Related topics: Halogenoalkanes · Aromatic Chemistry
The big idea
Every reaction you have learned is an arrow between two functional groups. Organic synthesis is just chaining those arrows to get from a cheap starting material to a target molecule, usually in up to four steps. The skill is knowing the map well enough to plan a route, and to spot when a step changes the length of the carbon chain.
The synthesis map
Almost the whole of organic chemistry connects into one web. The aliphatic map hangs off the alcohol, which is the busiest junction:
| From → to | Reagents & conditions | Type |
|---|---|---|
| alkane → halogenoalkane | Cl2 or Br2, UV light | free-radical substitution |
| alkene → halogenoalkane | HBr (or HCl), room temp | electrophilic addition |
| alkene → alcohol | steam, H3PO4 catalyst | hydration |
| halogenoalkane → alcohol | NaOH(aq), reflux | nucleophilic substitution |
| halogenoalkane → alkene | KOH in ethanol, reflux | elimination |
| halogenoalkane → nitrile (+1 C) | KCN in ethanol/water, reflux | nucleophilic substitution |
| halogenoalkane → amine | excess NH3, ethanol | nucleophilic substitution |
| alcohol → alkene | conc H2SO4 (or conc H3PO4), heat | dehydration (elimination) |
| 1° alcohol → aldehyde | K2Cr2O7/H2SO4, warm & distil | partial oxidation |
| 1° alcohol → carboxylic acid | K2Cr2O7/H2SO4, reflux | full oxidation |
| 2° alcohol → ketone | K2Cr2O7/H2SO4, reflux | oxidation |
| aldehyde → carboxylic acid | K2Cr2O7/H2SO4, reflux | oxidation |
| aldehyde/ketone → alcohol | NaBH4 | reduction |
| aldehyde/ketone → hydroxynitrile (+1 C) | KCN, then dilute acid (HCN also accepted) | nucleophilic addition |
| nitrile → amine | H2/Ni (or LiAlH4) | reduction |
| nitrile → carboxylic acid | dilute H2SO4, reflux | hydrolysis |
| carboxylic acid ↔ ester | alcohol + conc H2SO4 | esterification |
| acyl chloride / anhydride → carboxylic acid | water (vigorous, room temp) | nucleophilic addition–elimination |
| acyl chloride / anhydride → ester | alcohol, room temp | nucleophilic addition–elimination |
| acyl chloride / anhydride → primary amide | NH3, room temp | nucleophilic addition–elimination |
| acyl chloride / anhydride → N-substituted amide | primary amine, room temp | nucleophilic addition–elimination |
From carboxylic acids and derivatives (3.3.9) come the acyl chloride and the acid anhydride, the acid’s activated relatives. Water, alcohols, ammonia and primary amines attack either in rapid room-temperature nucleophilic addition–elimination, giving the acid, an ester, a primary amide or an N-substituted amide: the highest-yield way to make an ester.
The short aromatic map runs: benzene → (nitration) nitrobenzene → (Sn/conc HCl, then NaOH) phenylamine; and benzene → (Friedel–Crafts acylation, RCOCl/AlCl3) an aromatic ketone. See aromatic chemistry (3.3.10) for the mechanisms.
substitution addition elimination oxidation reduction hydrolysis / esterification addition–elimination
panel 1: the core map
Both nitrogen routes start from the halogenoalkane, and the KCN route adds a carbon.
panel 2: the oxidation ladder & the acid’s relatives
Ketones are the dead end: no further oxidation.
panel 3: the acyl chloride branch
Both are the acid’s activated relatives. You are given them, not asked to make them. The anhydride reacts more slowly and safely and is cheaper (why industry uses it for aspirin).
panel 4: the aromatic strand
Two steps take benzene to phenylamine; Friedel–Crafts acylation hangs a ketone off the ring.
Exam questions
Q1
[1 mark]
Which reagents convert a primary alcohol into a carboxylic acid?
Full oxidation to the acid needs reflux (to keep the aldehyde in the flask so it oxidises further). Distilling off gives the aldehyde; conc H2SO4 dehydrates to an alkene; NaBH4 reduces.
Source: AQA A-Level Chemistry past papers.
Planning a synthesis
A synthesis question gives you a start molecule and a target and asks for a route of up to four steps. Two habits make it reliable:
- Work from both ends. Trace arrows forwards from the starting material and backwards from the target until the two meet.
- Check the carbon count. If the target has one more carbon than the start, a nitrile step (KCN on a halogenoalkane, or KCN then dilute acid on a carbonyl) must appear somewhere: on this course, nothing else adds a carbon to an aliphatic chain.
Carrying a route out and purifying what it makes is Required practical 10. Practical skills are examined across all three papers.
Worked example: propene to propan-2-amine
Propene (3 C) → propan-2-amine (3 C). No chain change, so no nitrile needed, but watch the regiochemistry:
CH3CH=CH2 → HBr CH3CHBrCH3 → excess NH3 CH3CH(NH2)CH3
Step 1 follows Markovnikov’s rule: HBr adds to give the major product 2-bromopropane (via the more stable secondary carbocation), so the amine is propan-2-amine. (To lengthen a chain instead, you would route through a nitrile.)
Interactive: build the synthesis route
Get from the start molecule to the target. Build each step as a real reagent system: pick the reagent(s), and the conditions where they matter, then press React. Incomplete or wrong systems show what really happens (and cost a try); the finished route reads back as a written exam answer.
