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Planning a Synthetic Route

A concise revision guide to multi-step organic synthesis: the reaction map that joins every functional group, how to plan a route of up to 4 steps backwards from the target, worked routes for aliphatic and aromatic targets, and why chemists design routes with fewer steps, a high atom economy, no solvent and safer starting materials.

Paper 2 AQA
3.3.14 Organic synthesis
7405/2
1

Before You Start

A synthesis question joins reactions from the whole course. Nothing new is taught here: the skill is choosing and ordering reactions you already know.

2

The Reaction Map

Every functional group is joined to its neighbours by a reaction with a known reagent. Learn the map as a web of arrows: each arrow is one step, one reagent and one set of conditions.

FromToReagent and conditions
alkenehalogenoalkaneHBr at room temperature (electrophilic addition)
alkenealcoholsteam, H₃PO₄ catalyst, 300 °C, 60 atm (hydration)
halogenoalkanealcoholNaOH(aq), heat under reflux (nucleophilic substitution)
halogenoalkanenitrileKCN in ethanol, heat under reflux (+1 carbon)
halogenoalkaneprimary amineexcess concentrated NH₃ in ethanol, heated under pressure
halogenoalkanealkeneKOH in ethanol, heat under reflux (elimination)
alcoholhalogenoalkanePCl₅ (chloro-) or NaBr with concentrated H₂SO₄ (bromo-)
primary alcoholaldehydeK₂Cr₂O₇ / dilute H₂SO₄, distil the aldehyde as it forms
primary alcoholcarboxylic acidK₂Cr₂O₇ / dilute H₂SO₄, heat under reflux
secondary alcoholketoneK₂Cr₂O₇ / dilute H₂SO₄, heat under reflux
aldehyde or ketonealcoholNaBH₄ in water or ethanol (reduction)
aldehyde or ketonehydroxynitrileHCN with KCN (nucleophilic addition, +1 carbon)
nitrilecarboxylic aciddilute HCl, heat under reflux (hydrolysis) (extension: beyond the AQA specification)
nitrileprimary amineLiAlH₄ in dry ether, or H₂ with a nickel catalyst (reduction)
carboxylic acidacyl chloridePCl₅ (or SOCl₂) at room temperature
carboxylic acidesteralcohol with concentrated H₂SO₄ catalyst, heat under reflux
acyl chlorideester / amidealcohol / ammonia or amine, room temperature (addition-elimination)
estercarboxylic aciddilute acid under reflux, or NaOH(aq) under reflux then acidify
benzenenitrobenzeneconcentrated HNO₃ and concentrated H₂SO₄ at about 50 °C, below 60 °C (nitration)
nitrobenzenephenylamineSn and concentrated HCl under reflux, then NaOH(aq)
benzenealkyl- or acylbenzenehalogenoalkane or acyl chloride with an AlCl₃ catalyst (Friedel–Crafts)

The map drawn as structures: the two-carbon web from ethene, with the nitrile and hydroxynitrile branches, and the benzene row underneath, reagents on every arrow.

The same map as named boxes: use it to trace any route with your finger before writing it down.

Exam focus: Oxidising agent and conditions are both needed: K₂Cr₂O₇ / H₂SO₄ with distil for an aldehyde and with reflux for the acid. Reflux and distil are different answers.

3

How to Plan a Route

Work backwards from the target, one step at a time, until the starting material appears. Routes in questions have up to 4 steps.

  1. Compare the carbon skeletons: same number of carbons, one more, or a ring? One more carbon means a nitrile step somewhere.
  2. Name the functional group of the start and of the target and find the arrows that join them on the map.
  3. Ask what the target can be made from in one step; then what that intermediate can be made from; stop at the start.
  4. Write the reagent and conditions for every step and name the type of reaction.
  5. Check the route: no step over the limit, every intermediate named, side reactions and excess reagents considered.

Working backwards: ethanoic acid from ethene

Step 1.
Ethanoic acid comes from oxidising ethanal or ethanol
Ethanol comes from ethene by hydration
The skeleton is two carbons throughout
So no nitrile step is needed
Step 2.
Forwards: ethene + steam, H₃PO₄, 300 °C, 60 atm → ethanol
Ethanol + K₂Cr₂O₇ / H₂SO₄ under reflux → ethanoic acid
Answer.
Two steps
Via ethanal (distil, then reflux) is three steps and also correct
But the shorter route is the better answer when the question sets a limit

The five planning steps as a checklist, with the ethene to ethanoic acid example worked underneath.

Common mistake: Writing a reagent without conditions, or two changes in one arrow. One arrow is one reaction: an alcohol does not become an ester in a single step from an acyl chloride and an acid at once.

Check your understanding

Check: Reading the Reaction Map

Supply the missing reagent and conditions on a route, name the type of each reaction and decide whether the number of carbons changes.

4

Worked Routes 1: Aliphatic Targets

Three routes of two or three steps, each written as a question would expect it: intermediate, reagent, conditions and type of reaction.

