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A2 36

Organic synthesis

A Level · Organic chemistry · Paper 4 · builds on AS 21

🎯What you need to be able to do

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

📚The chemistry

The skills are the same as at AS (topic 21, which has the full AS reaction map); the toolkit is bigger. Below are the A Level additions, grouped by what they start from. Use both maps together.

The A Level reaction map

Arenes

  • → halogenoarene: Cl2/Br2 with AlCl3/AlBr3 — electrophilic substitution.
  • → nitroarene: conc. HNO3 + conc. H2SO4, 25–60 °C.
  • → alkylarene: RCl + AlCl3, heat — Friedel–Crafts alkylation.
  • → aryl ketone: RCOCl + AlCl3, heat — Friedel–Crafts acylation.
  • alkylarene → benzoic acid: hot alkaline KMnO4, then dilute acid.
  • → cyclohexane ring: H2, Pt or Ni, heat.
  • methyl side chain → C6H5CH2Cl: Cl2, UV light.
A map with benzene at the centre. Benzene goes to chlorobenzene with chlorine and aluminium chloride; to cyclohexane with hydrogen over nickel or platinum; to an aryl ketone with an acyl chloride and aluminium chloride; to nitrobenzene with concentrated nitric and sulfuric acids at 25 to 60 degrees; and to methylbenzene by Friedel-Crafts alkylation. Nitrobenzene is reduced by tin and concentrated hydrochloric acid to phenylamine, which gives a diazonium salt with sodium nitrite and hydrochloric acid below 10 degrees; that gives phenol on warming with water, and phenol gives an azo dye with a diazonium salt in alkali. Methylbenzene gives benzoic acid with hot alkaline potassium manganate(VII) then acid, and (chloromethyl)benzene with chlorine in UV light.
The aromatic half of the A Level reaction map, with the conditions for each step.

Nitrogen compounds

  • nitrobenzene → phenylamine: Sn + conc. HCl, reflux, then NaOH(aq).
  • phenylamine → diazonium salt: NaNO2 + dilute HCl, below 10 °C; → phenol on warming; → azo dye with a phenol in NaOH(aq).
  • nitrile or amide → amine: LiAlH4 (nitriles also H2/Ni).
  • halogenoalkane + primary amine → secondary amine: ethanol, sealed tube.
  • amide → acid + NH4+ (aqueous acid, heat) or carboxylate + NH3 (aqueous alkali, heat).

Acids and acyl chlorides

  • carboxylic acid → acyl chloride: PCl5, PCl3/heat, or SOCl2.
  • acyl chloride + water → acid; + alcohol or phenol → ester; + NH3 → amide; + amine → substituted amide. All at room temperature, releasing HCl.
  • methanoic acid → CO2 (mild oxidising agents); ethanedioic acid → CO2 (warm acidified KMnO4).

Phenols

  • → phenoxide: NaOH(aq) or Na.
  • → 2,4,6-tribromophenol: Br2(aq), room temperature.
  • → 2- and 4-nitrophenol: dilute HNO3, room temperature.
  • → phenyl ester: acyl chloride.

Strategy

Everything from AS still applies — compare skeletons, compare functional groups, work backwards — with three A Level additions:

  1. Order matters on a ring. A substituent already present decides where the next one goes. Put a 3-directing group in first if you need a 3-substituted product; a 2,4-directing group for 2- or 4-.
  2. Protecting and competing groups. A reagent may attack two groups. An acyl chloride reacts with both OH and NH2; NaOH hydrolyses both esters and amides; LiAlH4 reduces acids, esters, amides and carbonyls alike. Check every group in the molecule against every reagent.
  3. Building nitrogen in. The ring gets its nitrogen by nitration then reduction; a chain gets it from NH3, from a nitrile (reduced) or from an amide (reduced).

✏️Worked example

Paracetamol is HOC6H4NHCOCH3, with the OH and the NHCOCH3 groups at positions 1 and 4 of a benzene ring. (a) Name the functional groups in paracetamol. (b) Predict what happens when paracetamol is (i) added to bromine water; (ii) heated with aqueous sodium hydroxide. (c) Suggest a three-step synthesis of paracetamol from phenol, and name a possible by-product of the last step.

(a) A phenol (OH on the ring) and a secondary amide (–NHCO–).

(b)(i) The ring is activated by the OH group, so bromine water is decolourised and a substituted product forms. The 4-position is already taken, so bromine substitutes at the two positions next to the OH (which the OH activates), giving a dibromo derivative. (ii) Two groups react with hot NaOH: the amide is hydrolysed, giving the ethanoate ion and 4-aminophenol, and the phenol is deprotonated to the phenoxide. Products: the sodium salt of 4-aminophenol (as phenoxide) and sodium ethanoate.

