HookHow Boots turned a six-step drug into a three-step one

When Boots first synthesised ibuprofen in the 1960s, the route ran to six steps. Six separate reactions, each needing its own reagents, its own solvent and its own purification, and at every stage some of the material was thrown away as waste — so much that barely a third of the atoms in the starting materials ended up in the drug. Thirty years later a rival team at the BHC company designed a completely different route to exactly the same molecule that took just three steps, used a reusable catalyst, and put close to 80% of the starting atoms into the product. Same target, same painkiller on the shelf — but a route half as long and far less wasteful. That comparison won a US Presidential Green Chemistry award, and it is the entire spirit of this section.

Organic synthesis is the part of A-level chemistry where nothing new is taught and everything is tested at once. You are handed a starting molecule and a target molecule and asked to connect them, step by step, choosing reagents and conditions from the whole two-year toolkit of reactions. It is the exam's endgame: no single move is hard, but you have to see several moves ahead, spot when the carbon skeleton needs to grow, and keep the ring intact when you are working with benzene. This section pulls the aliphatic and aromatic reactions you already know into one map and teaches you to navigate it.

ModelThe aliphatic map — every functional group is a junction

Think of aliphatic synthesis as a network of stations connected by reactions you already know. An alkene is a hub: adding a hydrogen halide gives a halogenoalkane, adding steam with a phosphoric acid catalyst gives an alcohol, and adding hydrogen over nickel gives an alkane. A halogenoalkane is another hub: warm it with aqueous sodium hydroxide under reflux and nucleophilic substitution gives an alcohol; warm it with sodium hydroxide dissolved in ethanol and elimination gives an alkene instead; heat it with excess ammonia in ethanol and you get an amine.

The alcohol is the busiest junction of all. Oxidise a primary alcohol gently and you get an aldehyde, oxidise it hard and you get a carboxylic acid; dehydrate it and you get an alkene; react it with a carboxylic acid and you get an ester. Aldehydes and ketones can be reduced back to alcohols with sodium borohydride, and carboxylic acids lead on to esters, acyl chlorides and amides.

The single most important habit is to write reagents and conditions together, because the same two reactants give different products under different conditions. Aqueous NaOH substitutes; ethanolic NaOH eliminates. Distilling off the product as it forms stops oxidation at the aldehyde; refluxing drives it on to the acid. In an exam these conditions are worth as many marks as the reagents themselves, and they are where most candidates quietly lose them.

MechanismGrowing the carbon chain — the reactions that add a carbon

Most reactions keep the number of carbon atoms the same, so the moment a target has more carbons than the starting material, you need one of the two special reactions that lengthen the chain. Both use a cyanide as the source of the extra carbon.

The first is the reaction of a halogenoalkane with potassium cyanide (KCN dissolved in aqueous ethanol, heated under reflux). The \(\text{CN}^-\) ion is a nucleophile; it replaces the halogen and forms a nitrile, adding one carbon to the chain. That nitrile is then a springboard: reduce it (with \(\text{LiAlH}_4\), or hydrogen over a nickel catalyst) to make an amine, or hydrolyse it (reflux with dilute acid) to make a carboxylic acid. The second chain-lengthening reaction is the addition of hydrogen cyanide to an aldehyde or ketone, which gives a hydroxynitrile — again adding one carbon, this time next to a new \(-\text{OH}\) group.

Spotting the change in carbon number is the first thing to do in any route question. Count the carbons in the start and in the target. If the number is unchanged, you are only ever converting one functional group into another. If it has gone up by one, a KCN or HCN step must appear somewhere in your answer — there is almost no other way to do it at this level.

Worked example

Devise a two-step synthesis of butanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\), starting from 1-bromopropane, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\).

First, count carbons. 1-Bromopropane has three; butanoic acid has four. The chain must grow by one carbon, so a cyanide step is unavoidable, and since we are starting from a halogenoalkane the tool is KCN.

