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WACE Chemistry Unit 4 deep dive: Organic chemistry and chemical synthesis for the 2026 exam

WACEChemistryStudy guide18 min read

Revision deep dive for WACE Chemistry Unit 4: functional groups and IUPAC naming, isomerism, intermolecular forces, addition, substitution, oxidation and esterification reactions, reaction pathways, polymers, percentage yield, atom economy and green chemistry, and mass, infrared and NMR spectra, with worked examples and links to every Unit 4 dot point.

Jump to a section
  1. How Unit 4 fits the exam
  2. 1. Functional groups, homologous series and naming
  3. 2. Isomerism
  4. 3. Physical properties and intermolecular forces
  5. 4. The reaction toolkit
  6. 5. Reaction pathways
  7. 6. Polymers
  8. 7. Synthesis: yield, atom economy and green chemistry
  9. 8. Identifying compounds: spectra and crystallography
  10. Common mistakes
  11. Check your knowledge

How Unit 4 fits the exam

Unit 4 (organic chemistry and chemical synthesis) is the other half of the WACE Chemistry exam. It is heavy on structures, names and reactions, and its extended answer questions often combine a synthesis pathway, a yield calculation and a spectrum. This deep dive links every Unit 4 dot point on the site. For exam format and conventions, see the exam strategy guide; for Unit 3, see the equilibrium, acids and redox deep dive.

1. Functional groups, homologous series and naming

Dot points: functional groups and homologous series, organic structure and nomenclature, hydrocarbons: alkanes, alkenes and alkynes.

Family Functional group Suffix or prefix Example
Alkene C=C -ene but-2-ene
Haloalkane C-X chloro-, bromo- 2-chloropropane
Alcohol -OH -ol propan-1-ol
Aldehyde -CHO (end of chain) -al ethanal
Ketone C=O (inside chain) -one propanone
Carboxylic acid -COOH -oic acid ethanoic acid
Ester -COO- alkyl -oate methyl propanoate
Amine -NH2 -amine ethanamine
Amide -CONH2 -amide ethanamide
Naming a branched alcohol

Name HOCH2CH(CH3)CH2CH3\text{HOCH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3.

  1. Longest chain containing the carbon bonded to OH: 4 carbons, so butan.
  2. Number from the end nearest the OH: OH on carbon 1, methyl on carbon 2.
  3. Name: 2-methylbutan-1-ol.

2. Isomerism

Dot point: isomerism.

  • Structural isomers have the same molecular formula but different connectivity: butan-1-ol and butan-2-ol (position), or butan-1-ol and ethoxyethane (different functional groups), all C4H10O\text{C}_4\text{H}_{10}\text{O}.
  • Cis-trans (geometric) isomers need a C=C double bond with two different groups on each carbon. But-2-ene has cis and trans forms; but-1-ene does not, because carbon 1 carries two hydrogens.

3. Physical properties and intermolecular forces

Dot point: physical properties and intermolecular forces.

For molecules of similar size, boiling point rises with stronger intermolecular forces: dispersion only (alkanes) < dipole-dipole (aldehydes, ketones) < hydrogen bonding (alcohols, carboxylic acids). Butane (about −1 °C), propanal (about 48 °C) and propan-1-ol (about 97 °C) have similar molar masses but very different boiling points for exactly this reason. Within a homologous series, boiling point rises with chain length (stronger dispersion forces). Small alcohols and acids are water-soluble because they hydrogen bond with water; solubility falls as the non-polar chain grows.

4. The reaction toolkit

Dot points: addition reactions of alkenes, substitution reactions of haloalkanes, alcohols, oxidation of alcohols, carboxylic acids and esters, amines and amides.

Reaction Reagents and conditions Example
Addition of hydrogen H2\text{H}_2, nickel or platinum catalyst, heat ethene to ethane
Addition of halogen Br2\text{Br}_2 or Cl2\text{Cl}_2 ethene to 1,2-dibromoethane (bromine water decolourises)
Addition of hydrogen halide HBr or HCl but-2-ene to 2-bromobutane
Hydration steam, phosphoric acid catalyst, heat, pressure ethene to ethanol
Substitution of alkane halogen, UV light methane to chloromethane + HCl
Haloalkane to alcohol aqueous NaOH, heat chloroethane to ethanol
Oxidation of alcohol acidified Cr2O72−\text{Cr}_2\text{O}_7^{2-} or MnO4−\text{MnO}_4^-, heat propan-1-ol to propanoic acid
Esterification alcohol + carboxylic acid, conc. H2SO4\text{H}_2\text{SO}_4, reflux ethanol + ethanoic acid to ethyl ethanoate + water
Amide formation carboxylic acid (or derivative) + amine forms the amide (peptide) link

Alcohol class decides oxidation: primary gives aldehyde then carboxylic acid; secondary gives ketone; tertiary does not react. With acidified dichromate the observation is orange to green; with acidified permanganate, purple to colourless.