Exam questions
Q2
[4 marks]
Devise a two-step synthesis of butanoic acid (CH3CH2CH2COOH) from 1-bromopropane (CH3CH2CH2Br). Give the reagents and conditions for each step and name the intermediate.
Show answer
The target has one more carbon than the start, so route through a nitrile.
Step 1: KCN in aqueous ethanol, reflux → butanenitrile, CH3CH2CH2CN. 2 marks
Step 2: dilute H2SO4 (or HCl), reflux (hydrolysis) → butanoic acid. 2 marks
(The intermediate, butanenitrile, has gained the carbon from the CN group.)
Source: AQA A-Level Chemistry past papers.
Green chemistry & atom economy
The best route on paper is not always the best in industry. Chemists design processes to be greener, and the exam expects you to be able to say why:
- Fewer steps and high atom economy: less waste, and more of the starting atoms end up in the product, cutting cost and raw-material use.
- No solvent where possible: avoids the energy and hazard of using, recovering and disposing of large volumes of solvent.
- Non-hazardous starting materials: safer to handle and less harmful if released.
Atom economy
% atom economy = (Mr of desired product∑ Mr of all reactants) × 100
A reaction that makes only the wanted product has 100% atom economy. Addition reactions tend to be high; reactions that also throw off a small molecule (like substitution or elimination) are lower. High atom economy means less waste and better sustainability.
Precision points
- Dropping conditions when they carry a mark: some steps are just a reagent (HBr; NaBH4), but quote conditions where the question asks for them, or where they change the product (K2Cr2O7/H2SO4: distil for the aldehyde vs reflux for the acid).
- Missing a carbon-count change: if the target is one carbon longer, you must go via a nitrile.
- Confusing atom economy (about mass of atoms in the product) with percentage yield (about how much you actually made).
- Proposing a step that isn’t on the map: only use reactions from the AQA course.
Examiner reports describe multi-step synthesis as a major synoptic weakness: students underperform on very common reactions, on naming the mechanism, and on conditions and route selectivity. Depending on how a question is worded, each arrow can be worth up to three marks (reagent, any conditions, and the mechanism name), so read the command words and give exactly what is asked.
Capstone quiz: five past-paper questions
Five real AQA multiple-choice questions on this topic, in the style that opens Paper 3. Every one turns on the thing this topic is really about: where the carbons go. Pick one answer each; the reasoning appears once you commit.
Q1
[1 mark]
1-Bromopropane reacts with a solution of potassium cyanide in aqueous ethanol. What is the organic product of this reaction?
The cyanide ion is the nucleophile, so −CN replaces Br and its carbon joins the chain: three carbons in, four carbons out, giving butanenitrile. Propanenitrile is the classic trap: it keeps the nitrile group but forgets that the CN carbon counts, so the chain never lengthens. The two amines are what you would get only after a further reduction step.
Q2
[1 mark]
Which pair of reagents does not produce ethanol?
Hydrolysing an ester breaks the C–O single bond, so the alcohol comes from the alkyl group written after the oxygen. In CH3COOCH3 that group is CH3, so you get methanol. HCOOCH2CH3 is the tempting one because it also looks ester-like, but there the OCH2CH3 half releases ethanol. Read which side of the oxygen the ethyl group sits on.
Q3
[1 mark]
A two-step preparation of propylamine is shown: bromoethane → X → propylamine. What is X?
Bromoethane has two carbons and propylamine has three, so the intermediate must be where the extra carbon is gained. KCN gives propanenitrile, which is then reduced to propylamine. CH3CH2NH2 is the seductive answer: ethylamine is a perfectly sensible product of bromoethane with ammonia, but it still has only two carbons, so there is no way to reach a three-carbon amine in one more step.
Q4
[1 mark]
What is the product when 3-methylbutan-2-one reacts with acidified KCN?
This is the other chain-lengthening route: −CN adds across the C=O to give a hydroxynitrile one carbon longer. The naming trips people up because the nitrile carbon becomes C1, so the old carbonyl carbon is now C2. That is where the OH sits, alongside its methyl, with a second methyl on C3. The 3-hydroxy option is the trap for anyone numbering from the original ketone instead of from the CN.
Q5
[1 mark]
2-Methylbutylamine can be synthesised from an alkene. What is the identity of the alkene?
Work backwards. 2-Methylbutylamine is CH3CH2CH(CH3)CH2NH2; the CH2NH2 carbon arrived with the cyanide, so the halogenoalkane before it was 2-bromobutane, a four-carbon skeleton, which comes from but-2-ene plus HBr. The tempting error is to match the name: two options already say “2-methylbut”, but those alkenes have five carbons, and this route adds a carbon rather than losing one, so they overshoot.
Source: AQA A-Level Chemistry past papers.
3.3.14 Organic synthesis: Quick-reference summary
- The map: every reaction is an arrow between functional groups; the alcohol is the central junction of the aliphatic map.
- Chain length: only KCN (halogenoalkane → nitrile) and KCN then dilute acid (carbonyl → hydroxynitrile) add a carbon; nitriles then reduce to amines or hydrolyse to acids.
- Acyl chlorides (or acid anhydrides): water, an alcohol, NH3 or a primary amine give the acid, an ester, a primary amide or an N-substituted amide: rapid nucleophilic addition–elimination at room temperature.
- Planning: work forwards from the start and backwards from the target; watch for a step that changes the carbon count; give the reagent for every step, and conditions where they are asked for or where they change the product.
- Green chemistry: chemists design routes with high atom economy, fewer steps, no solvent where possible and non-hazardous starting materials.
- Atom economy = (Mr of desired productsum of Mr of all reactants) × 100.