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OxfordAQA A-Level Chemistry: Organic synthesis (9620) – Revision Notes

Revision notes for OxfordAQA 9620 organic synthesis: greener design, atom economy, the full reaction map and a method for routes of up to four steps.

Subject
Chemistry
Level
A LEVEL
Topic
Organic synthesis
Updated

Aligned to OxfordAQA A Level Chemistry (9620), Version 5.3 (first teaching 2019, first AS and A-level exams 2020). Official specification .

Syllabus page (what it covers and how it is assessed): OxfordAQA A Level Chemistry.

Syllabus points this page covers

9620 (A Level)

  • 32 Organic synthesis (whole topic)

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Use these condensed notes in the final run-up to your exams. The organic synthesis study guide has the full explanations and longer examples.

Their scope is section 3.3.14 Organic synthesis in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams from May/June 2020 onwards. Everything here is International A-level only, part of the A2 organic chemistry examined in Unit 4 (Organic 2 and Physical 2); Unit 5 (Practical and synoptic) covers all content. The atom economy formula is from 3.1.2.5.

Check yourself afterwards with the practice questions. Also see the 9620 chemistry course hub, the printable 9620 checklist and the free 10-minute diagnostics.

What 3.3.14 asks of you

Outcome In short
1 Explain the aim of processes with no solvent and non-hazardous starting materials
2 Explain the aim of fewer steps and high percentage atom economy
3 Plan a route of four steps or fewer, built only from 9620 reactions

No required practical is listed under 3.3.14. Required practical 4 (distil a product, 3.3.5.2) and Required practical 10 (make an organic solid and check how pure it is, 3.3.9.2) are the practical skills a synthesis draws on.

Key formulas

Quantity Formula
% atom economy Mr of the wanted product × 100, divided by the total Mr of all reactants
% yield actual amount of product ÷ theoretical maximum × 100
Overall yield step yields as decimals multiplied together, × 100

Outcome 1 in bullet points

No solvent

  • No energy or raw materials spent making and transporting it.
  • No separation step (often distillation) to remove it from the product.
  • No solvent waste and no recovery plant.
  • No flammable or toxic vapour to leak.

Non-hazardous starting materials

  • Less spent on sealed storage, transport rules, protective kit, training and emergency plans.
  • Lower risk to workers, neighbours and the environment if something goes wrong.
  • 9620 hazards to quote: KCN (very toxic; acid releases HCN gas), epoxyethane (extremely flammable, explosive with air, toxic, carcinogenic), chlorine and (chloromethyl)benzene (toxic, irritant).
  • Aspirin: ethanoic anhydride is chosen over ethanoyl chloride because it is cheaper, less corrosive, less violent with water and releases no HCl.

Outcome 2 in bullet points

  • Every step loses some product, so the overall yield drops with each step added.
  • Every step needs its own reagents, energy, separation, time and waste disposal.
  • High atom economy: most reactant atoms end up in the product, so less by-product and less waste.
  • Addition = 100% atom economy. Substitution and elimination are usually below 100% (an exception: an amine plus a halogenoalkane giving a quaternary ammonium salt as the only product).
  • A useful by-product (one that can be sold) softens a low atom economy.

