What organic reactions does VCE Chemistry cover?

VCE Chemistry Unit 3-4 organic covers: substitution reactions of alkanes with halogens; addition reactions of alkenes (with H2, halogens, hydrogen halides, water using Markovnikov's rule); oxidation of primary and secondary alcohols (to aldehydes/carboxylic acids and ketones respectively); esterification of carboxylic acids with alcohols (Fischer esterification); hydrolysis of esters (acid and base); addition polymerisation of alkenes; condensation polymerisation.

How do I approach pathway-design questions?

Use retrosynthesis. Work backwards from the target product: identify what functional group needs to be present; what is the precursor that could have produced it. Continue working backwards until you reach a feasible starting material. Then describe the synthesis forwards with conditions and named intermediates.

What conditions does VCAA expect?

Specific named conditions: UV light for alkane halogenation; warm aqueous NaOH for haloalkane to alcohol; acidified dichromate (Cr2O7^2-/H+) under distillation for primary alcohol to aldehyde, under reflux for primary alcohol to carboxylic acid; acidified dichromate or permanganate under reflux for secondary alcohol to ketone (tertiary alcohols not oxidised); concentrated H2SO4 catalyst under reflux for Fischer esterification; dilute H+ catalyst under reflux for ester hydrolysis.

What is Markovnikov's rule?

When HX or H2O adds to an unsymmetrical alkene, the H attaches to the carbon already carrying more hydrogens; X (or OH) attaches to the more substituted carbon. So propene + HBr gives 2-bromopropane (major) not 1-bromopropane. The rule reflects the relative stability of the carbocation intermediate.

How do I handle the Markovnikov vs anti-Markovnikov distinction?

Standard VCE Chemistry uses Markovnikov for HX and H2O addition. Anti-Markovnikov (peroxide-catalysed HBr) is outside the VCE syllabus. To get anti-Markovnikov products, use a different pathway: halogenate the alkane (free-radical substitution), then convert via further steps.

What are common pathway-design traps?

Tertiary alcohols do not oxidise (dead end for ketone synthesis from a tertiary precursor). Heat alone (without acid) is insufficient for esterification. Direct addition of HBr to propane is wrong; propane is an alkane, not an alkene. Use the correct reagents and conditions for each step. Always specify by-products (HX, water).

VICChemistry

VCE Chemistry organic synthesis pathways: the 2026 guide

A complete guide to VCE Chemistry organic synthesis pathways. The reaction toolkit, pathway diagrams, retrosynthesis, and worked syntheses for the Unit 3-4 organic content.

Generated by Claude Opus 4.816 min readVCAA-CHEM-ORGANICUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this guide is for
The reaction toolkit
Retrosynthesis
Worked example 1: Ethyl ethanoate from ethene
Worked example 2: Propan-2-ol from propane
Worked example 3: Pentan-2-one from pentan-2-ol
Pathway diagrams
Check your knowledge

What this guide is for

VCE Chemistry Unit 3-4 includes organic synthesis pathways: combining multiple reactions to convert one organic compound into another. The Section B exam often asks for a 5-step or 8-step synthesis with reagents and conditions. This guide covers the reaction toolkit, retrosynthesis, and three worked syntheses.

The reaction toolkit

Memorise each as both forward and reverse direction.

From	To	Reagents	Conditions
Alkane	Haloalkane	X2 (Br2, Cl2)	UV light
Alkene	Vicinal dihaloalkane	X2	room T
Alkene	Haloalkane	HX	room T; Markovnikov
Alkene	Alcohol	H2O	dilute H2SO4 catalyst, heat; Markovnikov
Alkene	Alkane	H2	Ni or Pt catalyst
Haloalkane	Alcohol	NaOH (aq)	warm aqueous, nucleophilic substitution
Primary alcohol	Aldehyde	acidified K2Cr2O7	distillation
Primary alcohol	Carboxylic acid	acidified K2Cr2O7, excess	reflux
Secondary alcohol	Ketone	acidified K2Cr2O7	reflux
Carboxylic acid + alcohol	Ester	concentrated H2SO4 catalyst	reflux
Ester + water	Carboxylic acid + alcohol	dilute H+ catalyst	reflux (acid hydrolysis)
Ester + NaOH	Carboxylate salt + alcohol	reflux	base hydrolysis (saponification)

Although the VCAA Study Design treats haloalkane to alcohol as a single "warm aqueous NaOH" step, two mechanisms underpin it. The contrast between SN2 (concerted, primary substrates, second-order) and SN1 (carbocation intermediate, tertiary substrates, first-order) clarifies why primary haloalkanes hydrolyse cleanly while tertiary ones often eliminate.