(c)

  • Step 1: dilute HNO3, room temperature (electrophilic substitution) → 2- and 4-nitrophenol; separate the 4-isomer. The OH is 2,4-directing and activating, which is why mild conditions suffice.
  • Step 2: Sn + concentrated HCl, reflux, then neutralise (reduction) → 4-aminophenol.
  • Step 3: ethanoyl chloride, room temperature (condensation, addition–elimination) → paracetamol + HCl.

By-products: in step 3, the acyl chloride can also react with the phenol OH, forming an ester (the O-ethanoyl compound), or with both groups. In step 1, 2-nitrophenol is formed alongside the wanted 4-isomer.

Four structures in a row: phenol; 4-nitrophenol, made with dilute nitric acid at room temperature by electrophilic substitution; 4-aminophenol, made by reduction with tin and concentrated hydrochloric acid then sodium hydroxide; and paracetamol, made with ethanoyl chloride at room temperature by addition-elimination. By-products are 2-nitrophenol in step 1 and an ester from the OH group in step 3.
Part (c): the OH directs the nitrogen to position 4, which is where paracetamol needs it.
Check it. Test each step against the target’s substitution pattern: the final NHCOCH3 is at position 4 relative to OH, so the nitrogen must enter at position 4 in step 1 — possible only because OH directs to 2 and 4. Had the route started by making phenylamine and then converting it to a phenol, there would be no second group to put the nitrogen back.
Assuming a reagent attacks only the group you are thinking about. Hot NaOH on paracetamol does two things; ethanoyl chloride on 4-aminophenol can attack the OH as well as the NH2. Synthesis and “predict the product” questions are designed around these second reactions — go through every group, every time.

📝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 phenol from benzene.
Step 1: conc. HNO3 + conc. H2SO4, 25–60 °C → nitrobenzene. Step 2: Sn + conc. HCl, reflux, then NaOH(aq) → phenylamine. Step 3: NaNO2 + dilute HCl below 10 °C → benzenediazonium chloride, then warm with water → phenol + N2.
2. Suggest a route from propanoic acid to propylamine, CH3CH2CH2NH2, without changing the carbon skeleton.
Step 1: SOCl2 (or PCl5) → propanoyl chloride, CH3CH2COCl. Step 2: NH3, room temperature → propanamide, CH3CH2CONH2. Step 3: LiAlH4 in dry ether → propylamine (the C=O is reduced to CH2). Three carbons throughout.
3. Compound W is H2NC6H4COOH (4-aminobenzoic acid). Predict its reaction with (a) HCl(aq); (b) NaOH(aq); (c) ethanoyl chloride; (d) methanol and conc. H2SO4.
(a) The amine is protonated: +H3NC6H4COOH Cl−. (b) The carboxylic acid is neutralised: H2NC6H4COO−Na+. (c) The amine is acylated: CH3CONHC6H4COOH + HCl. (d) The acid is esterified: H2NC6H4COOCH3 (a local anaesthetic, benzocaine, is the ethyl ester).
4. A student plans to make 4-nitrobenzoic acid by nitrating benzoic acid. Explain why this fails, and give a better route from methylbenzene.
COOH is 3-directing, so nitrating benzoic acid gives mainly 3-nitrobenzoic acid, not the 4-isomer. Better: nitrate methylbenzene first (CH3 is 2,4-directing; separate the 4-isomer, 4-nitromethylbenzene), then oxidise the side chain with hot alkaline KMnO4 followed by acid, giving 4-nitrobenzoic acid.
5. Give the reagents for converting benzene into phenylethanone and then into 1-phenylethanol, naming each reaction type.
Step 1: CH3COCl with AlCl3, heat — Friedel–Crafts acylation (electrophilic substitution) → C6H5COCH3. Step 2: NaBH4 (or LiAlH4) — reduction → C6H5CH(OH)CH3. The product has a chiral centre and forms as a racemic mixture.
6. In the route bromoethane → ethylamine (NH3, ethanol, pressure), name two by-products and suggest how to reduce them.
The ethylamine formed is itself a nucleophile and reacts with more bromoethane, giving diethylamine (secondary amine), then triethylamine and further substituted products; ammonium bromide and ethylammonium bromide also form. Use a large excess of ammonia, so that bromoethane is much more likely to meet NH3 than ethylamine. (Or avoid the problem entirely: make ethanenitrile from bromomethane and KCN, then reduce it with LiAlH4; nitrile reduction gives only the primary amine.)

🔗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 both reaction maps