AQA A-Level Chemistry: Organic synthesis (7405) – Revision Notes

These are condensed notes for the final weeks. For full explanations and worked examples, use the organic synthesis study guide.

They cover section 3.3.14 Organic synthesis of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The section is A-level only. It is examined on A-level Paper 2 (organic chemistry, section 3.3), and A-level Paper 3 can test any content. Atom economy is defined in 3.1.2.5.

Test yourself with the practice questions. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show what to revise first.

The three outcomes

1. Explain why chemists aim for processes that need no solvent and use non-hazardous starting materials.
2. Explain why chemists aim for production methods with fewer steps and a high percentage atom economy.
3. Use reactions in the specification to devise a synthesis of up to four steps.

Formulas

Quantity	How to calculate it
% atom economy	(Mr of desired product ÷ sum of Mr of all reactants) × 100
% yield	(actual amount of product ÷ maximum theoretical amount) × 100
Overall yield of a route	Multiply the step yields as decimals, then × 100

Why no solvent

• The solvent has to be made and transported, which uses energy and raw materials.
• It has to be separated from the product, often by distillation, which uses energy.
• It becomes waste unless it is recovered and recycled.
• Many organic solvents are flammable, toxic or volatile: fire risk, harm to workers, pollution.

Why non-hazardous starting materials

• Toxic, corrosive or flammable chemicals need special storage, transport, protective equipment and emergency plans. All of these cost money.
• They put workers and the local area at risk. Removing the hazard at the start removes the risk.
• KCN (3.3.8) is highly toxic and can release HCN gas with acid.
• For aspirin, industry prefers ethanoic anhydride to ethanoyl chloride (3.3.9.2): it is cheaper, less corrosive, reacts less vigorously with water and gives ethanoic acid, not HCl gas.

Why fewer steps and high atom economy

• Each step loses product, so overall yield falls with every extra step.
• Each step needs its own reagents, solvent, separation, energy and time.
• High atom economy means fewer reactant atoms become by-products, so less waste and better use of raw materials.

Must-know distinctions

Distinction	Key point
Atom economy vs yield	Atom economy comes from the equation. Yield comes from what you actually collect.
Addition vs substitution or elimination	Addition (one product): 100% atom economy. Substitution and elimination: always below 100%.
Anhydride vs acyl chloride	The chloride can have the higher atom economy, but the anhydride is cheaper and safer.
NaBH₄ vs other reductions	NaBH₄ reduces aldehydes and ketones only. Nitriles and nitro compounds need other reducing agents.
Aldehyde vs acid from a primary alcohol	Distil off as it forms for the aldehyde; reflux with excess oxidant for the acid.

Reaction map by starting group

Alkene

• • HBr → bromoalkane (major product via the more stable carbocation)
• • Br₂ → dibromoalkane
• • steam, acid catalyst → alcohol
• → addition polymer

Halogenoalkane

• NaOH(aq), warm → alcohol
• KCN in aqueous ethanol, reflux → nitrile (+1 C)
• excess NH₃ in ethanol, sealed, heat → primary amine
• KOH in ethanol, hot, reflux → alkene
• • amine → secondary or tertiary amine, or quaternary ammonium salt

Alcohol

• concentrated H₂SO₄ or H₃PO₄, heat → alkene
• primary: acidified K₂Cr₂O₇, distil → aldehyde; excess, reflux → carboxylic acid
• secondary: acidified K₂Cr₂O₇, reflux → ketone
• • carboxylic acid, acid catalyst → ester

Aldehyde or ketone

• NaBH₄(aq) → primary or secondary alcohol
• KCN then dilute acid → hydroxynitrile (+1 C)

Carboxylic acid derivatives

• Ester: hydrolysis in acid → acid + alcohol; in alkali → carboxylate salt + alcohol
• Acyl chloride or acid anhydride + water → acid; + alcohol → ester; + NH₃ → amide; + primary amine → N-substituted amide

Nitrogen compounds and benzene

• Nitrile → primary amine by reduction (for example LiAlH₄, or H₂ with Ni)
• Benzene + concentrated HNO₃ and H₂SO₄ → nitrobenzene
• Nitrobenzene → phenylamine by reduction (for example Sn and concentrated HCl, then NaOH)
• Benzene + acyl chloride, AlCl₃ catalyst → phenyl ketone

The specification does not name the reducing agents for nitriles or nitro compounds; those above are standard choices.

New C–C bonds: apart from addition polymerisation, only three reactions make one. KCN + halogenoalkane, KCN + aldehyde or ketone, and Friedel–Crafts acylation.

Method in steps

1  Draw start and target. List functional groups. Count carbons.
2  Carbon count changed? Pick a C–C forming reaction.
3  Work backwards one step at a time from the target.
4  Check where the group ends up (major product?).
5  For every step: reagent + conditions + intermediate name or structure.
6  Count the steps (four at most) and check each reaction is in the spec.