SN2 versus SN1: the concerted backside attack of SN2 contrasts with the stepwise ionisation of SN1 through a carbocation intermediate, which is why tertiary haloalkanes favour the SN1 path.

Retrosynthesis

Working backwards from the target:

What is the target compound? Identify its functional groups.
What is the immediate precursor? What reaction could have produced this functional group?
What is the precursor's precursor? Continue.
Reach a feasible starting material.
Write the synthesis forwards with reagents and conditions.

Example. Synthesise propyl ethanoate from propene.

Target: propyl ethanoate (CH3-CO-O-CH2-CH2-CH3). Ester.

Precursors: propan-1-ol + ethanoic acid (Fischer esterification).

Propan-1-ol from propene: addition of water (Markovnikov gives propan-2-ol, the wrong isomer). Need an alternative: propene + HBr (Markovnikov gives 2-bromopropane). Then nucleophilic substitution with NaOH to propan-2-ol. But we want propan-1-ol.

Anti-Markovnikov pathways are outside VCE scope. The cleanest route from propene to propan-1-ol involves intermediate steps that produce 1-bromopropane (radical addition with peroxides), outside standard VCE.

Practical route: assume propan-1-ol is available as starting material (or use a different starting point).

Ethanoic acid from ethanol (oxidation under reflux with acidified dichromate). Ethanol from ethene (addition of water, Markovnikov gives ethanol directly).

Pathway:

Ethene + H2O (dilute H2SO4, heat) → ethanol.
Half the ethanol: ethanol + acidified Cr2O7^2- (reflux) → ethanoic acid.
The remaining ethanol stays as ethanol... wait, we want propan-1-ol, not ethanol, for the propyl ester.

Re-read the target. Propyl ethanoate is the ester of propan-1-ol + ethanoic acid. So we need both propan-1-ol and ethanoic acid.

A cleaner alternative starting material: start with propan-1-ol and ethanol both available; oxidise ethanol to ethanoic acid; combine.

Worked example 1: Ethyl ethanoate from ethene

Target: ethyl ethanoate (CH3-CO-O-CH2-CH3).

Pathway:

Ethene (CH2=CH2) + H2O (dilute H2SO4, heat) → ethanol (CH3-CH2-OH).
Some ethanol kept; the rest oxidised: ethanol + acidified K2Cr2O7 (reflux) → ethanoic acid.
Ethanoic acid + ethanol (concentrated H2SO4 catalyst, reflux) → ethyl ethanoate + water (equilibrium).

This is the canonical "synthesis ethyl ethanoate from ethene" question.

Esterification reaction profile:

E_a

measures the climb to the tetrahedral transition state, and the small positive

\Delta H

explains why an excess of acid or removal of water is needed to push the equilibrium toward the ester.

Worked example 2: Propan-2-ol from propane

Target: propan-2-ol (CH3-CHOH-CH3).

Pathway:

Propane + Br2 (UV) → 2-bromopropane (Markovnikov-equivalent for substitution: secondary radical is more stable) + HBr.
2-bromopropane + NaOH (aq) → propan-2-ol + NaBr.

Worked example 3: Pentan-2-one from pentan-2-ol

Target: pentan-2-one (CH3-CO-CH2-CH2-CH3). Ketone.

Pentan-2-ol is a secondary alcohol. Oxidation:

Pentan-2-ol + acidified K2Cr2O7 (reflux) → pentan-2-one + H2O.

(The dichromate orange-to-green colour change is the visible indicator.)

