Chemistry • Year 12 • Module 7 • Lesson 19
Organic Reaction Pathways: Synthesis & Multi-Step Problems
Synthesise, evaluate and justify multi-step reaction pathways using scenario data and source critique. Band 5–6 extended reasoning.
1. Aspirin synthesis — data, scenario and multi-criteria evaluation
Scenario: Aspirin (acetylsalicylic acid) is one of Australia’s most widely dispensed pharmaceutical compounds. The CSL Behring manufacturing facility in Broadmeadows, Victoria, evaluates three synthesis routes for the acetylation of salicylic acid to produce aspirin. The key step in each route involves esterification of the phenolic –OH group of salicylic acid with an acylating agent. The table below summarises yield and process data for each route after pilot-scale trials.
| Route | Acylating agent | Catalyst | Temperature (°C) | Reaction time (min) | Crude yield (%) | Purity after workup (%) | Waste generated |
|---|---|---|---|---|---|---|---|
| 1 | Acetic anhydride | Conc. H₂SO₄ | 85 | 20 | 72 | 94 | Acetic acid, H₂SO₄ waste |
| 2 | Acetic anhydride | Phosphoric acid | 75 | 30 | 68 | 97 | Acetic acid, phosphate waste |
| 3 | Ethanoic acid (glacial) | Conc. H₂SO₄ | 85 | 60 | 48 | 91 | H₂SO₄ waste, water (from esterification) |
Source: hypothetical pilot data. Aspirin production chemistry adapted from Clayden et al., Organic Chemistry (2nd ed., 2012), Oxford.
Extended response (8 marks)
Evaluate which of the three synthesis routes is most appropriate for large-scale pharmaceutical manufacture of aspirin. In your response you must:
- Explain why Route 3 uses a longer reaction time than Routes 1 and 2, linking your answer to equilibrium principles and the nature of the acylating agent.
- Compare Routes 1 and 2 on three criteria drawn from the data (yield, purity, waste), and justify which is preferable for a pharmaceutical application where product purity is the primary concern.
- Use the concept of overall yield in multi-step synthesis to explain why even a 4% difference in purity (91% vs 95%) becomes significant when aspirin is produced over 100 steps in a broader synthesis chain.
- Reach an evidence-based judgement identifying the most appropriate route for CSL’s facility and explain one practical limitation of that route.
2. Source critique — student synthesis plan
Student Chemistry Study Notes — contains potential errors
“To synthesise propyl propanoate starting only from propan-1-ol, I only need two steps. In Step 1, I oxidise propan-1-ol with K₂Cr₂O₅/H₂SO₄ under reflux to get propanal. In Step 2, I react propanal directly with propan-1-ol under concentrated H₂SO₄ catalyst and reflux to get propyl propanoate. This works because esters are made from an aldehyde and an alcohol.”
HSC Year 12 Chemistry student, Band 4 exam response, October 2025 (hypothetical).
Source critique (7 marks)
Identify all scientific errors in the student’s synthesis plan. For each error: (a) state exactly what is wrong; (b) explain the correct chemistry; (c) provide the corrected step(s) that would give propyl propanoate from propan-1-ol only. Your response must also include a corrected, complete synthesis plan with all reagents, conditions and intermediate names.
Q1 — Aspirin synthesis evaluation: marking criteria (8 marks)
Criterion 1 — Route 3 reaction time (2 marks)
- 1 mark: Ethanoic acid is a weaker acylating agent than acetic anhydride. It must react via the equilibrium esterification mechanism (acid + phenol ⇌ ester + water), which is inherently slower because the reverse reaction also occurs, and the equilibrium position lies far to the left without yield-enhancement strategies.
- 1 mark: Acetic anhydride (Routes 1 and 2) drives the reaction to completion because the by-product (acetic acid) does not reversibly re-form the anhydride under these conditions — the reaction is essentially irreversible, so 100% conversion of reactant is theoretically achievable in a shorter time. Le Chatelier does not apply in the same way.
Criterion 2 — Comparison of Routes 1 vs 2 on three criteria (3 marks; 1 per criterion)
- Yield: Route 1 (72%) > Route 2 (68%) — Route 1 is higher yield by 4 percentage points; advantage to Route 1 in terms of raw material efficiency.
