Organic Synthesis

0. What Examiners Want

Where this topic appears. Recent AS papers have asked for: the reagents for each stage of a two-step conversion of a molecule with several groups (February/March 2023 Paper 1: citric acid to tricarballylic acid needs concentrated H₂SO₄, then H₂(g) with Ni); a synthesis from an alkene through HBr, KCN in ethanol and H⁺(aq) to a carboxylic acid, with the type of mechanism at each step (2024 Paper 2); and a three-step synthesis from propanone through a hydroxynitrile and a hydroxy acid to an alkene, with the reagents and reaction type for each step completed in a table (2022 Paper 2). This guide uses only the reactions in the AS syllabus (Topics 14 to 20). Reactions of benzene, phenol, amides, acyl chlorides and amino acids belong to A Level.

1. The AS Reaction Toolkit

The two maps hold every reaction you need for a synthesis question. Learn them as arrows: from a compound, to a compound, with a reagent, conditions and type. Three habits make the maps easier to remember.

• The alcohol is the hub. Alcohols are made from alkenes (steam), halogenoalkanes (NaOH(aq)), carbonyl compounds and acids (reduction) and esters (hydrolysis), and they turn into alkenes, halogenoalkanes, carbonyl compounds, acids and esters. Most routes pass through an alcohol.
• The carbonyl pair. A primary alcohol gives an aldehyde (distil) and then an acid (reflux). A secondary alcohol gives a ketone and stops there. NaBH₄ or LiAlH₄ reverses the first oxidation.
• The chain-lengthening pair. KCN with a halogenoalkane and HCN with a carbonyl compound are the only reactions that add a carbon. Both give a nitrile, which hydrolyses to an acid.

2. Identifying Functional Groups

To identify the groups in an unknown molecule, work through the tests in a sensible order and write down what you see each time. A test result tells you about one group only, so you combine several results.

Worked example. Compound X, C₄H₈O, gives an orange precipitate with 2,4-DNPH, so it contains C=O (an aldehyde or a ketone). It gives no silver mirror with Tollens’ reagent, so it is a ketone, not an aldehyde. It gives a pale yellow precipitate with alkaline iodine, so it contains CH₃CO–. The only C₄H₈O ketone is butanone, CH₃COCH₂CH₃. Reducing it with NaBH₄ gives butan-2-ol, CH₃CH(OH)CH₂CH₃, which also gives the yellow precipitate (it contains CH₃CH(OH)–) and is oxidised to orange-to-green by acidified K₂Cr₂O₇.

A few results need care:

• Bromine water is decolourised by alkenes only. Alkanes, alcohols and acids do not decolourise it.
• Sodium reacts with both alcohols and carboxylic acids (and water), so fizzing with Na does not tell them apart. Carbonate fizzes only with the acid.
• Acidified dichromate goes green with primary and secondary alcohols and with aldehydes. A tertiary alcohol and a ketone give no change.

3. Predicting Properties and Reactions

For a molecule with more than one functional group, treat each group on its own, using the reactions you know for it, and then check whether a reagent also attacks another group.

For HOCH₂CH=CHCOOH:

• With bromine water: the C=C reacts (orange to colourless); the OH and COOH do not.
• With sodium carbonate: only the COOH reacts, giving CO₂.
• With H₂(g) and Ni: the C=C is hydrogenated to give HOCH₂CH₂CH₂COOH.
• With LiAlH₄: the COOH is reduced to CH₂OH, giving HOCH₂CH=CHCH₂OH; the C=C is not reduced.
• With ethanol and concentrated H₂SO₄: the COOH forms an ester.
• With sodium metal: both the OH and the COOH react, so bubbles of H₂ are seen.

Worked example (choosing reagents). Citric acid has three COOH groups and one tertiary OH. It is converted into tricarballylic acid, which has the same three COOH groups and no OH, in two stages. Stage 1: concentrated H₂SO₄ removes H₂O (dehydration), giving a C=C. A tertiary alcohol cannot be oxidised, but it can be dehydrated. Stage 2: H₂(g) with a Ni catalyst adds H₂ across the C=C. LiAlH₄ would be wrong because it would also reduce the three COOH groups, and NaOH(aq) would only form salts.

Predicting properties. The functional groups also decide physical properties. OH and COOH groups form hydrogen bonds, so a molecule that has them has a higher boiling point and is more soluble in water than a hydrocarbon of similar M_r. A COOH group is acidic (it reacts with carbonates); an OH group is much less acidic and does not. A nitrile or halogenoalkane has no O–H or N–H bond to hydrogen-bond with itself.

4. Planning a Multi-Step Route

Work backwards from the product. Write down the starting compound and the target, name the functional group you need to form, and use the table above to see which compounds make it. Repeat until you reach the starting material, then write the route forwards with reagents and conditions.