Pathway diagrams

VCAA expects clear pathway diagrams. Standard format:

ethene
  | + H2O (dilute H2SO4, heat)
  v
ethanol
  | + acidified Cr2O7^2- (reflux)
  v
ethanoic acid + ethanol (kept separately)
  | + conc. H2SO4 catalyst, reflux
  v
ethyl ethanoate + water

Each arrow labelled with reagent above and conditions below.

Alkene to haloalkane to alcohol to carboxylic acid to ester: the canonical VCAA pathway with reagents over the arrows and conditions below.

Check your knowledge

A mix of nomenclature, retrosynthesis, mechanism and analytical-spectroscopy questions in the VCAA Unit 4 style. Aim to attempt under exam conditions before checking the solutions block.

Define the term Markovnikov's rule and state in one sentence why a tertiary carbocation is more stable than a primary carbocation. (2 marks)
Give the IUPAC name of $CH_3CH(OH)CH_2C(CH_3)_3$ and classify the alcohol as primary, secondary, or tertiary. (3 marks)
(a, 3) Design a two-step synthesis from ethene to ethanoic acid. State reagents and conditions for each step. (b, 2) Write a balanced equation for the formation of ethyl ethanoate from ethanoic acid, naming the catalyst and one technique used to drive the equilibrium toward product. (5 marks)
An unknown organic compound has the molecular formula $C_3H_6O_2$ . Its IR spectrum shows a strong absorption at $1740 \ \text{cm}^{-1}$ and no broad O-H absorption above $3000 \ \text{cm}^{-1}$ . Its $^{13}C$ NMR shows three signals. The mass spectrum shows $m/z = 74$ (parent ion) and a base peak at $m/z = 43$ . (a) Identify the compound. (b) Account for the $m/z = 43$ fragment. (5 marks)
(a, 2) Define structural isomer and stereoisomer. (b, 4) Draw and name all the structural isomers of $C_4H_9Br$ . (6 marks)
Plan a synthesis of propan-2-ol from propane using only the VCE reaction toolkit. Show each step with reagents and conditions, and identify any major side product. (5 marks)
The pesticide DDT is synthesised in industry from chlorobenzene and trichloroethanal. A modern analytical chemistry team in Melbourne wants to confirm an unknown peak in a wastewater HPLC trace from a former agricultural site near the Murray. Outline how mass spectrometry combined with NMR could be used to verify whether the peak is DDT. Give one safety consideration and one limit-of-detection consideration. (6 marks)
(a, 2) Distinguish between addition polymerisation and condensation polymerisation. (b, 2) Draw the repeating unit of poly(propene) and the repeating unit of the polyester formed from ethane-1,2-diol and benzene-1,4-dicarboxylic acid (PET). (c, 3) Compare the suitability of PET versus poly(propene) for a single-use water bottle, referring to chemical properties only. (7 marks)

