Chemistry • Year 12 • Module 7 • Lesson 19
HSC Exam Practice
Organic Reaction Pathways: Synthesis & Multi-Step Problems
Short answer
1.Short answer
Define retrosynthetic analysis and explain how it is applied at the start of a multi-step pathway problem.
Distinguish between the conditions that produce an aldehyde versus a carboxylic acid when a primary alcohol is oxidised with acidified potassium dichromate.
Identify the reagents and conditions required to convert a haloalkane to an alcohol. Explain why ethanolic NaOH must not be used for this conversion.
Outline why a secondary alcohol is described as a ‘dead end’ in a Module 7 synthesis route aimed at producing a carboxylic acid.
Describe the two-step synthesis of butan-2-ol from but-1-ene. For each step, state the reagents, conditions and the product formed. Explain why butan-2-ol, rather than butan-1-ol, is the expected product of direct hydration of but-1-ene.
Explain the role of concentrated sulfuric acid as a catalyst in esterification. In your answer, refer to both its catalytic function and its effect on equilibrium yield.
Data response
2.Data response — multi-step synthesis yield
A Sigma–Aldrich Australia chemist synthesises methyl pentanoate from 1-bromopentane in four steps. The table below shows the yield at each step of the synthesis.
| Step | Reaction | Step yield (%) |
|---|---|---|
| 1 | 1-Bromopentane → pentan-1-ol | 82 |
| 2 | Pentan-1-ol → pentanal | 71 |
| 3 | Pentanal → pentanoic acid | 88 |
| 4 | Pentanoic acid + methanol → methyl pentanoate | 64 |
Source: hypothetical synthesis data.
- Calculate the overall yield of methyl pentanoate from 1-bromopentane. Show your working. 2 marks
- Identify the step with the lowest yield and state the reagents and conditions required for that step, explaining why those conditions are used. 3 marks
- Predict the new overall yield if the Step 4 esterification yield were increased to 75% by using a 3:1 molar excess of methanol. State which principle explains this improvement. 2 marks
A student investigates the aspirin synthesis. The graph below shows the percentage yield of acetylsalicylic acid (aspirin) against reaction temperature for two catalyst conditions.
- Describe the trend in aspirin yield with increasing temperature for each catalyst. 2 marks
- At 80°C, H₂SO₄ gives a higher yield than H₃PO₄. Account for this difference using your knowledge of esterification conditions. 2 marks
Extended response
3.Extended response
Analyse the four-step synthesis of butyl butanoate from 1-bromobutane. In your response, write a balanced equation with full conditions for each step, name all intermediates, and justify the specific conditions used at each step — explaining why alternative conditions would produce a different product or a pathway dead end.
A Year 12 chemistry textbook states: “In organic synthesis, the more steps in a reaction pathway, the greater the overall yield, because more reactions provide more opportunities for the product to form.” Identify the scientific error in this statement, explain the correct chemistry, and discuss how this error would affect a student’s approach to choosing between two valid synthesis routes of different lengths.
Chemistry • Year 12 • Module 7 • Lesson 19
Answer Key
Retrosynthetic analysis is the process of working backwards from a target molecule to identify suitable precursors and the reagents needed at each step. Application: identify the functional group of the target product; identify which Module 7 reaction produces that functional group; identify the precursor one step back; repeat until the starting material is reached. (2 marks: 1 definition, 1 application method.)
Distillation gives an aldehyde: the primary alcohol is oxidised by K₂Cr₂O₅/H₂SO₄ and the volatile aldehyde product (lower boiling point than the alcohol) is removed from the flask as it forms, preventing further oxidation. Reflux with excess K₂Cr₂O₅/H₂SO₄ gives a carboxylic acid: the condenser returns vapour to the flask, keeping the intermediate aldehyde in contact with excess oxidant until complete oxidation occurs. Key: the apparatus (not just the reagent) determines the product. (3 marks: 1 aldehyde conditions, 1 carboxylic acid conditions, 1 explanation of why apparatus matters.)
Reagents: NaOH(aq). Conditions: reflux. Aqueous NaOH provides hydroxide ions in water, which act as nucleophiles, attacking the carbon bearing the halogen (S₂ substitution) and replacing it with –OH. Ethanolic NaOH must not be used because alkoxide/ethanol solvent promotes an elimination (E2) mechanism instead — the halogen is removed along with an adjacent hydrogen to form an alkene, not an alcohol. (3 marks: 1 correct reagent+conditions, 1 substitution mechanism, 1 why ethanolic gives elimination/alkene.)
Oxidation of a secondary alcohol (R–CHOH–R′) with K₂Cr₂O₅/H₂SO₄ produces a ketone (R–CO–R′). A ketone has no hydrogen bonded to the carbonyl carbon — there is no adjacent C–H bond that the oxidant can attack — so no further oxidation to a carboxylic acid is possible under Module 7 conditions. The pathway ends at the ketone; to reach a carboxylic acid, a primary alcohol must be the starting material instead. (2 marks: 1 ketone formed, 1 why oxidation cannot continue.)
Step 1: but-1-ene + H₂O (steam), H₂PO₄ catalyst, ~300°C, high pressure → butan-2-ol. (1 mark: correct reagents/conditions; 1 mark: butan-2-ol named.)
Step 2: Not required — hydration is a single step giving butan-2-ol directly. Accept “Step 2: but-1-ene + HBr (r.t., Markovnikov) → 2-bromobutane; then + NaOH(aq), reflux → butan-2-ol” if student interprets this as a two-step haloalkane route.