1. Compare the carbon chains. If the target has one more carbon than the start, the route must include KCN or HCN. If the chain length is the same, it must not.
2. Choose the shortest route. Every extra step lowers the overall yield. Ethene to ethanoic acid: (1) steam with H₃PO₄ catalyst gives ethanol; (2) acidified K₂Cr₂O₇ under reflux gives ethanoic acid. The route via bromoethane (HBr, then NaOH(aq), then oxidation) needs three steps and gives a lower yield.
3. Control the oxidation. Distil to stop at the aldehyde; reflux to reach the acid.
4. Use the right intermediate. A halogenoalkane is the starting point for amines and nitriles; an alcohol is the starting point for esters and carbonyl compounds.

Worked example. Convert bromoethane into propanoic acid. The product has three carbons and the start has two, so a KCN step is needed. (1) CH₃CH₂Br + KCN in ethanol, heat under reflux gives CH₃CH₂CN; (2) dilute HCl, heat under reflux gives CH₃CH₂COOH. In contrast, 1-bromopropane to propanoic acid keeps three carbons, so use (1) NaOH(aq), heat, giving propan-1-ol, then (2) acidified K₂Cr₂O₇ under reflux.

Worked example (ester). Convert ethene into ethyl ethanoate. (1) Steam with H₃PO₄ gives ethanol. (2) Some of the ethanol is oxidised with acidified K₂Cr₂O₇ under reflux to ethanoic acid. (3) Ethanoic acid and the remaining ethanol are heated with concentrated H₂SO₄ as catalyst to give CH₃COOCH₂CH₃ and water.

5. Analysing a Given Route

When a route is given, there are four things to say about each step: the reaction type, the reagent and conditions, any by-product and the effect on the carbon chain.

CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

CH₃CH₂Cl + KCN → CH₃CH₂CN + KCl

CH₃CH₂CN + 2H₂O + HCl → CH₃CH₂COOH + NH₄Cl

By-products come from two sources. The first is the small molecule that is always formed in a substitution, condensation or hydrolysis (HCl and POCl₃, KCl, NH₄Cl, H₂O). Addition reactions have none, so all of the atoms end up in the product. The second is a competing reaction that gives a minor product:

• Elimination competes with substitution of a halogenoalkane: a little alkene forms with KCN or NaOH, especially from a tertiary halogenoalkane or at high temperature.
• Further substitution: without excess NH₃ a halogenoalkane gives secondary and tertiary amines as well as the primary amine. Free-radical chlorination of an alkane gives a mixture of mono-, di- and further substituted products, which is why an alkene plus HX is a better route to a halogenoalkane.
• Over-oxidation: if the aldehyde is not distilled off it is oxidised to the carboxylic acid.
• Two isomers: HBr adds to propene to give mainly 2-bromopropane and a little 1-bromopropane (Markovnikov addition).
• Reversible reactions: esterification does not go to completion, so some acid and alcohol remain.

Worked example (from a past-paper style route). Propanone → (HCN, KCN catalyst, heat) (CH₃)₂C(OH)CN → (dilute HCl, heat under reflux) (CH₃)₂C(OH)COOH → (heated Al₂O₃) CH₂=C(CH₃)COOH. Step 1 is nucleophilic addition (no by-product, +1 C). Step 2 is hydrolysis (by-product NH₄Cl, no change in carbons). Step 3 is elimination (dehydration) (by-product H₂O, no change in carbons).

6. Worked Routes

Routes B and C show the key decision. Both start from 1-bromobutane (four carbons), but B keeps the chain length (butanoic acid, four carbons) while C adds a carbon (pentanoic acid, five). Read the target carefully: ‘butanoic acid from 1-bromobutane’ must not use KCN.

Percentage yield over two steps (using the Data Booklet). If route A gives 80% in step 1 and 70% in step 2, the overall yield is 0.80 × 0.70 = 56%. The M_r of ethylamine, C₂H₇N, is 2 × 12.0 + 7 × 1.0 + 14.0 = 45.0, so 1.00 mol of ethene (28.0 g) could give at most 45.0 g and the 56% yield gives 25.2 g. This is why the number of steps matters.

7. Quick Sheet and Checklist

Before you leave the question, check:

• Every step has a reagent and conditions (reflux, distil, in ethanol, excess, dilute or concentrated).
• The type of reaction is named for each step, with nucleophilic or electrophilic where relevant.
• Each intermediate is named or drawn, and the carbon count agrees from start to finish.
• The reducing agent suits the group (LiAlH₄ for COOH, NaBH₄ for C=O).
• The route is as short as it can be and uses only AS reactions.
• By-products are stated (HCl, NH₄Cl, H₂O, a second isomer) and test observations are written as colours, precipitates or gases.