways: Synthesis & Multi-Step Problems
Apply the four-step algorithm to pathway problems using real data, a reaction grid, and Australian industry contexts.
1. Reaction pathway grid — fill in the missing reagents and products
The table below shows a multi-step synthesis. Complete every shaded cell: either name the missing compound or write the reagents/conditions for the missing step. The carbon skeleton is 4 carbons throughout. 12 marks (1 mark per shaded cell)
| Step | Starting compound (name) | Reagents & conditions | Product name | Functional group class of product |
|---|---|---|---|---|
| 1 | 1-Chlorobutane | __________ reagent, __________ conditions | Butan-1-ol | __________ |
| 2 | Butan-1-ol | K₂Cr₂O₅/H₂SO₄, distillation | __________ | Aldehyde |
| 3 | __________ | K₂Cr₂O₅/H₂SO₄ (excess), reflux | Butanoic acid | Carboxylic acid |
| 4 | Butanoic acid + Butan-1-ol | __________ | __________ | Ester |
| 5 | __________ (name the ester from Step 4) | Conc. NaOH(aq), reflux, heat | __________ + __________ | Carboxylate salt + Alcohol |
| 6 | 1-Bromobutane | __________ | Butan-1-ol | Alcohol |
Note on Step 1: 1-Chlorobutane is a haloalkane. Which reagent and conditions convert a haloalkane to a primary alcohol?
2. Data table — step yield analysis in biodiesel synthesis
Biodiesel is produced in Australia from vegetable oil (a triglyceride ester) by a process called transesterification. A Bundaberg, Queensland, pilot facility tested four synthesis routes, varying the catalyst and temperature for the key esterification step. The table below shows the conditions and ester yield data for each route after 60 minutes of reaction. 10 marks
| Route | Catalyst | Temperature (°C) | Alcohol:Acid molar ratio | Ester yield after 60 min (%) |
|---|---|---|---|---|
| A | Conc. H₂SO₄ | 60 | 1:1 | 42 |
| B | Conc. H₂SO₄ | 60 | 3:1 | 67 |
| C | Conc. H₂SO₄ | 80 | 3:1 | 78 |
| D | NaOH (base cat.) | 60 | 3:1 | 88 |
Source: hypothetical pilot data, after Demirbas (2009) — Biodiesel: A Realistic Fuel Alternative for Diesel Engines, Springer.
2.1 Describe the trend in ester yield when the alcohol:acid molar ratio increases from 1:1 to 3:1 (compare Routes A and B). 2 marks
2.2 Using Le Chatelier’s principle, explain why a higher alcohol:acid ratio increases the ester yield in an equilibrium reaction. 3 marks
2.3 Compare Routes C and D. Route D uses a base catalyst (NaOH) instead of acid. Identify the most significant difference in yield and propose one reason why base catalysis may be more effective for this industrial application. 3 marks
2.4 A student claims that the Route D saponification product from NaOH-catalysed esterification is the same ester as the acid-catalysed routes. Assess this claim. 2 marks
3. Sequence the synthesis steps
The seven events below describe the synthesis of ethyl ethanoate starting from ethene, but they have been shuffled. Write the correct order (1–7) in the “Order” column. 6 marks — 1 per correctly placed step
| Step description (shuffled) | Your order (1–7) |
|---|---|
| Reflux ethanol with excess K₂Cr₂O₅/H₂SO₄ to produce ethanoic acid. | |
| Add steam across the C=C double bond of ethene using H₂PO₄ catalyst at 300°C to produce ethanol. | |
| The orange K₂Cr₂O₅ solution turns green, confirming oxidation has occurred. | |
| Add conc. H₂SO₄ catalyst and heat both compounds under reflux; write a reversible arrow in the equation. | |
| Identify the target: ethyl ethanoate = ethanoate part (from ethanoic acid) + ethyl part (from ethanol). | |
| Mix ethanoic acid with ethanol (both derived from original ethanol starting material). | |
| Collect the ester layer; yield is approximately 65% because the reaction is reversible. |
4. Case study — aroma compound synthesis for Australian food industry
Isoamyl acetate (3-methylbutyl ethanoate, CH₃COOCH₂CH₂CH(CH₃)₂) is responsible for the banana flavour used extensively in Australian food manufacturing, including confectionery produced in Sydney and Melbourne. CSL and Sigma–Aldrich Australia supply this compound for pharmaceutical excipients and flavour formulation. A synthetic route from 3-methylbutan-1-ol (isoamyl alcohol, derived from fermentation of wort in yeast metabolism) is outlined below. 7 marks
Synthesis outline: 3-methylbutan-1-ol → [Step A] → ethanoic acid + 3-methylbutan-1-ol → [Step B] → isoamyl acetate + H₂O
Note: ethanoic acid used in Step B is purchased separately (not synthesised from 3-methylbutan-1-ol in this route).