Explanation: Markovnikov’s rule states that in addition reactions, the functional group (here –OH or –X) attaches to the more substituted carbon (C2 of but-1-ene, which bears one alkyl group) rather than C1. Therefore the –OH is added at C2 giving butan-2-ol, not butan-1-ol. (4 marks: 1 reagents/conditions, 1 butan-2-ol as product, 1 Markovnikov rule stated, 1 C2 more substituted explanation.)
Catalytic function: Concentrated H₂SO₄ donates H♠ (proton) to the carboxylic acid, protonating the carbonyl oxygen. This makes the carbonyl carbon more electrophilic and susceptible to nucleophilic attack by the alcohol’s oxygen. H₂SO₄ is regenerated — it is a true catalyst. Effect on equilibrium yield: H₂SO₄ also acts as a dehydrating agent, absorbing the water produced in the esterification. By removing a product, it shifts the equilibrium position to the right by Le Chatelier’s principle, increasing the yield beyond what the equilibrium constant alone would dictate. (3 marks: 1 proton donation / activation, 1 true catalyst regenerated, 1 Le Chatelier / dehydration effect.)
(a) Overall yield = 0.82 × 0.71 × 0.88 × 0.64 = 0.82 × 0.71 = 0.5822; × 0.88 = 0.5123; × 0.64 = 32.8% (accept 32–33%). (2 marks: 1 method, 1 correct answer with working.)
(b) Step 2 has the lowest yield (71%). Step 2: Pentan-1-ol → pentanal. Reagents: K₂Cr₂O₅/H₂SO₄. Conditions: distillation. Why: pentanal has a lower boiling point than pentan-1-ol; distillation removes it as it forms, preventing further oxidation to pentanoic acid. The yield is limited because the distillation may not capture all pentanal before some is re-oxidised. (3 marks: 1 identifying Step 2, 1 correct reagents+conditions, 1 justification of distillation.)
(c) New overall yield = 0.82 × 0.71 × 0.88 × 0.75 = 38.4% (accept 38–39%). Principle: Le Chatelier’s principle — excess methanol shifts the esterification equilibrium to the right, increasing product formation. (2 marks: 1 correct calculation, 1 Le Chatelier.)
(a) For both catalysts, yield increases with temperature from 50°C to 80°C, then plateaus or slightly decreases at 90°C. The H₂SO₄ line gives consistently higher yields than H₃PO₄ at each temperature. (2 marks: 1 increase to plateau for both, 1 H₂SO₄ higher throughout.)
(b) Concentrated H₂SO₄ is both a stronger Brønsted acid (donates H♠ more readily, increasing reaction rate) and a more effective dehydrating agent than H₃PO₄. The better dehydrating action of H₂SO₄ removes more of the water product, shifting the equilibrium further right and producing a higher ester yield. (2 marks: 1 stronger proton donor / higher rate, 1 better dehydration / Le Chatelier.)
Marking criteria (8 marks):
Step 1 (2 marks): CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr. Conditions: NaOH(aq), reflux. Product: butan-1-ol. Justification: aqueous NaOH gives substitution (not elimination); ethanolic NaOH would give but-1-ene instead. (1 equation+intermediate, 1 justification.)
Step 2 (2 marks): CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O. Conditions: K₂Cr₂O₅/H₂SO₄, distillation. Product: butanal. Justification: distillation removes butanal (BP 75°C) as it forms; reflux would allow further oxidation to butanoic acid. (1 equation+intermediate, 1 justification.)
Step 3 (2 marks): CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH. Conditions: K₂Cr₂O₅/H₂SO₄ (excess), reflux. Product: butanoic acid. Justification: reflux keeps aldehyde in contact with excess oxidant until complete oxidation; distillation here would stop prematurely at butanal. (1 equation+intermediate, 1 justification.)
Step 4 (2 marks): CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O. Conditions: conc. H₂SO₄ catalyst, heat under reflux, reversible (⇌). Product: butyl butanoate. Justification: H₂SO₄ activates acid for attack by alcohol AND dehydrates water produced, shifting equilibrium right. Note: butan-1-ol from Step 1 is split — one portion oxidised (Steps 2–3), remainder used as alcohol in Step 4. (1 equation with ⇌ + product, 1 justification of H₂SO₄ and reversibility.)
Marking criteria (6 marks):
Error identification (1 mark): The statement is wrong. More steps do NOT increase overall yield; they decrease it. Each step has a yield less than 100%, so each additional step multiplies by a fraction, reducing the overall yield.
Correct chemistry (2 marks): Overall yield in multi-step synthesis = product of individual step yields. For example, 4 steps each at 80% yield gives 0.8⁴ = 41% overall. Adding a fifth step at 80% reduces this to 33%. The overall yield decreases exponentially as the number of steps increases — the shortest valid pathway always gives the highest theoretical yield. (1 mark for equation/calculation, 1 for exponential decrease principle.)
Effect on student approach (3 marks): A student who believed the error would choose a longer pathway over a shorter one, reasoning that ‘more steps = more product.’ (1) This would lead to lower product recovery, increasing reagent cost and waste. (2) The correct approach is to use retrosynthetic analysis to find the shortest valid pathway — preferring a 2-step route over a 4-step route even if the 4-step route uses individually familiar reactions, because the overall yield will be higher. (3) The student would also fail to appreciate why industrial chemists invest in developing high-yield individual steps — even a 5% improvement in one step has a compounding benefit across the whole synthesis chain.