A-Level Chemistry 9701 / Topic 36 / A2

Plan the route, step by step.

A Paper 4 synthesis question hands you a starting material and a target and asks for the route. Spot the functional groups, recall what each reagent does to them, chain the steps together, name each reaction type and work out what survives of your yield at the end.

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The key idea

Organic synthesis means getting from a starting material to a target using reactions you know. First identify every functional group in each molecule (arene, phenol, alcohol, halogenoalkane, alkene, aldehyde or ketone, carboxylic acid, ester, acyl chloride, amine, amide, nitrile) and predict how each reacts with the reagent given. Then work backwards from the target: what group must the previous compound have had? Each step needs a reagent, conditions and a reaction type (electrophilic substitution, nucleophilic addition, reduction, oxidation, hydrolysis, condensation). Nucleophilic addition to a planar C=O, such as HCN with ethanal, can create a chiral centre as a racemic mixture, but drugs are often needed as a single enantiomer. Yields multiply: the overall yield is the product of the step yields, so short routes are better.

Fig. 1: three A Level routes on one reaction map, with reagents for each numbered step

Section 01

Spot every functional group.

Highlight each functional group on the target molecules in the simulator until every group stands out at a glance.

Group	Typical reagent	What happens
arene ring	conc. HNO₃ + conc. H₂SO₄; Cl₂/AlCl₃	electrophilic substitution
phenol, ArOH	NaOH(aq); Br₂(aq)	forms a salt; white precipitate of 2,4,6-tribromophenol
alcohol, ROH	K₂Cr₂O₇/H⁺; Na	oxidised (primary, secondary); H₂ given off, but no reaction with NaOH(aq)
aldehyde, ketone	2,4-DNPH; HCN/KCN; NaBH₄	orange precipitate; nucleophilic addition; reduction
carboxylic acid	Na₂CO₃; SOCl₂	CO₂ fizzing; acyl chloride
acyl chloride	H₂O, ROH, ArOH, NH₃, RNH₂	acid, ester, amide (fast, room temperature, HCl fumes)
ester, amide	dilute acid or NaOH(aq), heat under reflux	hydrolysis
amine	dilute HCl; CH₃COCl	salt; amide
nitrile	dilute HCl, reflux; H₂/Ni or LiAlH₄	carboxylic acid; amine
halogenoalkane	NaOH(aq); KCN in ethanol; NH₃ in ethanol	alcohol; nitrile (one more carbon); amine
alkene	Br₂(aq); H₂/Ni; steam/H₃PO₄	electrophilic addition

Worked example: paracetamol, HOC₆H₄NHCOCH₃, contains a phenol and a secondary amide. With cold NaOH(aq) only the phenol reacts (a salt forms); hot NaOH(aq) also hydrolyses the amide to 4-aminophenoxide and ethanoate. Br₂(aq) substitutes the ring. There is no aldehyde or ketone, so 2,4-DNPH gives no precipitate.
Check every group: a reagent reacts with every group that responds to it, so a molecule with both an ester and an amide is hydrolysed at both by hot aqueous acid or alkali.

Stage 1 · Learn

Check what the sim just showed you

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Which two functional groups are present in paracetamol, HOC₆H₄NHCOCH₃?

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Which reagent converts benzoic acid into benzoyl chloride?

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A two-step synthesis has step yields of 80% and 50%. What is the overall yield?

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Why are many drugs now made as a single enantiomer?

Section 02

Chain the steps.

Build the set routes in the simulator step by step, then compare the slow, reversible acid-and-alcohol ester route with the faster acyl chloride route used in step 6 of Fig. 1.

Step	Reagents and conditions	Product	Reaction type
benzene →	conc. HNO₃ + conc. H₂SO₄, 25–60 °C	nitrobenzene	electrophilic substitution
→	Sn + conc. HCl, heat under reflux; then NaOH(aq)	phenylamine	reduction
→	NaNO₂ + HCl (HNO₂), below 10 °C	benzenediazonium chloride	diazotisation
→	phenol in NaOH(aq), below 10 °C	4-(phenylazo)phenol, an orange azo dye	coupling (electrophilic substitution)
methylbenzene →	hot alkaline KMnO₄, reflux; then dilute acid	benzoic acid	oxidation
→	SOCl₂ (or PCl₅), room temperature	benzoyl chloride	substitution
→	ethanol, or ethylamine, room temperature	ethyl benzoate, or N-ethylbenzamide	addition-elimination (condensation)
propanone →	HCN with a KCN catalyst, 10–20 °C	2-hydroxy-2-methylpropanenitrile	nucleophilic addition
→	dilute HCl(aq), heat under reflux	2-hydroxy-2-methylpropanoic acid	hydrolysis

Work backwards: a target amide comes from an acyl chloride, which comes from a carboxylic acid, which comes from oxidising an alkylarene, a primary alcohol or hydrolysing a nitrile.
Changing the carbon count: KCN with a halogenoalkane, or HCN with a carbonyl, adds one carbon (bromoethane → propanenitrile → propylamine with H₂/Ni). Hot alkaline KMnO₄ cuts any alkyl side chain on a ring back to one –COOH carbon.
Phenols need acyl chlorides: phenol does not esterify usefully with a carboxylic acid, so phenyl esters are made with an acyl chloride.

Section 03

Types, chirality and yield.

Use the chirality and yield tab to see where a planar intermediate gives a racemic product, then rehearse the step types behind each route: electrophilic substitution, nucleophilic addition and nucleophilic substitution.

Racemic products: CN⁻ attacks the planar C=O of ethanal equally from either face, so 2-hydroxypropanenitrile forms as a 50 : 50 racemic mixture. Propanone gives an achiral product because its central carbon carries two CH₃ groups.
Single-enantiomer drugs: only one enantiomer usually fits the target receptor or enzyme. The other may be inactive or harmful (thalidomide is the classic case). Making only the active form halves the dose, reduces side effects and avoids costly separation. Chemists use chiral catalysts, enzymes or naturally chiral starting materials, or separate the enantiomers.
Overall yield: multiply the step yields. Benzene to azo dye at 85%, 70% and 60%: 0.85 × 0.70 × 0.60 = 0.357, so 35.7% overall.
Worked mass: 15.6 g of benzene (M_r 78.0) is 0.200 mol, so at most 0.200 mol of dye (M_r 198.0) = 39.6 g. At 35.7% overall, 39.6 × 0.357 = 14.1 g.
Why fewer steps: five steps of 90% each give only 0.90⁵ = 59.0% overall; every extra step loses product.

Examiner trap

Do not add or average step yields: multiply them. Give conditions as well as reagents (reflux, temperature, catalyst) and say dilute or concentrated. Diazotisation must be kept below 10 °C or the diazonium salt decomposes. Sn/HCl reduction needs NaOH(aq) afterwards to free the amine from its salt. Check whether a carbonyl product is chiral before claiming a racemic mixture, and remember that one reagent attacks every group that responds to it.

Stage 2 · Exam

Exam-style questions

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Which two-step route converts propanone into 2-hydroxy-2-methylpropanoic acid?

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A four-step synthesis has step yields of 80%, 75%, 90% and 60%. What is the overall yield?

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Which reagent reacts with the –OH group of phenol but NOT with the –OH group of ethanol?

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In the route benzene → nitrobenzene → phenylamine, what are the types of reaction in the two steps?

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Six original Cambridge-style questions

Functional groups in complex molecules, multi-step routes with reagents and conditions, reaction types, chiral products and overall yield.