AQA A-Level Chemistry 7405 · 3.3.14 Organic Synthesis

Part 1: Planning routes and improving yield

All 3 parts available · worked answers and exam guidance included. Reviewed 2 October 2026.

Plan AQA organic syntheses of up to four steps, track carbon atoms and compare yield, atom economy and practical choices.

Start with the target, then work backwards

Identify the target functional group and the carbon skeleton separately. Ask which reaction makes the final bond or functional group, then which starting material that reaction needs. Repeat until you reach the supplied reactant. Finally read the route forwards and check every intermediate, reagent and condition.

Changing an alcohol into an aldehyde does not lengthen its carbon chain. Replacing a halogen by CN adds one carbon; reducing that nitrile keeps this new carbon. A reagent name alone does not demonstrate a route: draw or write each intermediate unambiguously.

A compact reaction toolkit

These are route-planning reminders. The individual topic notes explain mechanisms and limitations. Separate steps may need isolation or a change of conditions; listing mutually incompatible reagents together is not a valid one-pot method.

Useful transformations within the AQA course
TransformationReagents and conditionsPlanning consequence
Halogenoalkane → alcoholAqueous NaOH or KOH; heatNucleophilic substitution; same carbon count
Halogenoalkane → alkeneEthanolic KOH; heatElimination; may give positional isomers
Halogenoalkane → nitrileKCN in aqueous ethanol; heat under refluxSubstitution; adds one carbon
Halogenoalkane → primary amineExcess ethanolic ammonia; heat in suitable pressure apparatusFurther alkylation is a competing reaction
Nitrile → primary amineHydrogen with a nickel catalyst under suitable conditions, or LiAlH4 in dry ether followed by work-upThe nitrile carbon becomes CH2; NaBH4 is not the standard reagent
Primary alcohol → aldehyde / acidAcidified potassium dichromate(VI); distil aldehyde / reflux with excess oxidant for acidControl oxidation and whether product stays in the flask
Secondary alcohol → ketoneAcidified potassium dichromate(VI); heatTertiary alcohols resist this usual oxidation
Aldehyde or ketone → alcoholNaBH4 in a suitable aqueous/alcoholic mediumReduction; carbon skeleton unchanged
Alcohol → alkene → halogenoalkaneConcentrated H2SO4 or H3PO4 and heat; then hydrogen halideElimination followed by electrophilic addition
Carboxylic acid + alcohol → esterConcentrated H2SO4 catalyst; heat under refluxReversible; subsequent separation and purification needed
Acyl chloride + primary amine → N-substituted amideSuitable controlled conditions; excess amine or another baseNucleophilic addition–elimination; neutralise the HCl formed
Benzene → nitrobenzene → phenylamineConcentrated HNO3/H2SO4 at about 50–55 °C; then Sn/HCl and heat, followed by NaOHElectrophilic substitution, reduction, then release the free amine
Benzene → aromatic ketoneAcyl chloride and anhydrous AlCl3; controlled conditionsFriedel–Crafts acylation; new carbon–carbon bond

Worked four-step route: ethanol to propylamine

The target contains three carbons but ethanol contains two. Plan a nitrile intermediate to supply the extra carbon. This is one chemically valid four-step route; it is not a claim that industry would choose it.

Step 1: dehydrate ethanol with concentrated phosphoric acid and heat to form ethene. Step 2: add HBr to ethene to form bromoethane. Step 3: heat bromoethane with KCN in aqueous ethanol under reflux to form propanenitrile. Step 4: reduce propanenitrile with H2/Ni under suitable conditions to give propylamine (propan-1-amine).

CH3CH2OH → CH2=CH2 + H2O
CH2=CH2 + HBr → CH3CH2Br
CH3CH2Br + CN− → CH3CH2CN + Br−
CH3CH2CN + 2H2 → CH3CH2CH2NH2

Diagram placeholder

Diagram placeholder — four-step synthesis map

Labels to include:

  • Ethanol: two carbon atoms
  • Ethene: two carbon atoms
  • Bromoethane: two carbon atoms
  • Propanenitrile: highlight the new CN carbon
  • Propylamine: the highlighted carbon is now CH2NH2
  • Put a separate reagent and condition above every arrow

Draw the five structures in sequence. Keep the original two-carbon fragment in one colour and the cyanide carbon in another. The map must show four reaction arrows, not treat reagent addition as a new intermediate.

Check compatibility and competing products

A molecule with two functional groups can react at both. An oxidant intended to oxidise a primary alcohol can also oxidise an aldehyde elsewhere. Strong acid or base can hydrolyse an ester. Check the whole structure before choosing a reagent.

A racemic intermediate may remain racemic unless a later step separates or selectively transforms its enantiomers. A constitutional isomer mixture may need separation. Do not give a single pure product merely because it is the one you want.

For a synthesis ending in an amide, distinguish the carbonyl-containing acyl fragment from the amine fragment. N-ethylpropanamide needs a propanoyl group and an ethylamine group; ethanoyl chloride plus propylamine makes a different amide.

Yield multiplies; atom economy answers a different question

For consecutive steps with one-to-one stoichiometry, multiply fractional yields. If stoichiometric ratios change, calculate moles at each stage instead of blindly multiplying product masses.

A route with yields of 80.0%, 75.0% and 90.0% gives 0.800 × 0.750 × 0.900 = 0.540, or 54.0% overall. Starting with 0.200 mol of the limiting reactant can therefore produce 0.108 mol of the final product for this one-to-one route.

Atom economy = Mr of the desired product, including its stoichiometric coefficient, ÷ total Mr of all reactants with their coefficients × 100. It is calculated from a balanced equation, not the measured mass collected. High atom economy does not guarantee a high experimental yield.

When comparing routes, consider number of steps, yield, selectivity, atom economy, energy, separation, solvent use and reagent hazards. Avoid unnecessary solvents and hazardous substances where a workable alternative exists. A shorter route is not automatically better if it needs difficult purification or a much more hazardous reagent.

Explain why your route works

In AQA June 2022 Paper 2 Q10, the route involves amine chemistry and structural deduction. The report discusses difficulties with reagents and further reactions; use that context to practise connecting a named transformation to the actual structure.

For each arrow, check the functional-group change, carbon count, charge and conditions. State a necessary work-up, such as adding alkali after reducing nitrobenzene in acid, rather than presenting an ammonium salt as the free amine.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Q1. Why does bromoethane react with cyanide to give a three-carbon product?Show answer

The carbon atom of CN− forms a bond to the carbon that originally carried Br. CH3CH2CN therefore contains three carbons.

Q2. Give the reagents for converting a primary alcohol into an aldehyde rather than the carboxylic acid.Show answer

Heat with acidified potassium dichromate(VI), using controlled oxidation and distilling off the aldehyde as it forms. Reflux with excess oxidant favours further oxidation to the acid.

Q3. A three-step one-to-one route has yields 90%, 80% and 75%. What is its overall yield?Show answer

0.90 × 0.80 × 0.75 × 100 = 54%. Do not average the three percentages.

Q4. Which two organic reactants can make N-ethylpropanamide by acylation?Show answer

Propanoyl chloride and ethylamine. Use excess amine or another suitable base to deal with the HCl formed. The product is CH3CH2CONHCH2CH3.

Q5. A route has fewer steps but a lower yield and uses a hazardous solvent. Is it necessarily greener?Show answer

No. Compare material consumption and waste, atom economy, yield, solvent hazards and recovery, energy and purification. Step count alone cannot settle the comparison.

Sources

Sources and examiner guidance (reviewed 2 October 2026)

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