Whiteboard Chemistry with Joe White

Organic Synthesis

The whole course as one map: every functional-group interconversion in one place, how to chain them into a multi-step synthesis, and why chemists design routes that are shorter, greener and higher in atom economy.

AQA 7404/7405 Paper 2 A-level only
C=C C–Br C–OH –CHO –COOH
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:

Key aliphatic interconversions (reagent & conditions)
From → toReagents & conditionsType
alkane → halogenoalkaneCl2 or Br2, UV lightfree-radical substitution
alkene → halogenoalkaneHBr (or HCl), room tempelectrophilic addition
alkene → alcoholsteam, H3PO4 catalysthydration
halogenoalkane → alcoholNaOH(aq), refluxnucleophilic substitution
halogenoalkane → alkeneKOH in ethanol, refluxelimination
halogenoalkane → nitrile (+1 C)KCN in ethanol/water, refluxnucleophilic substitution
halogenoalkane → amineexcess NH3, ethanolnucleophilic substitution
alcohol → alkeneconc H2SO4 (or conc H3PO4), heatdehydration (elimination)
1° alcohol → aldehydeK2Cr2O7/H2SO4, warm & distilpartial oxidation
1° alcohol → carboxylic acidK2Cr2O7/H2SO4, refluxfull oxidation
2° alcohol → ketoneK2Cr2O7/H2SO4, refluxoxidation
aldehyde → carboxylic acidK2Cr2O7/H2SO4, refluxoxidation
aldehyde/ketone → alcoholNaBH4reduction
aldehyde/ketone → hydroxynitrile (+1 C)KCN, then dilute acid (HCN also accepted)nucleophilic addition
nitrile → amineH2/Ni (or LiAlH4)reduction
nitrile → carboxylic aciddilute H2SO4, refluxhydrolysis
carboxylic acid ↔ esteralcohol + conc H2SO4esterification
acyl chloride / anhydride → carboxylic acidwater (vigorous, room temp)nucleophilic addition–elimination
acyl chloride / anhydride → esteralcohol, room tempnucleophilic addition–elimination
acyl chloride / anhydride → primary amideNH3, room tempnucleophilic addition–elimination
acyl chloride / anhydride → N-substituted amideprimary amine, room tempnucleophilic 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

Cl₂ or Br₂UV lightfree-radicalsubstitutionHBrelectrophilic additionKOH in ethanol, refluxeliminationNaOH(aq)refluxnucleophilicsubstitutionsteamH₃PO₄ catalysthydrationconc H₂SO₄heateliminationKCN in ethanol/water,refluxnucleophilic substitution+1 CH₂ / Nireductionexcess NH₃in ethanolnucleophilic substitutionalkaneC–C and C–H onlyhalogenoalkaneC–BralkeneC=CnitrileC≡NalcoholC–OHamineC–NH₂

Both nitrogen routes start from the halogenoalkane — and the KCN route adds a carbon.

panel 2 — the oxidation ladder & the acid’s relatives

1°, reflux: straight to the acid1°: K₂Cr₂O₇ / H₂SO₄warm & distilNaBH₄reductionK₂Cr₂O₇ / H₂SO₄refluxoxidation2°: K₂Cr₂O₇ / H₂SO₄,refluxoxidationNaBH₄reduction✗ not oxidised furtherKCN, then dilute acidnucleophilic addition+1 C+1 C↓ alcohol +conc H₂SO₄esterification↑ dilute acid,refluxhydrolysisdilute H₂SO₄,refluxhydrolysisalcohol (1° / 2°)C–OHaldehyde–CHOcarboxylic acid–COOHketoneC–CO–Cester–COO–hydroxynitrile–CH(OH)CNnitrileC≡N — from panel 1

The nitrile made in panel 1 hydrolyses to the acid. Ketones are the dead end — no further oxidation.

panel 3 — the acyl chloride branch

all four arrows: nucleophilic addition–elimination (acylation)water(vigorous, room temp)an alcoholroom tempNH₃room tempa primary amineroom tempacyl chloride / acid anhydride–COCl / (CH₃CO)₂Ocarboxylic acid–COOHester–COO–primary amide–CONH₂N-substituted amide–CONHR

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

conc HNO₃ / conc H₂SO₄50 °Celectrophilic substitutionSn / conc HCl,then NaOHreductionRCOCl / AlCl₃(Friedel–Crafts acylation)electrophilic substitutionbenzeneC₆H₆nitrobenzeneC₆H₅NO₂phenylamineC₆H₅NH₂aromatic ketoneC₆H₅COR

Two steps take benzene to phenylamine; Friedel–Crafts acylation hangs a ketone off the ring.

Learn the map and most synthesis questions become a matter of tracing arrows.
🧪 Exam-style questions
Q1 [1 mark]

Which reagents convert a primary alcohol into a carboxylic acid?

  1. NaBH4
  2. K2Cr2O7/H2SO4, warmed and distilled off
  3. K2Cr2O7/H2SO4, heated under reflux
  4. conc H2SO4, heated
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.
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:

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.)

🧪 Exam-style 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

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 specification.

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.

🎯 Build the mark scheme — plan a route

Devise a synthesis of ethanol from ethene, and state one reason chemists prefer routes with high atom economy. Award yourself a mark for each:

  • Ethene + steam, H3PO4 catalyst → ethanol (direct hydration, one step)…
  • …or ethene + HBr → bromoethane, then NaOH(aq), reflux → ethanol (two steps).
  • Quote the conditions each route needs (catalyst; reflux).
  • High atom economy = less waste / cheaper / more sustainable (more of the atoms end up in the product).
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 product ÷ sum of Mr of all products) × 100.

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