Organic synthesis
Contents: 8 sections
What this topic actually tests
Nothing here is new chemistry. Every reaction has appeared in chapters 14 to 20. What is being tested is whether you can plan, and the questions come in two shapes: build a route from A to B, or read a given route and say what each step is.
The reliable method is to work through it in a fixed order rather than searching for a route by inspection.
- Count the carbons in the starting material and the target. If they differ, a chain-lengthening step is compulsory, and that decides most of the route on its own.
- List the functional groups at each end.
- Work backwards from the target, asking what makes this group.
- Fill in the reagents and conditions, which is where the marks are.
The two chain-lengthening steps
Only two reactions on the AS syllabus add a carbon, so if the target has more carbons than the starting material, one of them must appear:
| Reaction | Reagent and conditions | Effect |
|---|---|---|
| Halogenoalkane to nitrile | KCN in ethanol, heat under reflux | Adds one carbon |
| Aldehyde or ketone to hydroxynitrile | HCN with a trace of NaCN | Adds one carbon |
Spotting the carbon count first is the single most useful habit in this topic, because it turns an open-ended search into a much narrower one.
The map of reactions
Almost every AS route is built from these steps.
From an alkene
- Add HBr, room temperature, to give a halogenoalkane
- Add steam with a phosphoric acid catalyst to give an alcohol
- Add H₂ with a nickel catalyst to give an alkane
- Add cold dilute acidified KMnO₄ to give a diol
From a halogenoalkane
- NaOH(aq), warm, to give an alcohol
- KCN in ethanol, reflux, to give a nitrile, adding a carbon
- Excess NH₃ in ethanol, sealed tube, to give an amine
- NaOH in ethanol, hot, to give an alkene by elimination
From an alcohol
- Acidified K₂Cr₂O₇, distil, to give an aldehyde from a primary alcohol
- Acidified K₂Cr₂O₇, reflux, to give a carboxylic acid from a primary alcohol
- Acidified K₂Cr₂O₇, reflux, to give a ketone from a secondary alcohol
- Concentrated H₂SO₄ or hot Al₂O₃ to give an alkene by dehydration
- Carboxylic acid with concentrated H₂SO₄ to give an ester
- PCl₅ to give a halogenoalkane
From a carbonyl compound
- NaBH₄ to give an alcohol
- HCN with a trace of NaCN to give a hydroxynitrile, adding a carbon
- Tollens' or Fehling's, for an aldehyde only, to give a carboxylic acid
From a nitrile
- Dilute acid, reflux, to give a carboxylic acid
- LiAlH₄ in dry ether, or H₂ with nickel, to give an amine
From a carboxylic acid
- Alcohol with concentrated H₂SO₄ to give an ester
- PCl₅ or SOCl₂ to give an acyl chloride
- LiAlH₄ to give a primary alcohol
Worked route
Convert propan-1-ol into butanoic acid.
Step 0: count. Propan-1-ol has three carbons; butanoic acid has four. A carbon must be added, so KCN or HCN is compulsory. KCN needs a halogenoalkane, so the route must pass through one.
Step 1. Propan-1-ol to 1-bromopropane. Reagent: PCl₅, or HBr made in situ from NaBr and concentrated H₂SO₄.
Step 2. 1-bromopropane to butanenitrile. Reagent: KCN in ethanol, heated under reflux. This is the carbon-adding step, and the product has four carbons.
Step 3. Butanenitrile to butanoic acid. Reagent: dilute hydrochloric acid, heated under reflux.
Three steps, and step 0 decided the shape of all of them.
Reading a given route
For the second kind of question, name the type of reaction as well as the reagent. The types on the syllabus are addition, substitution, elimination, oxidation, reduction, hydrolysis, condensation and polymerisation.
A quick way to classify:
- A double bond disappears and nothing leaves: addition
- A group is swapped for another: substitution
- A small molecule is lost and a double bond appears: elimination
- Oxygen gained or hydrogen lost: oxidation
- Hydrogen gained or oxygen lost: reduction
- Water splits a bond: hydrolysis
Writing the conditions
Most lost marks in this topic are conditions rather than reagents, because the same reagent does different things under different conditions. Four pairs are worth memorising as pairs:
| Same reagent | Aqueous or distil | Ethanolic or reflux |
|---|---|---|
| NaOH with a halogenoalkane | Substitution to an alcohol | Elimination to an alkene |
| K₂Cr₂O₇ with a primary alcohol | Distil for the aldehyde | Reflux for the acid |
| Cl₂ with NaOH | Cold dilute gives chlorate(I) | Hot concentrated gives chlorate(V) |
| Ester hydrolysis | Acid gives the acid, reversible | Alkali gives the salt, complete |
If a question gives you a reagent and asks for the product, check the conditions before answering, because in each of these the conditions carry the answer.
Common mistakes
- Starting to plan without counting the carbons, and missing that a chain-lengthening step is needed.
- Giving a reagent without its conditions. Both are usually required.
- Using KCN on an alcohol. It reacts with a halogenoalkane, so the alcohol has to be converted first.
- Trying to oxidise a tertiary alcohol, or a ketone, in a route. Neither happens.
- Forgetting that direct esterification is reversible and low-yielding, when an acyl chloride would be the better answer for a good yield.
- Naming a reagent correctly but the reaction type wrongly, when the question asks for both.
Check you have it
Question 1
The diagram shows a synthetic route to produce 1-methylcyclohexanol. What is reagent Y?
Answer: A.
Follow the route. HBr adds to 1-methylcyclohexene, and Markovnikov's rule puts the bromine on the more substituted carbon, the one already carrying the methyl, because that gives the more stable tertiary carbocation. Product X is 1-bromo-1-methylcyclohexane.
The target replaces that Br with an OH in the same position, so the second step is nucleophilic substitution, and the reagent for hydrolysing a halogenoalkane is aqueous sodium hydroxide with heat.
C, ethanolic NaOH, is the same base in a different solvent, and the solvent is what decides the outcome. In ethanol the hydroxide acts as a base rather than a nucleophile, removing a hydrogen from the neighbouring carbon and eliminating HBr to give an alkene. That would simply undo the first step.
B and D are oxidising agents, and neither converts a halogenoalkane into anything. Cold dilute KMnO4 would make a diol from the original alkene, and hot concentrated KMnO4 would break its ring open.
Aqueous means substitution and ethanolic means elimination, with the same reagent in both. That single pair is worth memorising, because Cambridge offers both in the same list most years.
What the syllabus asks for on this topicSyllabus points
Syllabus points
- Devise multi-step synthetic routes for preparing organic molecules using the reactions on the syllabus.
- Select appropriate reagents and conditions for each step.
- Analyse a given synthetic route in terms of the type of reaction and the reagents used.
- Identify the functional groups in a molecule and predict how they will react.
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