The 9620 reaction map

Start Reagent and conditions Product Section
Alkene HBr Bromoalkane; the major isomer comes from the more stable carbocation 3.3.4.2
Alkene Br₂ Dibromoalkane 3.3.4.2
Alkene Concentrated H₂SO₄ Alkyl hydrogensulfate 3.3.4.2
Alkene Addition polymerisation Poly(alkene) 3.3.4.3
Ethene O₂, silver catalyst (partial oxidation) Epoxyethane 3.3.4.4
Epoxyethane Water; or an alcohol Ethane-1,2-diol; or an ether-alcohol 3.3.4.4
Halogenoalkane NaOH(aq), warm Alcohol 3.3.3.1
Halogenoalkane KCN in aqueous ethanol, reflux Nitrile (+1 C) 3.3.3.1
Halogenoalkane Excess NH₃ in ethanol, heat, sealed Primary amine 3.3.3.1, 3.3.11.1
Halogenoalkane KOH in ethanol, hot Alkene 3.3.3.2
Halogenoalkane Amine 2° or 3° amine, quaternary ammonium salt 3.3.11.3
Alcohol Concentrated H₂SO₄ or H₃PO₄, heat Alkene 3.3.5.2
Primary alcohol Acidified K₂Cr₂O₇, distil as formed / excess, reflux Aldehyde / carboxylic acid 3.3.5.1
Secondary alcohol Acidified K₂Cr₂O₇, reflux Ketone 3.3.5.1
Aldehyde or ketone NaBH₄(aq) Primary or secondary alcohol 3.3.8.1
Aldehyde or ketone KCN, then dilute acid Hydroxynitrile (+1 C) 3.3.8.1
Carboxylic acid + alcohol Acid catalyst, heat Ester 3.3.9.1
Ester Acid or alkali, heat Acid (or its salt) + alcohol 3.3.9.1
Vegetable oil Methanol, catalyst Biodiesel (methyl esters) + glycerol 3.3.9.1
Acyl chloride or anhydride Water / alcohol / NH₃ / primary amine Acid / ester / amide / N-substituted amide 3.3.9.2
Benzene Conc. HNO₃ + conc. H₂SO₄ Nitrobenzene 3.3.10.2
Benzene Fuming or conc. H₂SO₄ Benzenesulfonic acid 3.3.10.2
Benzene Acyl chloride, AlCl₃ Phenyl ketone 3.3.10.2
Benzene Chloroalkane, AlCl₃ Alkylbenzene 3.3.10.2
Benzene Cl₂, UV light C₆H₆Cl₆ (ring addition) 3.3.10.2
Methylbenzene Cl₂, UV light (Chloromethyl)benzene (side chain) 3.3.10.2
Nitrile Reduction (H₂/Ni or LiAlH₄) Primary amine 3.3.11.1
Nitro compound Reduction (Sn + conc. HCl, then NaOH) Aromatic amine 3.3.11.1
Diacid + diol or diamine Condensation Polyester or polyamide 3.3.12.1

Reagents in brackets for the two reductions, and the silver catalyst, are standard choices; 9620 itself leaves them unnamed.

Must-know distinctions

Pair The difference
Atom economy vs yield Atom economy is calculated from the equation; yield from the mass you collect
Cl₂ + UV on benzene vs on methylbenzene Benzene: addition to the ring. Methylbenzene: substitution on the side chain
Ring Cl vs side-chain Cl Ring Cl is not replaced by nucleophiles; side-chain Cl reacts like a halogenoalkane
NaOH(aq) vs KOH in ethanol Aqueous favours substitution (alcohol); ethanolic favours elimination (alkene)
Excess NH₃ vs excess halogenoalkane Excess NH₃ gives mainly the primary amine; otherwise further substitution follows
Nitrile reduction vs NH₃ route to an amine Nitrile route adds a carbon and gives one amine only

Carbon–carbon bond forming steps

Outside polymerisation, four reactions in 9620 join carbon to carbon:

  1. Halogenoalkane + KCN (nitrile).
  2. Aldehyde or ketone + KCN, then dilute acid (hydroxynitrile).
  3. Friedel–Crafts acylation (acyl chloride + AlCl₃).
  4. Friedel–Crafts alkylation (chloroalkane + AlCl₃).

Not in the 9620 content

  • Steam hydration of an alkene.
  • Turning an alcohol directly into a halogenoalkane.
  • Making an acyl chloride or an acid anhydride.
  • Electrophilic substitution beyond a single substituent (the specification limits this to monosubstitution).

A route that needs one of these is not acceptable as a 9620 synthesis.

Method in steps

1  Draw start and target; mark groups; count carbons.
2  More carbons in the target? Choose a C–C step from the list.
3  Work back from the target one reaction at a time.
4  Check the position of each group (major product, ring vs side chain).
5  Each step: reagent + conditions + product name or structure.
6  Count steps (no more than four) and check every reaction is in 9620.