5. Reaction pathways

Dot point: organic reaction pathways.

Designing a two-branch synthesis

Show how ethyl ethanoate can be made using ethene as the only organic starting material.

  1. Hydration: CH2=CH2+H2O→CH3CH2OH\text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH} (steam, phosphoric acid catalyst).
  2. Oxidation of part of the ethanol: CH3CH2OH→CH3COOH\text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOH} (acidified dichromate, heat under reflux).
  3. Esterification: CH3COOH+CH3CH2OH⇌CH3COOCH2CH3+H2O\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O} (conc. sulfuric acid catalyst, reflux).

Full marks need the reagent and condition for each step, and a structure or name for each intermediate.

6. Polymers

Dot point: polymers.

  • Addition polymers: monomers with C=C join end to end; the double bond opens. Propene gives polypropene with repeating unit −CH2−CH(CH3)−-\text{CH}_2{-}\text{CH}(\text{CH}_3)-.
  • Condensation polymers: each monomer has two reactive groups; water is released at each link. A diacid with a diol gives a polyester; a diacid with a diamine gives a polyamide; amino acids give proteins joined by peptide (amide) links.

7. Synthesis: yield, atom economy and green chemistry

Dot points: percentage yield and atom economy, green chemistry principles, chemical synthesis and analysis.

Efficiency of a synthesis

percentage yield=actual masstheoretical mass×100atom economy=M(desired product)∑M(all reactants)×100\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100 \qquad \text{atom economy} = \frac{M(\text{desired product})}{\sum M(\text{all reactants})} \times 100

Yield and atom economy

Yield. 12.0 g of ethanol (M=46.07M = 46.07) reacts with excess ethanoic acid and 15.0 g of ethyl ethanoate (M=88.10M = 88.10) is collected.

n(ethanol)=12.046.07=0.260n(\text{ethanol}) = \dfrac{12.0}{46.07} = 0.260 mol, so the theoretical mass of ester is 0.260×88.10=22.90.260 \times 88.10 = 22.9 g, and the yield is 15.022.9×100=65.4\dfrac{15.0}{22.9} \times 100 = 65.4 percent.

Atom economy. Making ethanol by fermentation, C6H12O6→2C2H5OH+2CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2, has atom economy 2×46.07180.16×100=51.1\dfrac{2 \times 46.07}{180.16} \times 100 = 51.1 percent. Hydration of ethene has 100 percent atom economy because ethanol is the only product. Green chemistry weighs this against the fact that fermentation uses a renewable feedstock at low temperature.

8. Identifying compounds: spectra and crystallography

Dot points: mass spectrometry, infrared spectroscopy, NMR spectroscopy, x-ray crystallography.

  • Mass spectrometry: the molecular ion peak gives the molar mass (ethanol, 46); fragments give clues (a peak at 15 suggests CH3+\text{CH}_3^+).
  • Infrared: a broad band around 3200 to 3550 cm−1^{-1} means an alcohol O-H; a very broad band around 2500 to 3300 cm−1^{-1} with a strong peak near 1700 cm−1^{-1} means a carboxylic acid; a strong peak near 1700 cm−1^{-1} without the broad O-H means a carbonyl (aldehyde, ketone or ester). Use the data booklet values in the exam.
  • NMR: the number of signals equals the number of distinct chemical environments. Propan-1-ol has 3 carbon environments; propan-2-ol, being symmetrical, has 2.
  • X-ray crystallography: the diffraction pattern of a crystal is used to determine the 3D arrangement of atoms, including bond lengths and angles.

Common mistakes

Where Unit 4 marks go missing
  • Numbering the chain from the wrong end, or not choosing the longest chain that contains the functional group.
  • Claiming tertiary alcohols oxidise, or giving the wrong colour change for the oxidant used.
  • Forgetting water as a product of esterification and condensation polymerisation.
  • Confusing percentage yield with atom economy.
  • Drawing a repeating unit of an addition polymer with the double bond still in it.

Check your knowledge

  1. Name the product of oxidising butan-2-ol with acidified dichromate. (Answer: butanone.)
  2. Which of propanone, propan-1-ol and butane has the highest boiling point? (Answer: propan-1-ol, because of hydrogen bonding.)
  3. Name the ester formed from methanol and propanoic acid. (Answer: methyl propanoate.)

Then try the Unit 4 practice quiz.

Sources & how we know this

  • chemistry
  • wace
  • wace-chemistry
  • unit-4
  • organic-chemistry
  • synthesis
  • spectroscopy
  • polymers
  • year-12
  • 2026

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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