Route (a): ethene to ethanoic acid through ethanal

Step 1. Ethene + steam, H₃PO₄ catalyst, 300 °C and 60 atm → ethanol (hydration, electrophilic addition).

Step 2.
Ethanol + K₂Cr₂O₇ / dilute H₂SO₄, distil the product as it forms → ethanal (oxidation)
Distilling removes the aldehyde before it oxidises further
Answer.
Ethanal + K₂Cr₂O₇ / dilute H₂SO₄, heat under reflux → ethanoic acid (oxidation)
Three steps, two carbons throughout

Route (b): 1-bromopropane to butanoic acid or butylamine

Step 1.
The target has four carbons, the start three: a nitrile step is needed
1-Bromopropane + KCN in ethanol, heat under reflux → butanenitrile (nucleophilic substitution)
Answer.
Butanenitrile + dilute HCl under reflux → butanoic acid + NH₄Cl (hydrolysis) (extension: beyond the AQA specification)
Or butanenitrile + 4[H] with LiAlH₄ or H₂/Ni → butylamine (reduction)
Two steps either way

Route (c): propan-2-ol to 2-hydroxy-2-methylpropanoic acid

Step 1. Propan-2-ol + K₂Cr₂O₇ / dilute H₂SO₄, heat under reflux → propanone (oxidation of a secondary alcohol to a ketone).

Step 2. Propanone + HCN with KCN → 2-hydroxy-2-methylpropanenitrile, (CH₃)₂C(OH)CN (nucleophilic addition, one carbon added).

Answer.
Hydrolyse with dilute HCl under reflux → 2-hydroxy-2-methylpropanoic acid, (CH₃)₂C(OH)COOH (extension: beyond the AQA specification)
Three steps: three carbons become four

Routes (a), (b) and (d) drawn as structures with the reagent and conditions over every arrow and the reaction type beneath it.

Exam focus: When a route branches, as in (b), the question usually asks for one product: read which, and give only that final step.

Check your understanding

Quick Check: Which Proposed Route Works?

Pick the one route in each set that would work.

5

Worked Routes 2: Aromatic Targets and Choosing the Alcohol

Aromatic routes start from benzene. The ring is kept and its substituent is changed step by step; the order of steps decides where new groups go.

Route (d): benzene to N-phenylethanamide

Step 1. Benzene + concentrated HNO₃ and concentrated H₂SO₄ at about 50 °C (below 60 °C) → nitrobenzene (nitration, electrophilic substitution).

Step 2.
Nitrobenzene + Sn and concentrated HCl, heat under reflux, then NaOH(aq) → phenylamine (reduction
The alkali frees the amine from its salt)
Answer.
Phenylamine + ethanoyl chloride at room temperature → N-phenylethanamide + HCl (nucleophilic addition-elimination)
Three steps, the ring untouched after step 1
  • Nitrate first, then reduce: phenylamine itself is too reactive to nitrate cleanly, and NH₂ directs new groups to the 2 and 4 positions. See 3.3.10 Aromatic Chemistry.
  • Protecting the amine as an amide (step 3) is used industrially before further ring substitution: the amide is less activating.
  • Keep the nitration at about 50 °C, below 60 °C, to avoid dinitration.

Route (f): but-1-ene to butanone, and which alcohol forms

Step 1.
But-1-ene + steam with H₃PO₄ at 300 °C and 60 atm → butan-2-ol as the major product (hydration)
H adds to the CH₂ carbon, which has more hydrogens
Step 2.
The secondary carbocation CH₃CH₂CH⁺CH₃ is more stable than the primary one
So butan-2-ol forms rather than butan-1-ol
Alternative: HBr → 2-bromobutane, then NaOH(aq) under reflux
Answer.
Butan-2-ol + K₂Cr₂O₇ / dilute H₂SO₄, heat under reflux → butanone (oxidation of a secondary alcohol)
Butan-1-ol would give butanal or butanoic acid instead

Remember: Which alcohol an unsymmetrical alkene gives decides the whole route: a secondary alcohol leads to a ketone, a primary alcohol to an aldehyde or acid.

Check your understanding

Check: Writing Multi-Step Routes

Plan routes of two to 4 steps from a given start to a given target, name every intermediate and give the reagent, conditions and reaction type for each step.

6

Unfamiliar Reactions Given in the Question

Questions often supply a reaction you have not met, with its reagents and conditions. Treat it as one more arrow on the map and use it exactly as given.

  • Read what the new reagent does to the functional group: which bond breaks, which forms, and whether the carbon count changes.
  • Use the reagents and conditions printed in the question, word for word; do not swap in a familiar reagent.
  • Apply the reaction to the molecule in the question, not to the example given: the same change on a different skeleton.
  • Check the other groups in the molecule: would they react with the same reagent? A question may ask you to protect one.
Step 1.
Example: “Compound P reacts with a reagent that converts an aldehyde into an alkene with one more carbon.”
Identify the change: C=O becomes C=CH₂
The carbon count rises by one
Answer.
Apply it to the given aldehyde, draw the alkene
Continue the route from the alkene with familiar reactions (hydration, HBr addition and so on)

Exam focus: Marks for an unfamiliar step go to applying the pattern correctly to the new molecule and keeping the reagents and conditions as printed.

