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S6 Chemistry: Organic Synthesis Routes (Paper 3: Challenge)
S6 Chemistry: Organic Synthesis Routes (Paper 3: Challenge)
Ten items designing short routes and handling multi-step yields, all from molar masses, with answers and notes.
Direct answers Answers in order: aqueous sodium hydroxide under reflux, oxidation, ethanol with concentrated sulphuric acid, one extra carbon and an acid, prove the carbonyl then separate aldehyde from ketone, three steps, 9.86 g, 40.0 per cent, 80.0 per cent, a ketone.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Chemistry: Organic Synthesis Routes |
| Level | S6 |
02 Questions (10 items)
- FoundationWhich reagent converts 2-bromopropane into propan-2-ol?
- FoundationConverting propan-2-ol into propanone is which type of reaction?
- FoundationWhat must be added to turn propanoic acid into ethyl propanoate?
- IntermediateBromoethane is treated with potassium cyanide, then heated with dilute sulphuric acid. What is the net result?
- IntermediateWhy does a synthesis check use dinitrophenylhydrazine before Tollens reagent?
- IntermediateWhat is the shortest number of steps from ethene to ethyl ethanoate?
- Challenge12.0 g of propan-2-ol is oxidised to propanone in 85 per cent yield. What mass is obtained? (60 and 58)
- Challenge21.8 g of bromoethane gives only 5.92 g of propanoic acid by the nitrile route. What is the overall yield? (109 and 74)
- Challenge0.10 mol of propanoic acid with excess ethanol gives 8.16 g of ethyl propanoate. What is the yield? (102)
- Error redoError replay: a student writes that a secondary alcohol is oxidised to a carboxylic acid. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Aqueous sodium hydroxide under reflux | The aqueous route favours nucleophilic substitution, so hydroxide replaces bromine to give the secondary alcohol. |
| 2 | Oxidation with acidified dichromate | A secondary alcohol is oxidised to a ketone, and the ketone resists further oxidation by acidified dichromate. |
| 3 | Ethanol and concentrated sulphuric acid with heat | The acid and the alcohol lose water under sulphuric acid catalysis to form the ester, and the step is reversible. |
| 4 | The chain gains one carbon and a carboxylic acid is formed | The cyanide contributes an extra carbon, and hydrolysis turns it into a carboxyl group, so C2 becomes a C3 acid. |
| 5 | The first proves a carbonyl group, the second separates aldehyde from ketone | The two tests are complementary, and skipping one can mistake a ketone for an aldehyde. |
| 6 | Three: hydration to ethanol, oxidation to ethanoic acid, then esterification | Each step changes one functional group, so the shortest route is three conversions. |
| 7 | 9.86 g | 12 over 60 gives 0.20 mol, a theoretical 11.6 g, and 0.85 of that is 9.86 g. |
| 8 | 40.0% | 0.20 mol should give 14.8 g, and 5.92 over 14.8 is 0.400. |
| 9 | 80.0% | The theoretical mass is 10.2 g and the actual 8.16 g, so the ratio is 0.800. |
| 10 | A ketone, propanone | The hydroxyl carbon of a secondary alcohol carries one hydrogen, so oxidation stops at the carbonyl group. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
For a multi-step yield, write the theoretical moles of every step in one row before multiplying.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.