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AS 21

Organic synthesis

AS Level · Organic chemistry · Papers 1 and 2 · extended in A2 36

🎯What you need to be able to do

  • For a molecule with several functional groups, identify the groups using the reactions in the syllabus, and predict its properties and reactions.
  • Devise multi-step synthetic routes using the reactions in the syllabus.
  • Analyse a given route: the type of reaction, the reagents and conditions for each step, and possible by-products.

📚The chemistry

This topic has no new reactions. It asks you to use the ones from topics 14–19 together, which is why it carries so many marks on Paper 2 and why it rewards having every reaction at your fingertips. The map below is the whole AS toolkit on one page. Learn it as reagent + conditions + type, not just starting material and product.

The AS reaction map

Alkanes

  • → halogenoalkane: Cl2 or Br2, UV light — free-radical substitution (gives a mixture).
  • → shorter alkane + alkene: heat, Al2O3 — cracking.

Alkenes

  • → alkane: H2, Pt or Ni, heat — addition (hydrogenation).
  • → alcohol: H2O(g), H3PO4 catalyst — electrophilic addition.
  • → halogenoalkane: HX(g), room temperature — electrophilic addition (Markovnikov).
  • → dihalogenoalkane: X2, room temperature — electrophilic addition.
  • → diol: cold dilute acidified KMnO4 — oxidation.
  • → ketones / carboxylic acids / CO2: hot concentrated acidified KMnO4 — oxidative cleavage.
  • → poly(alkene): addition polymerisation.

Halogenoalkanes

  • → alcohol: NaOH(aq), heat — nucleophilic substitution.
  • → nitrile (one more C): KCN in ethanol, heat — nucleophilic substitution.
  • → amine: excess NH3 in ethanol, heat under pressure — nucleophilic substitution.
  • → alkene: NaOH in ethanol, heat — elimination.

Alcohols

  • → halogenoalkane: HX, or KCl + conc. H2SO4/H3PO4, or PCl3 + heat, or PCl5, or SOCl2 — substitution.
  • primary → aldehyde: acidified K2Cr2O7, distil — oxidation.
  • primary → carboxylic acid: acidified K2Cr2O7, reflux — oxidation.
  • secondary → ketone: acidified K2Cr2O7 — oxidation.
  • → alkene: heated Al2O3 or conc. acid — dehydration (elimination).
  • → ester: carboxylic acid, conc. H2SO4, heat — condensation.
  • → alkoxide + H2: Na(s).

Aldehydes and ketones

  • → alcohol: NaBH4 or LiAlH4 — reduction.
  • → hydroxynitrile (one more C): HCN, KCN catalyst, heat — nucleophilic addition.
  • aldehyde → carboxylic acid: acidified K2Cr2O7, reflux — oxidation.

Carboxylic acids, esters and nitriles

  • acid → ester: alcohol, conc. H2SO4, heat.
  • acid → primary alcohol: LiAlH4 — reduction.
  • acid → salt: reactive metal (+ H2), alkali (+ H2O), or carbonate (+ H2O + CO2).
  • ester → acid + alcohol: dilute acid, heat (reversible); or dilute NaOH, heat (salt + alcohol), then acidify.
  • nitrile → carboxylic acid: dilute acid, heat; or dilute NaOH, heat, then acidify — hydrolysis.

Molecules with several functional groups

Exam molecules often contain two or three functional groups. Treat each group independently: go through the molecule group by group and ask what each reagent does to each one. Most reagents are selective, and knowing which ones are is the skill being tested:

  • NaBH4 reduces C=O in aldehydes and ketones, but not C=C, COOH or esters. LiAlH4 also reduces COOH.
  • Bromine water reacts with C=C only.
  • Tollens’ and Fehling’s react with aldehydes only.
  • Sodium carbonate reacts with COOH only; sodium metal with any O–H (alcohol or acid).
  • Acidified dichromate oxidises primary and secondary alcohols and aldehydes, leaving C=C, ketones and tertiary alcohols alone.
  • Hot concentrated KMnO4 is not selective: it also oxidises alcohols and aldehydes, and cleaves C=C.

Devising a route

Work backwards from the target as well as forwards from the start:

  1. Compare the carbon skeletons. One carbon more? The route needs a nitrile (from KCN or HCN). Same number? You only need to change functional groups.
  2. Compare the functional groups and their positions. If the group has to move to a different carbon, go through an alkene: eliminate, then add back with Markovnikov orientation.
  3. Find a link. Halogenoalkanes and alcohols are the great intermediates: from either you can reach almost every other AS functional group.
  4. For each step, give the reagent, the conditions (reflux, distil, UV, pressure, solvent) and the structure of the intermediate.

Analysing a route

When you are given a route, name each step by type — addition, substitution, elimination, oxidation, reduction, hydrolysis, condensation — and by mechanism where there is one. Then look for by-products: the other isomer from Markovnikov addition, the alkene that competes with substitution, the carboxylic acid that forms if an aldehyde is not distilled off, the further-substituted products of free-radical substitution or of amine formation. Each one lowers the yield of the product you want.

