OxfordAQA A-Level Chemistry: Organic synthesis (9620)

Section 3.3.14 Organic synthesis of 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, is the subject of this guide. All of it is International A-level only: it belongs with sections 3.3.7 to 3.3.16, the A2 organic chemistry examined in Unit 4 (Organic 2 and Physical 2), while Unit 5 (Practical and synoptic) covers all content. Each of its three outcomes is taught below with worked examples.

Pair this page with the revision notes for recall and the practice questions for exam-style work. Course links: the OxfordAQA A-Level Chemistry hub, the printable 9620 checklist and the free 10-minute diagnostics, which flag weak topics.

Coverage

3.3.14 outcome	In plain terms	Stage
1	Say why good processes avoid solvents and start from non-hazardous materials	International A-level only
2	Say why short routes with a high percentage atom economy are preferred	International A-level only
3	Plan a route of no more than four steps using only 9620 reactions	International A-level only

No required practical appears under 3.3.14. The hands-on side of synthesis is in Required practical 4 (distilling a product, 3.3.5.2) and Required practical 10 (making an organic solid and checking its purity, 3.3.9.2). 3.1.2.5 gives the atom economy formula.

Outcome 1: solvent-free processes and safer starting materials

The cost of a solvent

A solvent only carries the reactants. None of it ends up in the product, yet a factory pays for it at every stage:

• Making and shipping it takes energy and raw materials.
• It must be separated from the product afterwards, usually by distillation, which takes more energy.
• Whatever is not recovered is waste.
• Many organic solvents are volatile and flammable, and some are toxic, so leaks bring fire risk, harm to workers and air pollution.

When one of the reactants is itself a liquid in which the others dissolve, no separate solvent is needed. The hydration of epoxyethane is a good example: water is both a reactant and the medium.

The cost of a hazardous starting material

A hazardous chemical needs sealed storage, controlled transport, protective equipment, training and emergency planning. Each adds cost and is a point where an accident can hurt people or the environment. A safer starting material removes the risk instead of managing it.

Examples you meet in this specification:

• Potassium cyanide (3.3.8.1 asks for “the hazards of using KCN”). It is very toxic, and acid releases hydrogen cyanide gas from it.
• Epoxyethane production (3.3.4.4 asks you to understand its hazards). Epoxyethane is extremely flammable, forms explosive mixtures with air, and is toxic and carcinogenic.
• (Chloromethyl)benzene, the side-chain chlorination product of methylbenzene, is toxic, irritates the eyes and skin, and may cause cancer. Chlorine, used to make it, is a toxic gas.
• Ethanoic anhydride rather than ethanoyl chloride for aspirin (3.3.9.2): the anhydride is cheaper, less corrosive, reacts less violently with water and gives no hydrogen chloride gas.

Outcome 2: short routes with high atom economy

The atom economy formula

From 3.1.2.5:

% atom economy = Mr of the wanted product × 100 ÷ (total Mr of every reactant)

Take the masses from the balanced equation, multiplying each reactant by its coefficient. A high value means most reactant atoms finish in the wanted product, so less is thrown away. Additions (one product) give 100%. Substitutions and eliminations usually make a by-product as well, so they usually give less than 100%.

Atom economy is a property of the equation. Percentage yield is a property of the experiment: how much product you actually collect compared with the maximum. Do not mix them up.

Worked example 1: two routes to ethane-1,2-diol

Ethane-1,2-diol (antifreeze) can be made from ethene by two routes that use only reactions in this specification. Relative atomic masses: C = 12.0, H = 1.0, O = 16.0, Br = 79.9, Na = 23.0.

Route A (two additions)
  Step 1  2CH₂=CH₂ + O₂ → 2C₂H₄O                  partial oxidation, silver catalyst
  Step 2  C₂H₄O + H₂O → HOCH₂CH₂OH
  Overall C₂H₄ + ½O₂ + H₂O → C₂H₆O₂
  Sum of reactants = 28.0 + 16.0 + 18.0 = 62.0
  % atom economy = 62.0 ÷ 62.0 × 100 = 100%

Route B (addition, then substitution)
  Step 1  CH₂=CH₂ + Br₂ → BrCH₂CH₂Br
  Step 2  BrCH₂CH₂Br + 2NaOH → HOCH₂CH₂OH + 2NaBr   NaOH(aq), warm
  Sum of reactants = 28.0 + 159.8 + 2(40.0) = 267.8
  % atom economy = 62.0 ÷ 267.8 × 100 = 23.2%

Route B wastes about three quarters of the reactant mass as sodium bromide. It also uses bromine, which is toxic and corrosive, and hydroxide can cause some elimination as a side reaction. Route A wins on atom economy and waste, but its first step is the hazardous epoxyethane process, so it needs careful temperature control and leak prevention. A good evaluation names both sides and then makes a judgement.

Why fewer steps

Each step loses product to side reactions, incomplete reaction and transfers. The overall yield is the product of the step yields written as decimals. Every extra step also needs its own reagents, energy, separation, time and waste handling.

Worked example 2: yield over three steps

A chemist makes hexan-2-one from 25.2 g of hex-1-ene (Mr 84.0) by this route:

Step 1  hex-1-ene + HBr → 2-bromohexane (major product)          yield 75%
Step 2  2-bromohexane + NaOH(aq), warm → hexan-2-ol               yield 82%
Step 3  hexan-2-ol + acidified K₂Cr₂O₇, reflux → hexan-2-one       yield 70%

Each step is 1 : 1. Calculate the mass of hexan-2-one (Mr 100.0) obtained.