Step 1 — extend the chain. Reflux 1-bromopropane with potassium cyanide dissolved in aqueous ethanol. The \(\text{CN}^-\) substitutes the bromine to give butanenitrile: \[\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{KBr}\]

Step 2 — make the acid. Reflux the nitrile with dilute hydrochloric acid to hydrolyse it to the carboxylic acid: \[\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{NH}_4\text{Cl}\]

The four-carbon target is reached in two steps, and — the marks people drop — every reagent is named with its condition: KCN in aqueous ethanol under reflux, then dilute acid under reflux.

ModelThe aromatic map — substitute the ring, don't add to it

Benzene follows its own rules, and the first is that it reacts by electrophilic substitution, not addition. Its delocalised ring of electrons is stable, so reactions swap a hydrogen on the ring for something else and keep the ring intact rather than breaking it open the way an alkene would. Learn the ring reactions as a short menu.

Nitration — warm benzene with a mixture of concentrated nitric and concentrated sulfuric acids at about 50 °C — puts a \(-\text{NO}_2\) group on the ring to give nitrobenzene. Friedel–Crafts acylation — an acyl chloride with an aluminium chloride catalyst — puts a \(-\text{COR}\) group on the ring to give an aromatic ketone. Friedel–Crafts alkylation — a halogenoalkane with an aluminium chloride catalyst — puts an alkyl group on. And halogenation — a halogen with a halogen-carrier catalyst such as \(\text{AlCl}_3\) or \(\text{FeBr}_3\) — puts on a halogen.

The reaction that unlocks aromatic synthesis, though, is a reduction. Nitrobenzene can be reduced to phenylamine (an aromatic amine) by heating with tin and concentrated hydrochloric acid, then adding sodium hydroxide to liberate the free amine. This nitrate-then-reduce sequence is the standard way to get a nitrogen onto a benzene ring, and phenylamine is the gateway to dyes and to many aromatic products. Because the \(-\text{NO}_2\) group can only be put on by nitration and then turned into \(-\text{NH}_2\) by reduction, that pair of steps appears in a huge fraction of aromatic route questions.

CaseA worked aromatic route — benzene to a paracetamol relative

Aromatic routes are usually shorter than aliphatic ones — often just two or three steps — but each step has strict conditions that must be quoted exactly. The classic three-move sequence is nitrate the ring, reduce the nitro group to an amine, then react the amine with an acylating agent. That last acylation is the same addition–elimination chemistry you met with amines and acyl chlorides, now used to finish a synthesis.

Worked example

Starting from benzene, devise a synthesis of N-phenylethanamide, \(\text{C}_6\text{H}_5\text{NHCOCH}_3\) (the acetanilide that is a close chemical relative of paracetamol).

Step 1 — nitration. Warm benzene with a mixture of concentrated nitric acid and concentrated sulfuric acid at about 50 °C. An \(-\text{NO}_2\) group substitutes onto the ring: \[\text{C}_6\text{H}_6 \rightarrow \text{C}_6\text{H}_5\text{NO}_2\]

Step 2 — reduction to the amine. Heat nitrobenzene under reflux with tin and concentrated hydrochloric acid, then add excess sodium hydroxide to release the free amine, phenylamine: \[\text{C}_6\text{H}_5\text{NO}_2 \rightarrow \text{C}_6\text{H}_5\text{NH}_2\]

Step 3 — acylation. React phenylamine with ethanoyl chloride (or, more gently, ethanoic anhydride). The amine's lone pair attacks the acyl group and an amide forms: \[\text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{HCl}\]

Three steps, benzene to product. The examiner is checking three things: that you nitrated first (you cannot put \(-\text{NH}_2\) straight onto benzene), that you reduced with Sn/concentrated HCl and then NaOH, and that you knew an acyl chloride or anhydride — not a carboxylic acid — is what acylates an amine cleanly.

DataChoosing between routes — steps, yield and atom economy

Route questions rarely have a single right answer, and the higher-mark versions ask you to justify a choice. Three criteria decide it. Number of steps: every extra step means another separation and another chance to lose material, so a shorter route is almost always preferable — the ibuprofen story in the introduction is exactly this, six steps beaten by three. Overall yield: yields multiply, so a four-step route where each step is 70% efficient delivers only \(0.70^4 \approx 0.24\), just 24% of the theoretical product, even though no single step looks bad.

