Organic Synthesis (A Level)

0. What Examiners Want

This guide covers Topic 36 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. It uses every reaction in Topics 15–18 and 25–35, and section 5 adds the coursebook material on chirality and single-enantiomer drugs, which explains why synthesis matters in the pharmaceutical industry.

1. Identifying Functional Groups

Most molecules in Paper 4 carry more than one functional group. To identify them you use test reagents, each of which gives a visible result with one group (or a few). To predict the reactions you ask of each group in turn: what does it do with this reagent?

Each test shows one group, or a small set of groups. Several tests give the same result for different groups, so combine them.

Predicting the reactions of a molecule with several groups. Treat each group separately and ask whether the reagent can attack it. Some reagents are selective: LiAlH₄ reduces C=O, –COOH and –CN but not C=C or a benzene ring; NaBH₄ reduces aldehydes and ketones but not a carboxylic acid; Na₂CO₃ reacts with a carboxylic acid but not with a phenol; H₂/Ni reduces C=C and C=O.

One molecule, three groups: the reagents that attack each group are listed below. A reagent such as Br₂(aq) can attack more than one group.

Worked example. The compound HOC₆H₄CH=CHCOOH is shown above. Describe how you would show that it contains a carboxylic acid group but that the phenol group is also present. [4]

Add Na₂CO₃(aq): effervescence of CO₂ shows the –COOH group [1]; a phenol alone would not react [1]. Add neutral FeCl₃(aq) to a separate sample: a violet colour shows the phenol [1]. Br₂(aq) would not distinguish, because the C=C and the phenol ring both decolourise or react with it [1].

Worked example. State the product when HOC₆H₄CH=CHCOOH reacts with (i) H₂ and Ni, (ii) LiAlH₄ in dry ether, (iii) excess aqueous NaOH. [3]

(i) HOC₆H₄CH₂CH₂COOH (the C=C is reduced; the ring is not reduced under mild conditions) [1]. (ii) HOC₆H₄CH=CHCH₂OH (only the –COOH is reduced) [1]. (iii) ⁻OC₆H₄CH=CHCOO⁻ (both acidic groups give sodium salts) [1].

2. The Reaction Maps

A synthesis uses reactions you already know. The two maps below join them up: each numbered arrow carries a reagent and condition listed underneath the map. Learn them as a network, not as a list; the best way to do so is to redraw each map from memory.

Aliphatic map: 16 numbered steps. Arrows 13–16 add a carbon (nitrile) or build an amine.

Aromatic map: ring substitution (1–3, 6), reduction (4, 7), diazotisation (5), oxidation of the side chain (8), and the reactions of the diazonium salt (9, 10).

3. Devising a Synthetic Route

To devise a route you work backwards from the target and use as few steps as possible, because some material is lost at each step and the yield of every reaction is below 100%. The overall yield is the product of the individual yields.

Compare, decide whether carbon must be added, work backwards, check the order, then write every step in full.

Changing the carbon skeleton. If the target has more carbon atoms than the starting material you must add one. The syllabus gives three ways: the nitrile route (halogenoalkane → nitrile, which hydrolyses to an acid or reduces to an amine), HCN addition to an aldehyde or ketone (a hydroxynitrile, which hydrolyses to a hydroxy acid), and the Friedel–Crafts reaction on a benzene ring. Shortening a chain is rare: only the iodoform reaction removes a carbon.

Each row is one scheme. The first three add one carbon to an aliphatic chain; the last two add an alkyl or an acyl group to a benzene ring.

Four routes of two or three steps. Check that every reagent is on the reaction maps.

Worked example. Devise a route from ethene to propylamine, CH₃CH₂CH₂NH₂. Give reagents and conditions, and name the intermediates. [5]

Propylamine has one more carbon than ethene, so the nitrile route is needed. Step 1: HBr, room temperature, gives bromoethane [2: reagent, intermediate]. Step 2: KCN in ethanol, heat under reflux, gives propanenitrile, CH₃CH₂CN [2]. Step 3: LiAlH₄ in dry ether (or H₂/Ni) gives propylamine [1].

Worked example. Devise a route to convert benzene into phenol. [5]

Step 1: conc. HNO₃ + conc. H₂SO₄, 25–60 °C: nitrobenzene [1]. Step 2: Sn + conc. HCl, heat under reflux, then NaOH(aq): phenylamine [2]. Step 3: NaNO₂ + HCl(aq), below 10 °C: benzenediazonium chloride [1]. Step 4: warm with water: phenol (N₂ is lost) [1].