4.1 Identify the functional group class of 3-methylbutan-1-ol and state its IUPAC suffix. 1 mark
4.2 Step A is not shown in the overall outline above. If 3-methylbutan-1-ol were to be oxidised to the corresponding carboxylic acid (3-methylbutanoic acid), write the reagents and conditions required, and explain which piece of apparatus is used and why. 3 marks
4.3 Step B is the esterification of ethanoic acid with 3-methylbutan-1-ol. Write the balanced equation for this reaction using full structural names, state all conditions, and explain why the yield of isoamyl acetate is less than 100%. 3 marks
Q1 — Reaction pathway grid
Step 1 reagents/conditions: NaOH(aq), reflux (nucleophilic substitution of C–Cl bond by OH¯). Functional group class: alcohol.
Step 2 product: Butanal (an aldehyde).
Step 3 starting compound: Butanal.
Step 4 reagents/conditions: Conc. H₂SO₄ catalyst, heat under reflux. Reversible reaction (⇌). Product: butyl butanoate.
Step 5 starting compound: Butyl butanoate. Products: sodium butanoate (CH₃CH₂CH₂COO▿Na▿) + butan-1-ol. (Saponification is irreversible.)
Step 6 reagents/conditions: NaOH(aq), reflux.
Q2 — Data table: biodiesel synthesis
2.1 Increasing the molar ratio from 1:1 to 3:1 increases the ester yield from 42% to 67% (an increase of 25 percentage points) at 60°C with H₂SO₄ catalyst. The trend is a positive relationship between alcohol excess and yield.
2.2 Esterification is a reversible reaction: acid + alcohol ⇌ ester + water. Adding excess alcohol increases the concentration of a reactant. By Le Chatelier’s principle, the system responds by shifting the equilibrium position to the right (toward products) to reduce the increased alcohol concentration. This produces more ester before a new equilibrium is reached, raising the yield.
2.3 Route D (NaOH, 60°C, 3:1) gives 88% yield vs Route C (H₂SO₄, 80°C, 3:1) at 78% — a 10 percentage point advantage at a lower temperature. Base catalysis is more effective here because NaOH can saponify free fatty acids in the oil feedstock, reducing competing side reactions, and because the base catalyst can operate at lower temperature, reducing energy costs for the facility. (Note: NaOH-catalysed transesterification is the standard industrial biodiesel process.)
2.4 The claim is partially correct. The forward reaction product is the same ester (butyl butanoate or the equivalent fatty acid ester). However, in NaOH-catalysed transesterification (saponification in reverse), the product is actually a carboxylate salt (RCOO▿Na▿) + glycerol rather than the free ester + water. The conditions fundamentally differ: acid catalysis gives a reversible equilibrium yielding free ester; NaOH-catalysed transesterification is irreversible, producing the sodium carboxylate salt — so the products are not identical.
Q3 — Sequence the synthesis steps (ethene to ethyl ethanoate)
Correct order: 5 → 2 → 3 → 1 → 6 → 4 → 7
Rationale: First identify the target product and its components (5), then produce ethanol by hydration of ethene (2), confirm oxidation by colour change observation (3), then oxidise ethanol to ethanoic acid under reflux with excess K₂Cr₂O₅ (1), combine both reagents (6), run esterification under H₂SO₄/reflux (4), collect the ester product with ~65% yield (7).
Q4 — Aroma compound synthesis (isoamyl acetate)
4.1 3-Methylbutan-1-ol is an alcohol. IUPAC suffix: –ol.
4.2 Reagents: K₂Cr₂O₅/H₂SO₄ (excess). Conditions: heat, reflux condenser. Reflux is used (not distillation) because the goal is full oxidation to the carboxylic acid (3-methylbutanoic acid): the condenser returns vapour to the flask, keeping the intermediate aldehyde in contact with excess oxidant until complete oxidation occurs. Distillation would prematurely remove the aldehyde intermediate. K₂Cr₂O₅ colour change: orange → green.
4.3 Balanced equation: ethanoic acid + 3-methylbutan-1-ol ⇌ 3-methylbutyl ethanoate (isoamyl acetate) + water. Conditions: concentrated H₂SO₄ catalyst, heat under reflux. Yield is less than 100% because esterification is a reversible equilibrium reaction — at equilibrium, both the forward (ester formation) and reverse (hydrolysis) reactions occur simultaneously. The system reaches a position where both reactants and products coexist; the yield is determined by the equilibrium constant (typically ~65% for simple esters without yield-enhancement strategies).