- Purity: Route 2 (97%) > Route 1 (94%) — Route 2 is 3 percentage points purer after workup; advantage to Route 2. For a pharmaceutical application requiring high purity (safety, regulatory compliance), Route 2 is preferable on this criterion.
- Waste: Both generate acetic acid and an acid catalyst waste stream. Route 1’s H₂SO₄ waste is more corrosive and more difficult to neutralise safely than phosphate waste from Route 2; advantage to Route 2 in industrial waste management.
- Justified preference: Route 2 for pharmaceutical manufacture because purity is the primary concern — even a 3% difference in purity translates to significant impurity load at the commercial scale, and pharmaceutical regulatory frameworks (TGA, ARTG) require specific purity thresholds.
Criterion 3 — Overall yield compounding (1 mark)
- Overall yield = product of individual step yields. Over 100 hypothetical steps, Route 1 (94% purity/step) yields 0.94¹²⁰ = approximately 0.24% of theoretical yield; Route 2 (97%) yields 0.97¹²⁰ = approximately 5.2%. A 3% per-step purity advantage compounds to a 21-fold improvement in overall yield at 100 steps — the difference is transformative at industrial scale.
Criterion 4 — Judgement with limitation (2 marks)
- 1 mark: Route 2 is the most appropriate — lower temperature (75°C), highest purity (97%), manageable waste stream, and acceptable yield (68%).
- 1 mark: Practical limitation — phosphoric acid catalyst is less commercially available in high-concentration form than H₂SO₄, increasing reagent cost; or: longer reaction time (30 min vs 20 min) reduces throughput in a batch process. Accept any chemically valid limitation specific to Route 2.
Q2 — Source critique: marking criteria (7 marks)
Error 1 (2 marks): Reflux instead of distillation in Step 1 gives carboxylic acid, not aldehyde
- 1 mark: The student states reflux with K₂Cr₂O₅/H₂SO₄ gives propanal (an aldehyde). This is wrong. Reflux with excess oxidant gives propanoic acid (a carboxylic acid), not propanal. To stop at the aldehyde, distillation must be used to remove the volatile propanal (BP 49°C) before further oxidation can occur.
- 1 mark: Corrected Step 1: propan-1-ol + K₂Cr₂O₅/H₂SO₄, distillation → propanal. OR: propan-1-ol + K₂Cr₂O₅/H₂SO₄ (excess), reflux → propanoic acid (if student proceeds with acid in the corrected plan).
Error 2 (2 marks): Esters are NOT made from an aldehyde and an alcohol — wrong reagent class
- 1 mark: The student claims propyl propanoate can be made from propanal and propan-1-ol. This is chemically incorrect. Esterification requires a carboxylic acid (R–COOH) and an alcohol. An aldehyde (R–CHO) does not undergo esterification with an alcohol under these conditions.
- 1 mark: The aldehyde must first be oxidised to propanoic acid (K₂Cr₂O₅/H₂SO₄ excess, reflux) before esterification can proceed.
Error 3 (1 mark): Student claims ‘only two steps’ — actually three steps are required
- 1 mark: The correct synthesis from propan-1-ol requires three steps: (1) oxidation to propanal (distillation), (2) further oxidation to propanoic acid (reflux), (3) esterification of propanoic acid + propan-1-ol. The student’s two-step plan skips the essential intermediate carboxylic acid.
Corrected synthesis plan (2 marks)
- Step 1: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O | K₂Cr₂O₅/H₂SO₄, distillation. Intermediate: propanal.
- Step 2: CH₃CH₂CHO + [O] → CH₃CH₂COOH | K₂Cr₂O₅/H₂SO₄ (excess), reflux. Intermediate: propanoic acid.
- Step 3: CH₃CH₂COOH + CH₃CH₂CH₂OH ⇌ CH₃CH₂COOCH₂CH₂CH₃ + H₂O | conc. H₂SO₄ catalyst, reflux. Product: propyl propanoate. Reversible (⇌), yield ~65%.