7

Green Chemistry: Designing a Better Route

Chemists design production methods to be safer, cheaper and less wasteful. Four aims appear in questions, and two of them have calculations.

AimWhy
fewer stepseach step loses product; the overall yield is the product of the step yields, so fewer steps keep more
high percentage atom economymore of the reactant atoms end up in the desired product, so less waste and better use of resources
no solventless waste to treat, no volatile organic compounds released, lower cost and less energy to recover the solvent
non-hazardous starting materialssafer for workers and the environment, less protective equipment and containment, cheaper storage and disposal

Worked example 1: yield over several steps

Step 1.
A three-step route has a yield of 80 % at each step
Overall yield = 0.80 × 0.80 × 0.80 = 0.512
Answer.
About 51 % of the theoretical amount is obtained
A one-step route at 80 % keeps 80 %: fewer steps raise the overall yield

Worked example 2: atom economy of two routes to ethanol

Step 1.
Atom economy = (Mr of desired product) / (sum of Mr of all reactants) × 100
Hydration: C₂H₄ + H₂O → C₂H₅OH, (46) / (28 + 18) = 100 %: an addition reaction
Step 2.
Fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Desired product 2 × 46 = 92
Reactants 180
Answer.
(92) / (180) × 100 = 51.1 %
Almost half the mass of glucose leaves as carbon dioxide
So the atom economy is low even though the yield may be high

The four aims with the two worked calculations: 0.8³ = 51 % overall yield and 100 % against 51.1 % atom economy.

Remember: Yield and atom economy are different: yield compares what is obtained with the theoretical maximum; atom economy compares the wanted product with everything made by the equation.

Check your understanding

Check: Routes, Yields and Unfamiliar Steps

Apply a reaction given in a question to a new molecule, complete a route with an unfamiliar step in it, and compare two routes to the same target.

8

Common Exam Points

  • One arrow is one reaction: give the reagent AND the conditions for every step, and name the type of reaction.
  • Count carbons first: one more carbon in the target means a nitrile step; the same count means functional group changes only.
  • Oxidation of a primary alcohol: distil for the aldehyde, reflux for the carboxylic acid; a secondary alcohol gives a ketone under reflux.
  • Aromatic routes: nitrate benzene at about 50 °C (below 60 °C), reduce nitrobenzene with Sn and concentrated HCl then NaOH, acylate phenylamine with an acyl chloride.
  • Routes have up to 4 steps; a shorter correct route is preferred when the question sets a limit.
  • Unfamiliar reactions: use the reagents and conditions as printed and apply the same change to the molecule in the question.
  • Overall yield = product of the step yields (0.8³ = 51 %); atom economy = (Mr of product) / (total Mr of reactants) × 100.
  • Atom economy of ethanol by hydration of ethene 100 %, by fermentation 51.1 %; aim for fewer steps, no solvent and non-hazardous starting materials.
  • Show every intermediate by name or structure; a route with a missing intermediate loses the mark for that step.

FAQs

Use these quick answers to check how to plan a route of up to 4 steps from the reaction map, the order of steps in aromatic syntheses, and why fewer steps and a high atom economy matter.

How do I start planning a synthetic route?

Compare the target with the starting material: which functional groups change, and does the carbon skeleton stay the same, gain a carbon or need a ring? Then work backwards step by step, asking which compound could make the target, until you reach the start.

How do I make ethanoic acid from ethene?

Two steps. Hydrate ethene with steam over a phosphoric(V) acid catalyst at about 300 °C and 60 atm to give ethanol. Then heat the ethanol under reflux with excess acidified potassium dichromate(VI), which oxidises it through ethanal to ethanoic acid.

In what order do I carry out the steps from benzene to N-phenylethanamide?

Nitrate first: concentrated nitric and sulfuric acids at about 50 °C give nitrobenzene. Reduce the nitro group with tin and concentrated hydrochloric acid under reflux, then add sodium hydroxide to free phenylamine. Finally react the amine with ethanoyl chloride to give N-phenylethanamide.

What do I do when a question gives me a reaction I have never seen?

Treat it as one step with exactly the reagents and conditions the question states. Work out which functional group changes by comparing the formulae before and after, then fit the step into your route like any other. Examiners test your planning, not the new reaction.

Why do chemists design routes with fewer steps?

Every step loses some product, so the yields multiply: three steps at 80 % each give only 0.8³ = 51 % overall. Fewer steps also mean less solvent, energy and waste. High atom economy, avoiding solvents and safer starting materials are the other green aims.

Dr. Mohammed Al-Fatah

Written by:
Dr. Mohammed Al-Fatah

Chemistry specialist revision notes for A Level Chemistry.

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