Solutions

Q1: Markovnikov's rule states that when a hydrogen halide $HX$ adds across an unsymmetrical alkene, the hydrogen attaches to the carbon already bearing the most hydrogens, and $X$ attaches to the more substituted carbon. The rule reflects the fact that the more stable (more substituted) carbocation intermediate forms preferentially; a tertiary carbocation is stabilised by the inductive donation of three neighbouring alkyl groups (and by hyperconjugation), spreading the positive charge across the molecule.
Q2: Longest chain containing the OH is four carbons. Number so the OH gets the lowest locant: OH on C2, three methyl groups on C4 (a tert-butyl substituent). However, the longest chain that includes OH is 4 carbons (butan-2-ol skeleton) with a substituent. The compound is 4,4-dimethylpentan-2-ol. The OH is bonded to a carbon that itself carries one H and two C substituents, so it is a secondary alcohol.
Q3: (a) Step 1: hydration of ethene to ethanol. $CH_2{=}CH_2 + H_2O \xrightarrow{H_2SO_4, \text{ heat}} CH_3CH_2OH$ . Step 2: oxidation of ethanol to ethanoic acid using acidified potassium dichromate under reflux. $CH_3CH_2OH + 2[O] \xrightarrow{K_2Cr_2O_7/H_2SO_4, \text{ reflux}} CH_3COOH + H_2O$ . (b) $CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O$ with concentrated $H_2SO_4$ as catalyst. Removing water (Dean-Stark trap) or using a large excess of one reagent drives Le Chatelier's equilibrium toward ester formation.
Q4: (a) $C_3H_6O_2$ has degree of unsaturation 1 (consistent with a C=O). The IR at $1740 \ \text{cm}^{-1}$ is the carbonyl stretch in the ester region, and the absence of broad O-H rules out carboxylic acid. Three $^{13}C$ signals match methyl ethanoate $CH_3COOCH_3$ (three distinct carbons: methyl, carbonyl, methoxy). $M = 74$ matches. The compound is methyl ethanoate. (b) Loss of methoxide ( $OCH_3$ , mass 31) from the parent ion ( $m/z = 74$ ) gives the acylium cation $[CH_3CO]^+$ at $m/z = 43$ , the characteristic base peak for methyl esters.
Q5: (a) Structural isomers share a molecular formula but differ in the connectivity of atoms. Stereoisomers share a molecular formula and connectivity but differ in the three-dimensional arrangement of atoms (e.g., cis/trans or enantiomers). (b) Four structural isomers of $C_4H_9Br$ : 1-bromobutane ( $CH_3CH_2CH_2CH_2Br$ , primary), 2-bromobutane ( $CH_3CH_2CHBrCH_3$ , secondary), 1-bromo-2-methylpropane ( $(CH_3)_2CHCH_2Br$ , primary), and 2-bromo-2-methylpropane ( $(CH_3)_3CBr$ , tertiary).
Q6: Step 1 (free-radical halogenation): $CH_3CH_2CH_3 + Br_2 \xrightarrow{UV} CH_3CHBrCH_3 + HBr$ . The secondary radical predominates because secondary C-H bonds are weaker, so 2-bromopropane is the major product (1-bromopropane is the minor side product). Step 2 (nucleophilic substitution / hydrolysis): $CH_3CHBrCH_3 + NaOH(aq) \xrightarrow{\text{warm}} CH_3CH(OH)CH_3 + NaBr$ . The overall sequence converts propane to propan-2-ol via the 2-bromopropane intermediate, with 1-bromopropane as the main side product after step 1.
Q7: DDT has formula $C_{14}H_9Cl_5$ , $M = 354.5$ . (i) GC-MS: inject the sample onto a non-polar capillary column; DDT elutes at a characteristic retention time. The mass spectrum shows the parent-ion cluster at $m/z \approx 354$ with the distinctive $^{35}Cl/^{37}Cl$ isotope pattern (five chlorines give a multi-peaked envelope spanning 354 to 364 with intensity ratios calculable from $(3:1)^5$ ). (ii) $^1H$ NMR: confirms two aromatic doublets (para-substituted ring protons) at $\delta \approx 7.3$ and a single methine signal near $\delta \approx 5.0$ . Safety: DDT is persistent and bioaccumulative; handle in a fume hood with PPE. Limit of detection: solid-phase extraction or large-volume injection is required because environmental concentrations are typically at the parts-per-trillion level, below direct injection sensitivity.
Q8: (a) Addition polymerisation joins monomers across C=C bonds with no by-product, conserving all atoms. Condensation polymerisation joins monomers with the loss of a small molecule (usually water) at each linkage. (b) Poly(propene) repeating unit: $-(CH_2CH(CH_3))-$ . PET repeating unit: $-(O{-}CH_2CH_2{-}O{-}CO{-}C_6H_4{-}CO)-$ . (c) Both are stable to dilute acid and base at ambient temperatures. PET has polar ester linkages giving stronger intermolecular dipole-dipole and dispersion forces, hence higher tensile strength and a higher glass-transition temperature, so it holds shape under carbonation pressure. Poly(propene) is non-polar (dispersion forces only), more flexible, and gas-permeable. For a carbonated water bottle PET is preferred because $CO_2$ permeation through poly(propene) would flatten the drink within days.