AQA A-Level Chemistry: Organic synthesis (7405)

This guide teaches 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 whole section is A-level only. It is examined on A-level Paper 2, which covers organic chemistry (section 3.3), and A-level Paper 3 can test any content. The section has three learning outcomes, and this guide teaches all three with worked examples.

For quick recall, use the revision notes. To test yourself, use 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 where your gaps are.

What this unit covers

Spec point	What you must be able to do	Status
3.3.14, outcome 1	Explain why chemists aim to design processes that do not need a solvent and that use non-hazardous starting materials	A-level only
3.3.14, outcome 2	Explain why chemists aim to design production methods with fewer steps and a high percentage atom economy	A-level only
3.3.14, outcome 3	Use reactions in the specification to devise a synthesis, with up to four steps, for an organic compound	A-level only

Section 3.3.14 lists no required practical of its own. The practical side of synthesis appears elsewhere: Required practical 5 (distillation of a product from a reaction, in 3.3.5.3) and Required practical 10 (preparing a pure organic solid and testing its purity, and a pure organic liquid, in 3.3.9.2). The atom economy formula itself is in 3.1.2.5.

Outcome 1: no solvent and non-hazardous starting materials

Why avoid a solvent

A solvent does not end up in the product. It is there only to dissolve the reactants. That brings costs:

• It must be made, bought and transported, using energy and raw materials.
• It must be removed from the product, often by distillation, which needs energy.
• It becomes waste unless it is recovered and recycled, which also needs energy and equipment.
• Many organic solvents are flammable, toxic or volatile. They add fire risk, harm to workers and pollution if vapour escapes.

A process without a solvent has fewer separation steps, less waste and lower risk. It is often cheaper too.

Why use non-hazardous starting materials

A hazardous substance is one that is toxic, corrosive, flammable or harmful to the environment. Every hazardous chemical must be stored, transported and handled with controls such as protective equipment, special containers and emergency plans. Each control costs money, and each is a point where something can go wrong for workers or the local area. Swapping to a less hazardous starting material removes those risks at the source.

Two examples from the specification:

• Potassium cyanide. 3.3.8 asks you to know “the hazards of using KCN”. It is highly toxic, and with acid it can release hydrogen cyanide gas. A route that avoids KCN avoids that risk.
• Ethanoic anhydride and ethanoyl chloride. 3.3.9.2 asks for “the industrial advantages of ethanoic anhydride over ethanoyl chloride in the manufacture of the drug aspirin”. Ethanoic anhydride is cheaper, less corrosive and reacts less vigorously with water. It gives ethanoic acid as the by-product instead of corrosive hydrogen chloride gas.

Outcome 2: fewer steps and high atom economy

Percentage atom economy

From 3.1.2.5:

% atom economy = (Mr of desired product ÷ sum of Mr of all reactants) × 100

Use the balanced equation, include every reactant (with its coefficient), and put only the wanted product on top. Atom economy measures how many of the reactant atoms end up in the product you want. A low value means a lot of by-product, which is waste unless it can be sold or used.

Reaction types differ:

• Addition reactions (for example HBr or steam adding to an alkene, or addition polymerisation) have one product, so the atom economy is 100%.
• Substitution and elimination reactions always give a by-product, so the atom economy is below 100%.

Atom economy is not the same as percentage yield. Yield compares the amount you actually get with the maximum possible. A reaction can have a high yield and a low atom economy, or the other way round.

Worked example 1: atom economy and the aspirin choice

Aspirin (C₉H₈O₄) is made from 2-hydroxybenzoic acid (C₇H₆O₃). Compare the two acylating agents. Use Ar: H = 1.0, C = 12.0, O = 16.0, Cl = 35.5.

With ethanoic anhydride:
C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH
Mr:  138.0 + 102.0          180.0
% atom economy = 180.0 ÷ 240.0 × 100 = 75.0%

With ethanoyl chloride:
C₇H₆O₃ + CH₃COCl → C₉H₈O₄ + HCl
Mr:  138.0 + 78.5           180.0
% atom economy = 180.0 ÷ 216.5 × 100 = 83.1%

The chloride route has the higher atom economy, yet industry uses the anhydride. The reasons are the ones in outcome 1: the anhydride is cheaper, safer to handle and gives no HCl gas. In a “discuss” or “evaluate” answer, weigh atom economy against cost and hazard. Don’t treat it as the only factor.

Why fewer steps

Every step loses some product. The overall yield of a route is the product of the step yields (as decimals). More steps also mean more reagents, more solvent, more separations, more energy, more time and more waste.

Worked example 2: overall yield

Route A has four steps, each with an 80% yield. Route B makes the same product in two steps, each with an 80% yield. Starting from 2.00 mol of starting material, how much product does each give? (Assume a 1 : 1 mole ratio at every step.)

Route A: 0.80 × 0.80 × 0.80 × 0.80 = 0.4096 → overall yield 41.0%
         2.00 × 0.4096 = 0.819 mol
Route B: 0.80 × 0.80 = 0.64 → overall yield 64.0%
         2.00 × 0.64 = 1.28 mol

Route B gives more product and uses fewer reagents and separations. Multiply yields; never add them.

Outcome 3: devising a synthesis of up to four steps

You may use only reactions in the specification. That is a real limit. For example, the content column has no reaction that turns an alcohol directly into a halogenoalkane (a chloroalkane preparation appears only as a skills opportunity), and no way to make an acyl chloride. Plan around such gaps.