✏️Worked example

Compound W is HOCH2CH=CHCHO. (a) Name the functional groups in W. (b) Predict what you would observe with (i) bromine water; (ii) Tollens’ reagent; (iii) sodium metal. (c) Give the structure of the product when W reacts with NaBH4. (d) Give the structure of the product when W is heated under reflux with excess acidified potassium dichromate(VI). (e) Propose a two-step route from propan-1-ol to propan-2-ol, and name a by-product.

(a) A primary alcohol (CH2OH), an alkene (C=C) and an aldehyde (CHO).

(b)(i) The C=C decolourises bromine water, orange → colourless. (ii) The aldehyde gives a silver mirror. (iii) The O–H of the alcohol reacts: effervescence of hydrogen.

(c) NaBH4 reduces only the aldehyde, to a primary alcohol; the C=C is untouched: HOCH2CH=CHCH2OH (but-2-ene-1,4-diol).

(d) Under reflux with excess oxidising agent, both the primary alcohol and the aldehyde are oxidised all the way to carboxylic acid groups; dichromate leaves the C=C: HOOCCH=CHCOOH (butenedioic acid).

(e) The OH must move from C1 to C2 on the same carbon skeleton, so go through an alkene. Step 1: dehydrate propan-1-ol with heated Al2O3 (or concentrated H3PO4) to propene (elimination). Step 2: steam with an H3PO4 catalyst (electrophilic addition). Markovnikov addition puts the OH on the middle carbon, giving propan-2-ol as the major product, via the more stable secondary carbocation. By-product: propan-1-ol, from the minor route through the primary carbocation.

Check it. Count carbons and check each step changes only what it should. In (c) and (d) the chain is still four carbons with the C=C in place; in (e) three carbons throughout. Then check selectivity: if you ever find NaBH4 reducing a C=C, or dichromate cleaving one, go back to the reagent list above.
Reagents that do too much. Using hot concentrated KMnO4 in (d) would also cleave the C=C, and using H2/Ni in (c) would reduce the C=C as well as the aldehyde. In synthesis the reagent must do the job and nothing else — that is what the selectivity list is for.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Devise a three-step synthesis of propanoic acid from ethene.
Ethene has 2 carbons, the target 3, so a nitrile is needed. Step 1: HBr(g), room temperature (electrophilic addition) → bromoethane. Step 2: KCN in ethanol, heat under reflux (nucleophilic substitution) → propanenitrile, CH3CH2CN. Step 3: dilute HCl, heat under reflux (hydrolysis) → propanoic acid.
2. Suggest a route from 1-bromopropane to propanone, naming the intermediates, and explain why it cannot be done in two steps.
Propanone needs the functional group on C2, but the bromine is on C1, so the group must move. Step 1: NaOH in ethanol, heat (elimination) → propene. Step 2: steam with an H3PO4 catalyst (Markovnikov addition) → propan-2-ol. Step 3: acidified K2Cr2O7, heat → propanone. Two steps are not enough: substituting the Br (with NaOH(aq)) keeps the OH on C1, and oxidising propan-1-ol gives propanal or propanoic acid, never a ketone. Moving a group along the chain always goes through the alkene.
3. Compound V is CH3COCH2CH2COOH. Predict the result with (a) sodium carbonate; (b) 2,4-DNPH; (c) Fehling’s solution; (d) alkaline aqueous iodine.
V contains a ketone and a carboxylic acid. (a) Effervescence (CO2) from the COOH. (b) Orange precipitate from the ketone C=O (the COOH does not react). (c) No change — stays blue; there is no aldehyde. (d) Yellow precipitate of CHI3, because of the CH3CO– group.
4. In the route: butan-2-ol → (step 1) → 2-bromobutane → (step 2) → 2-methylbutanenitrile, give the reagent and type of each step, and name a by-product of step 2.
Step 1: HBr(g), or KBr with concentrated H3PO4 (the bromide version of the KCl method) — substitution of OH by Br. Step 2: KCN in ethanol, heat — nucleophilic substitution; the product CH3CH(CN)CH2CH3 has five carbons, named 2-methylbutanenitrile. By-product of step 2: CN− can also act as a base, so some elimination occurs, giving but-1-ene and but-2-ene.
5. How would you convert ethanol into ethyl ethanoate using ethanol as the only organic starting material?
Split the ethanol into two portions. Oxidise one portion to ethanoic acid: acidified K2Cr2O7 in excess, heat under reflux. Then esterify: heat the ethanoic acid with the second portion of ethanol and a few drops of concentrated H2SO4. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O.
6. Explain why free-radical substitution is a poor first step in a synthesis of pure 1-chloropropane from propane.
Free-radical substitution is unselective. Chlorine radicals abstract hydrogen from either carbon, so both 1-chloropropane and 2-chloropropane form; further substitution gives dichloro- and polychloropropanes; and termination steps give hexane and other alkanes. The yield of 1-chloropropane is low and it must be separated from a mixture. A better route starts from propan-1-ol, using SOCl2 or PCl5, which gives a single product.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — organic reaction summaries by functional group, useful to test yourself against this page’s map