Worked example. 2.80 g of ethene is converted into ethylamine in two steps with yields of 80% and 75%. Calculate the mass of ethylamine formed. (M_r: C₂H₄ = 28.0, C₂H₅NH₂ = 45.0) [3]

Amount of ethene = 2.80 ÷ 28.0 = 0.100 mol [1]. Overall yield = 0.80 × 0.75 = 0.60, so amount of amine = 0.060 mol [1]. Mass = 0.060 × 45.0 = 2.70 g [1].

4. Analysing a Given Route

When a route is given you must say, for each step, the type of reaction, the reagent and condition, and (when asked) the possible by-products. A systematic approach is to compare the formula before and after, find which group changed, and name the reaction from that change.

The table lists the type and a possible by-product for each of the three steps.

Worked example. Propene is converted to propanone in three steps: (1) HBr; (2) NaOH(aq), heat; (3) K₂Cr₂O₇/H₂SO₄, heat. Name the type of reaction in each step and give a possible by-product of step 1. [5]

Step 1: electrophilic addition [1]. Step 2: nucleophilic substitution (hydrolysis) [1]. Step 3: oxidation [1]. By-product of step 1: 1-bromopropane [1], formed through the less stable primary carbocation, so it is the minor product [1].

Worked example. In the route benzene → methylbenzene → 4-nitromethylbenzene → 4-methylphenylamine, explain why 2-nitromethylbenzene is also formed in step 2, and state the type of reaction in step 3. [3]

–CH₃ is 2,4-directing, so the NO₂⁺ electrophile also attacks the 2-position [1]; the isomers must be separated [1]. Step 3 is reduction [1].

5. Chirality and Single-Enantiomer Drugs

Many drugs are chiral. A chiral centre is a carbon bonded to four different groups, and the molecule exists as two non-superimposable mirror images, enantiomers. They have identical physical and chemical properties, except that they rotate plane-polarised light in opposite directions, and they differ in their biological effect because enzymes and receptors are chiral.

The two enantiomers of lactic acid. A 50 : 50 mixture is racemic and optically inactive.

An ordinary laboratory synthesis that creates a new chiral centre from a planar group gives a racemic mixture (50 : 50) because the reagent attacks either face of the planar group equally. For a drug, one enantiomer may be effective and the other harmful: one enantiomer of naproxen treats arthritis while the other can damage the liver, and one enantiomer of a tuberculosis drug causes blindness. A pure enantiomer therefore means a lower dose (about half), better therapeutic activity, fewer side-effects for the patient and less risk of legal action for the manufacturer. About 80% of new patented drugs are single enantiomers.

Resolution wastes half the product; the chiral pool and chiral catalysts avoid the racemic mixture.

• Optical resolution. The racemic mixture reacts with a pure chiral compound (chiral auxiliary). The new compounds differ in solubility, so they are separated by fractional crystallisation, and the wanted one is converted back (for example with dilute alkali). The process is slow, uses extra reagents and discards half the material. HPLC with an optically active stationary phase can also separate the enantiomers.
• Chiral pool. Start from a natural compound that already has the required configuration (L-amino acids, carbohydrates) and keep it unchanged through the route.
• Chiral catalyst or enzyme. A catalyst with a chiral shape (for example a ruthenium organometallic catalyst in the manufacture of naproxen) gives only one enantiomer. Enzymes have a stereospecific active site; they are often fixed on inert supports so they can be reused. They need fewer steps, so the process is greener.

Worked example. Ethanal reacts with HCN, with a trace of KCN, to give 2-hydroxypropanenitrile, which is hydrolysed to lactic acid. Explain why the lactic acid formed is optically inactive. [3]

The C=O group is planar, so the ⁻CN nucleophile can attack the δ+ carbon from either side with equal probability [1]. The two enantiomers are formed in equal amounts (a racemic mixture) [1]. The rotations of plane-polarised light cancel, so the product is optically inactive [1].

Worked example. State two benefits of using a single enantiomer rather than a racemic mixture of a drug. [2]

A lower dose is needed because all the material is active, so cost falls [1]; the other enantiomer cannot cause side-effects, so patients are safer and the manufacturer faces less risk of legal action [1].

6. Quick Sheet and Checklist

• [ ] I can identify each functional group in a molecule from the tests and predict what each reagent does to each group.
• [ ] I can redraw the aliphatic and aromatic reaction maps with reagents and conditions.
• [ ] I can devise a route of two to four steps, working backwards, with intermediates, reagents and conditions.
• [ ] I know three ways to add a carbon atom and the one syllabus reaction that removes one.
• [ ] I can state the type of reaction for each step of a given route and a possible by-product.
• [ ] I can explain why the order of steps matters on a benzene ring.
• [ ] I can explain chirality, enantiomers, racemic mixtures and why drugs need a single enantiomer, with three ways to make one.