The reaction map

From	To	Reagent and conditions	Spec
Alkene	Halogenoalkane	HBr	3.3.4.2
Alkene	Dibromoalkane	Br₂	3.3.4.2
Alkene	Alcohol	Steam with an acid catalyst (for example phosphoric acid)	3.3.5.1
Alkene	Addition polymer	Addition polymerisation	3.3.4.3
Halogenoalkane	Alcohol	Aqueous NaOH or KOH, warm (reflux)	3.3.3.1
Halogenoalkane	Nitrile (one more C)	KCN in aqueous ethanol, heat under reflux	3.3.3.1
Halogenoalkane	Primary amine	Excess NH₃ in ethanol, heat in a sealed container	3.3.3.1, 3.3.11.1
Halogenoalkane	Alkene	KOH in ethanol, hot, under reflux	3.3.3.2
Alcohol	Alkene	Concentrated H₂SO₄ or H₃PO₄, heat	3.3.5.3
Primary alcohol	Aldehyde	Acidified K₂Cr₂O₇, distil off as it forms	3.3.5.2
Primary alcohol	Carboxylic acid	Excess acidified K₂Cr₂O₇, reflux	3.3.5.2
Secondary alcohol	Ketone	Acidified K₂Cr₂O₇, reflux	3.3.5.2
Aldehyde / ketone	Primary / secondary alcohol	NaBH₄ in aqueous solution	3.3.8
Aldehyde / ketone	Hydroxynitrile (one more C)	KCN, then dilute acid	3.3.8
Carboxylic acid + alcohol	Ester	Acid catalyst (for example concentrated H₂SO₄), heat	3.3.9.1
Ester	Acid (or its salt) + alcohol	Hydrolysis in acid or alkaline conditions	3.3.9.1
Acyl chloride or acid anhydride	Acid, ester, amide or N-substituted amide	Water, alcohol, NH₃ or a primary amine	3.3.9.2
Nitrile	Primary amine	Reduction, for example LiAlH₄, or H₂ with a nickel catalyst	3.3.11.1
Benzene	Nitrobenzene	Concentrated HNO₃ and concentrated H₂SO₄	3.3.10.2
Nitrobenzene	Phenylamine	Reduction, for example Sn and concentrated HCl, then NaOH	3.3.11.1
Benzene	Phenyl ketone	Acyl chloride with an AlCl₃ catalyst	3.3.10.2
Amine + halogenoalkane	Secondary, tertiary amine, quaternary salt	Nucleophilic substitution	3.3.11.3

The reagents for the two reductions are not named in the specification; the ones above are standard.

Method

1. Compare start and target. Write both structures. List the functional groups and count the carbons.
2. Carbon count changed? Apart from addition polymerisation, only three reactions in the map make a new C–C bond: KCN with a halogenoalkane, KCN with an aldehyde or ketone, and Friedel–Crafts acylation.
3. Work backwards from the target. Ask: “What could this be made from in one step?”
4. Check the position of the functional group. Addition to an unsymmetrical alkene goes through the more stable carbocation, so the group ends up on the more substituted carbon in the major product.
5. Write every step with reagent, conditions and the name or structure of each intermediate.

Worked example 3: changing the chain length

Devise a three-step synthesis of propan-1-amine, CH₃CH₂CH₂NH₂, from ethene.

Target has 3 C, start has 2 C, so a C–C bond must form. Working back: a primary amine with an extra carbon comes from reducing a nitrile; the nitrile comes from a halogenoalkane and KCN.

Step 1  CH₂=CH₂ + HBr → CH₃CH₂Br                 (bromoethane)
Step 2  CH₃CH₂Br + KCN → CH₃CH₂CN + KBr           KCN in aqueous ethanol, reflux
                                                  (propanenitrile)
Step 3  CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂            reduction, e.g. LiAlH₄

Bromoethane with ammonia would give ethylamine, with only 2 C. That is the trap.

Worked example 4: moving a functional group

Convert propan-1-ol into propan-2-ol in two steps.

Step 1  CH₃CH₂CH₂OH → CH₃CH=CH₂ + H₂O             concentrated H₂SO₄, heat (elimination)
Step 2  CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃            steam, acid catalyst (major product)

The secondary carbocation is more stable than the primary one, so the OH ends up on carbon 2. The minor product, propan-1-ol, must be separated.

Worked example 5: a four-step route

Make poly(phenylethene) from benzene.

Step 1  C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl          AlCl₃ catalyst (phenylethanone)
Step 2  C₆H₅COCH₃ + 2[H] → C₆H₅CH(OH)CH₃          NaBH₄ (aq) (1-phenylethanol)
Step 3  C₆H₅CH(OH)CH₃ → C₆H₅CH=CH₂ + H₂O         concentrated H₂SO₄, heat (phenylethene)
Step 4  n C₆H₅CH=CH₂ → –[CH(C₆H₅)–CH₂]ₙ–          addition polymerisation

Each step comes from a different section: 3.3.10, 3.3.8, 3.3.5 and 3.3.4.

Common errors

• Adding or averaging step yields. Multiply them.
• Leaving a reactant out of the atom economy denominator, or forgetting a coefficient such as 2NH₃.
• Saying addition reactions have “high” atom economy. With one product it is exactly 100%.
• Using a reaction outside the specification, such as converting an alcohol straight to a halogenoalkane.
• Giving a reagent with no conditions, such as “KCN” without “in aqueous ethanol, heat under reflux”.
• Using NaBH₄ on a nitrile. NaBH₄ in the specification reduces aldehydes and ketones only.
• Forgetting that HBr or steam adds to give a major and a minor product with an unsymmetrical alkene.
• Using “excess” in the wrong place: excess NH₃ for a primary amine; excess oxidising agent for an acid.

Where to go next

Revise the individual reactions in halogenoalkanes, alkenes and alcohols, then identify products with organic analysis. Practise with the organic synthesis questions, and see the exam preparation guide.

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 the atom economy statement in 3.1.2.5.