Worked reminders

Atom economy of a cyanide step. Relative atomic masses: C = 12.0, H = 1.0, K = 39.1, N = 14.0, Cl = 35.5.

C₆H₅CH₂Cl + KCN → C₆H₅CH₂CN + KCl
Sum of reactants = 126.5 + 65.1 = 191.6
% atom economy  = 117.0 ÷ 191.6 × 100 = 61.1%

Spot the broken route. A student plans: benzene + Cl₂ in UV light → chlorobenzene → warm NaOH(aq) → an alcohol. Both steps fail. UV light makes chlorine add to benzene, giving C₆H₆Cl₆, not chlorobenzene. And even if chlorobenzene were available, the ring C–Cl bond is strengthened by overlap of a chlorine lone pair with the delocalised ring, so hydroxide does not substitute it. A workable 9620 alternative to put an OH group near a ring is methylbenzene + Cl₂ (UV), then NaOH(aq), which gives phenylmethanol, C₆H₅CH₂OH.

Revise the single reactions in halogenoalkanes, alkenes, alcohols and aromatic chemistry. Confirm a product with organic analysis.

Quick self-test

  1. State two benefits of a process that uses no solvent.
  2. Give one reason for avoiding hazardous starting materials, with a 9620 example of such a material.
  3. Calculate the % atom economy for making ethylamine: CH₃CH₂Cl + 2NH₃ → CH₃CH₂NH₂ + NH₄Cl. (Relative atomic masses: C = 12.0, H = 1.0, N = 14.0, Cl = 35.5)
  4. A route has three steps with yields of 92%, 88% and 61%. Calculate the overall yield.
  5. Route P has two steps (80% and 55%). Route Q has one step (50%). Which gives more product?
  6. Name the four reactions in 9620, other than polymerisation, that form a new C–C bond.
  7. Give the reagent and condition to convert methylbenzene into (chloromethyl)benzene.
  8. Give the reagent and catalyst to make ethylbenzene from benzene.
  9. State the reagent that converts epoxyethane into ethane-1,2-diol.
  10. Give the reagent that converts a ketone into a secondary alcohol.

Answers

  1. Any two: no energy to make, transport or remove it; no solvent waste; no flammable or toxic vapour.
  2. Hazardous materials need costly controls and put people at risk; for example KCN, which is very toxic.
  3. Sum = 64.5 + 2(17.0) = 98.5; % atom economy = 45.0 ÷ 98.5 × 100 = 45.7%.
  4. 0.92 × 0.88 × 0.61 = 0.4939, so 49.4%.
  5. Route P: 0.80 × 0.55 = 0.44, so 44%. Route Q (50%) gives more, and with fewer steps.
  6. Halogenoalkane + KCN; aldehyde or ketone + KCN then dilute acid; Friedel–Crafts acylation; Friedel–Crafts alkylation.
  7. Chlorine, in UV light.
  8. Chloroethane, with an AlCl₃ catalyst.
  9. Water.
  10. NaBH₄ in aqueous solution.

Where marks are usually lost

  • Using a reaction that is not in 9620, such as steam hydration or alcohol to halogenoalkane.
  • Writing AlCl₃ when the side chain of methylbenzene is to be chlorinated; UV light is needed.
  • Planning substitution of a chlorine atom attached directly to a benzene ring.
  • Multiplying step yields as percentages (80 × 55) instead of decimals, or adding them.
  • Dropping a coefficient from the atom economy denominator, such as the 2 in 2NH₃.
  • Reagent with no condition: “KOH” alone does not show whether substitution or elimination is wanted.
  • Ignoring the major product: HBr on propene gives mainly 2-bromopropane, not 1-bromopropane.
  • Giving “safer” and “less dangerous” as two separate reasons; they make one point.
  • Forgetting that KCN with an aldehyde or unsymmetrical ketone gives a racemic mixture.

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.14 Organic synthesis (International A-level only), with percentage atom economy from 3.1.2.5.

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