Atom economy is the third, and it is why the KCN route to a carboxylic acid is efficient (almost every atom of the nitrile ends up in the product) while a route that builds and then discards a large leaving group is wasteful. A high atom economy means less of the starting material is thrown away as by-products, which matters for cost, for waste disposal and for the environment.

The practical exam habit that ties all of this together is to work backwards from the target — a way of thinking chemists call retrosynthesis. Look at the target's functional group and ask what single reaction makes it, then what makes that intermediate, and keep stepping back until you reach the given starting material. Working backwards keeps you from wandering down long forward routes, and it naturally finds the shortest path — the one the mark scheme is usually looking for.

VocabularyKey terms the mark scheme pays for

Functional group interconversion

Changing one functional group into another (e.g. alcohol to halogenoalkane) while keeping the carbon skeleton the same — the individual moves that make up a synthetic route.

Chain lengthening (nitrile route)

Adding one carbon by reacting a halogenoalkane with KCN in aqueous ethanol to form a nitrile, which can then be reduced to an amine or hydrolysed to a carboxylic acid.

Reagents and conditions

Both are needed to specify a step: the same reactants can give different products (aqueous vs ethanolic NaOH; distil vs reflux), so conditions carry marks in their own right.

Nitration

Electrophilic substitution of benzene using concentrated HNO₃ and concentrated H₂SO₄ at about 50 °C, placing an –NO₂ group on the ring to form nitrobenzene.

Reduction of nitrobenzene

Conversion of –NO₂ to –NH₂ by heating with tin and concentrated HCl, then adding NaOH — the standard way to make an aromatic amine (phenylamine).

Friedel–Crafts acylation

Electrophilic substitution of benzene by an acyl chloride with an AlCl₃ halogen-carrier catalyst, attaching a –COR group to form an aromatic ketone.

Retrosynthesis

Planning a route by working backwards from the target: identifying the reaction that makes it, then the reaction that makes that intermediate, until you reach the starting material.

Atom economy

The proportion of reactant atoms that end up in the desired product; higher atom economy means less waste and is a key criterion when choosing between synthetic routes.

TrapsMisconceptions that cost marks

“You can turn a halogenoalkane straight into a pure primary amine with ammonia.”

Actually: Reacting a halogenoalkane with ammonia gives a messy mixture, because the amine formed reacts further to give secondary and tertiary amines. To make a clean primary amine — and to add a carbon — the nitrile route (KCN, then reduction) is the reliable choice.

“Any oxidation of a primary alcohol gives a carboxylic acid.”

Actually: The conditions decide the product. Distilling the aldehyde off as it forms stops the oxidation at the aldehyde; only refluxing with excess oxidising agent takes it all the way to the carboxylic acid. Naming the reagent without the condition loses the mark.

“Benzene reacts by addition, like an alkene, in synthesis.”

Actually: Benzene's delocalised ring is stable, so it undergoes electrophilic substitution — a hydrogen on the ring is swapped for a new group and the ring survives. Writing an addition across the ring is a fundamental error in an aromatic route.

ExamWhat examiners want

Count the carbons first. If the target has more carbons than the starting material, a chain-lengthening step (KCN with a halogenoalkane, or HCN with a carbonyl) must appear — this is the single most common thing candidates miss, and without it the route is impossible. If the carbon count is unchanged, you are only interconverting functional groups.

Write reagents and conditions for every step, not just reagents. Aqueous NaOH under reflux (substitution) versus NaOH in ethanol under reflux (elimination); gentle oxidation with distillation (aldehyde) versus reflux (carboxylic acid); Sn and concentrated HCl then NaOH for reducing nitrobenzene. Each condition is independently creditable and the mark scheme lists them.

For aromatic targets, remember you cannot put –NH₂ directly onto benzene: nitrate the ring, then reduce. Keep to electrophilic substitution and preserve the ring. Where a question asks you to compare or justify a route, argue from number of steps (fewer is better), overall yield (yields multiply, so 0.70⁴ ≈ 24%) and atom economy (less waste). Plan by working backwards from the target so you find the shortest path the mark scheme rewards, and quote a balanced equation for a step whenever you are asked to.