Worked reminder: splitting the starting material

Make ethyl ethanoate using bromoethane as the only organic starting material.

Step 1  CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr         NaOH(aq), warm
Step 2  Oxidise part of the ethanol:
        CH₃CH₂OH + 2[O] → CH₃COOH + H₂O            excess acidified K₂Cr₂O₇, reflux
Step 3  CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O    conc. H₂SO₄ catalyst, heat

Both halves of the ester come from the same starting compound. Three steps.

Evaluating a process: a checklist

When a question asks you to compare two routes or evaluate a process, test each route against the same points and say which route wins on each:

• Number of steps. Fewer steps usually means a higher overall yield and less time and energy.
• Atom economy. Calculate it from the balanced equation for each route. Higher means less waste.
• Yield. A route with a lower atom economy can still be better if its yield is much higher.
• Solvent. Does the route need one? Can it be recycled?
• Hazards. Are any reagents toxic, corrosive or flammable (KCN, acyl chlorides, concentrated acids)?
• By-products. Can they be sold or used, or are they waste? A useful by-product reduces the cost of a low atom economy.
• Conditions. High temperature or pressure costs energy and equipment.

Finish with a judgement that names the route you prefer and the deciding reason. A list of points with no judgement does not answer “evaluate”.

Related topics

The single reactions are revised in halogenoalkanes, alkenes and alcohols. To confirm what a synthesis has made, use organic analysis.

Quick self-test

1. Give two reasons why chemists prefer a process with no solvent.
2. Calculate the % atom economy for making ethyl ethanoate from ethanoic acid and ethanol. (Ar: H = 1.0, C = 12.0, O = 16.0)
3. State the % atom economy for ethanal + HCN → CH₃CH(OH)CN.
4. A three-step route has step yields of 90%, 80% and 50%. Calculate the overall yield.
5. Apart from addition polymerisation, name the three reactions in the specification that form a new C–C bond.
6. Give the reagent and conditions to make butanoic acid from butan-1-ol.
7. Outline a two-step route from 1-bromobutane to butan-2-ol.
8. Name the organic product of ethanoyl chloride with methanol.
9. Give the reagents for a two-step synthesis of phenylamine from benzene.
10. Explain why ethanal with KCN, then dilute acid, gives an optically inactive product.
11. Give two reasons why a company might choose a route with the lower atom economy.
12. Outline a two-step route from propan-1-ol to 2-hydroxybutanenitrile.

Answers

1. Energy is needed to remove the solvent; it becomes waste unless recycled; many are flammable or toxic. (Any two.)
2. Mr: ethanoic acid 60.0, ethanol 46.0, ethyl ethanoate 88.0. 88.0 ÷ 106.0 × 100 = 83.0%.
3. 100%: addition gives a single product.
4. 0.90 × 0.80 × 0.50 = 0.36, so 36.0%.
5. KCN with a halogenoalkane; KCN with an aldehyde or ketone; Friedel–Crafts acylation.
6. Excess acidified potassium dichromate(VI), heat under reflux.
7. KOH in ethanol, hot, reflux → but-1-ene; then steam with an acid catalyst → butan-2-ol (major product).
8. Methyl ethanoate (with HCl).
9. Concentrated HNO₃ and concentrated H₂SO₄ → nitrobenzene; then Sn and concentrated HCl, followed by NaOH.
10. The C=O group is planar. CN⁻ attacks from either side with equal probability, so equal amounts of the two enantiomers form: a racemic mixture, which is optically inactive.
11. Any two: its yield is higher; it has fewer steps; its reagents are cheaper or less hazardous; it needs no solvent; its by-product can be sold.
12. Acidified potassium dichromate(VI), distil off the propanal as it forms; then KCN followed by dilute acid → CH₃CH₂CH(OH)CN. The chain grows from 3 to 4 carbons in step 2.

Where marks are usually lost

• Adding or averaging step yields instead of multiplying them.
• Missing a reactant or a coefficient (such as 2NH₃) from the atom economy denominator.
• Calling an addition reaction “high” atom economy instead of exactly 100%.
• Using a conversion that is not in the specification, such as alcohol straight to halogenoalkane.
• Giving a reagent with no conditions (“KCN” with no solvent or reflux; “K₂Cr₂O₇” with no acid).
• Choosing NaBH₄ to reduce a nitrile or a nitro compound.
• Ignoring the major and minor products when HBr or steam adds to an unsymmetrical alkene.
• Using ammonia when the chain must grow by one carbon: that needs KCN first.
• Answering “why no solvent?” with “it is cheaper” and no reason why.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.14 Organic synthesis (A-level only), with atom economy from 3.1.2.5.