Amount of hex-1-ene  = 25.2 ÷ 84.0 = 0.300 mol
Overall fraction     = 0.75 × 0.82 × 0.70 = 0.4305
Amount of hexan-2-one = 0.300 × 0.4305 = 0.12915 mol
Mass                 = 0.12915 × 100.0 = 12.9 g

Adding a fourth step with an 80% yield would cut the overall yield from 43.1% to 34.4%. Why three steps here? This specification has no reaction that adds water directly to an alkene, so the OH group has to go on through the bromoalkane.

Outcome 3: planning routes of up to four steps

Only reactions in this specification count. Some reactions you may have seen elsewhere are not in the 9620 content, for example steam hydration of an alkene, or converting an alcohol directly into a halogenoalkane. A route that relies on them is not a route here.

Reactions that matter most for planning

Reaction	Effect on the molecule	Section
KCN with a halogenoalkane	Adds one carbon (nitrile)	3.3.3.1
KCN, then dilute acid, with an aldehyde or ketone	Adds one carbon (hydroxynitrile)	3.3.8.1
Friedel–Crafts acylation or alkylation (AlCl₃ catalyst)	Joins a carbon chain to a benzene ring	3.3.10.2
Chlorine with methylbenzene (free radical)	Puts Cl on the side chain: (chloromethyl)benzene	3.3.10.2
Partial oxidation of ethene	Gives epoxyethane, which opens with water or alcohols	3.3.4.4
Reduction of nitriles or nitro compounds	Makes primary amines	3.3.11.1

The full reaction map is in the revision notes.

A planning method

1. Draw the starting material and the target. Mark the functional groups and count the carbons.
2. If the carbon count goes up, choose a C–C forming reaction from the table above.
3. Work backwards from the target: “What one reaction in the specification could make this?”
4. Check positions. HBr adds to an unsymmetrical alkene to give the major product via the more stable carbocation. Chlorine on a benzene ring is far less reactive than chlorine on a side chain.
5. For each step, write the reagent, the conditions and the name or structure of the product.

Ring chlorine versus side-chain chlorine

3.3.10.2 asks you to explain the relative reactivities of chlorine in the ring and in the side chain. In (chloromethyl)benzene, C₆H₅CH₂Cl, the chlorine sits on an sp³ carbon outside the ring and reacts as a primary halogenoalkane does: hydroxide, cyanide and ammonia substitute it. In chlorobenzene the chlorine is bonded directly to the ring. A lone pair on chlorine overlaps with the delocalised ring electrons, which strengthens the C–Cl bond, and the electron-rich ring repels nucleophiles. So nucleophilic substitution does not happen under normal conditions. For synthesis this means: to substitute, put the chlorine on the side chain.

Worked example 3: lengthening a side chain

Devise a three-step synthesis of 2-phenylethanamine, C₆H₅CH₂CH₂NH₂, from methylbenzene.

The target has one carbon more than methylbenzene, so a C–C bond has to be made. Reasoning backwards: reduce a nitrile to get the amine; make the nitrile with cyanide; and give cyanide a side-chain halogen to replace.

Step 1  C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl          ultraviolet light (free-radical substitution)
        product: (chloromethyl)benzene
Step 2  C₆H₅CH₂Cl + KCN → C₆H₅CH₂CN + KCl         KCN in aqueous ethanol, heat under reflux
        product: phenylethanenitrile
Step 3  C₆H₅CH₂CN + 4[H] → C₆H₅CH₂CH₂NH₂         reduction, for example H₂ with a nickel catalyst
        product: 2-phenylethanamine

The specification states that nitriles are reduced to primary amines but does not name the reducing agent; H₂ with nickel, or LiAlH₄, are standard choices. Using UV light, not AlCl₃, in step 1 matters: it puts the chlorine on the side chain, where step 2 can replace it.

Worked example 4: a four-step route to an ester

Phenylmethyl ethanoate, CH₃COOCH₂C₆H₅, has a jasmine-like smell. Devise a four-step synthesis from benzene.

Working back: the ester comes from phenylmethanol and an ethanoyl source; phenylmethanol comes from (chloromethyl)benzene and hydroxide; that comes from methylbenzene; methylbenzene comes from benzene by alkylation.

Step 1  C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl              AlCl₃ catalyst (Friedel–Crafts alkylation)
Step 2  C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl           ultraviolet light
Step 3  C₆H₅CH₂Cl + NaOH → C₆H₅CH₂OH + NaCl       NaOH(aq), warm
Step 4  C₆H₅CH₂OH + (CH₃CO)₂O → CH₃COOCH₂C₆H₅ + CH₃COOH
                                                  ethanoic anhydride, warm

Step 4 could instead use ethanoic acid with a concentrated sulfuric acid catalyst. That esterification is reversible, so the yield is lower, but its atom economy is higher (89.3% against 71.4% with the anhydride). That trade-off is what outcome 2 is about.

Chirality in a synthesis

If a step makes a chiral centre from a planar group, such as KCN adding to an aldehyde or an unsymmetrical ketone, the product is a racemic mixture. When only one enantiomer is wanted (for example in a drug), the other must be separated or the route redesigned. See optical isomerism.

Common errors

• Treating “UV light” and “AlCl₃” as interchangeable. With methylbenzene, UV light chlorinates the side chain; with benzene itself, UV light gives ring addition.
• Planning a nucleophilic substitution on chlorine bonded directly to a benzene ring.
• Using steam hydration or an alcohol-to-halogenoalkane step. Neither is in the 9620 content.
• Adding step yields, or averaging them. Multiply.
• Leaving a coefficient out of the atom economy denominator, such as 2NaOH in route B above.
• Using NaBH₄ to reduce a nitrile. In this specification NaBH₄ reduces aldehydes and ketones.

Where to go next

Revise single reactions in halogenoalkanes, alkenes, alcohols, aldehydes and ketones and aromatic chemistry, and atom economy in amount of substance.

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.