June 2026
The 2025 VCE Chemistry exam showed that organic chemistry cannot be answered by recognising names alone.
Students needed to reason from structure.
Across the paper, they were asked about ester formation, protein hydrolysis, separation techniques, functional group tests, boiling point differences, IR spectra, mass spectra, melting point range, triglyceride unsaturation and ¹³C NMR. These questions rewarded students who could connect molecular structure to chemical behaviour.
Which functional group is present?
Which bond is broken?
Which product forms?
Which test would distinguish these two compounds?
Which separation technique suits the physical properties?
How many carbon environments does the molecule have?
What does a molecular ion peak show?
What does a broad, lower melting point range indicate?
These questions were not asking students to recite organic chemistry.
They asked students to use it.
Ester formation required correct assembly
Question 22 asked for the major organic product when propan-1-ol reacts with butanoic acid in the presence of sulfuric acid.
This was an esterification reaction.
The alcohol provides the alkyl group attached to the oxygen. The carboxylic acid provides the carbonyl-containing section of the ester.
Propan-1-ol is:
CH₃CH₂CH₂OH
Butanoic acid is:
CH₃CH₂CH₂COOH
The ester formed is:
CH₃CH₂CH₂COOCH₂CH₂CH₃
Water is also produced.
This question rewarded students who could build the ester from the two reactants. A common organic chemistry error is reversing the acid and alcohol fragments or choosing a structure with the wrong number of carbons.
Esterification is not just “acid plus alcohol gives ester”.
The product structure must preserve the correct parts of each reactant.
Hydrolysis required recognising the bond being broken
Question 23 asked about protein hydrolysis.
The correct answer was that the amide links in the protein break, and water is a reactant.
This tested the difference between condensation and hydrolysis.
In condensation reactions, smaller molecules join together and water is produced. In hydrolysis, water is used to break a larger molecule into smaller molecules.
Proteins contain amide links between amino acid residues. During hydrolysis, those amide links are broken, producing smaller peptides or amino acids.
This distinction matters because students often memorise that “water is involved” without identifying whether it is a reactant or product.
In hydrolysis, water is consumed.
In condensation, water is formed.
That difference is central to biomolecule chemistry.
Fractional distillation was chosen because of boiling points
Question 25 asked which method would be most suitable to separate volatile polar compounds with similar boiling points from ethanol.
The correct answer was fractional distillation.
The report explained that fractional distillation is designed to separate volatile compounds with similar boiling points. Simple distillation requires significantly different boiling points, and solvent extraction is more useful when compounds have distinctly different polarities.
This was not simply a question about naming separation techniques.
Students needed to match the physical properties of the mixture to the method. The compounds were volatile, polar and had similar boiling points to ethanol. That made fractional distillation the appropriate method.
Organic separation questions often work this way.
The method follows the property.
Molecular formulas required reading skeletal structures
Question 26 gave the skeletal structures of geranial and linalool, two natural compounds found in essential oils.
Students needed to determine their molecular formulas.
The correct formulas were:
Geranial: C₁₀H₁₆O
Linalool: C₁₀H₁₈O
This question rewarded students who could interpret skeletal structures accurately.
In skeletal formulas, carbon atoms are represented by line ends and vertices. Hydrogen atoms attached to carbon are usually not shown and must be inferred based on carbon’s valency. Functional group atoms such as oxygen are shown explicitly.
Students who count only the visible atoms will undercount hydrogens or misread the structure.
Skeletal structure interpretation is one of the most important skills in VCE organic chemistry.
Functional group tests had to distinguish the compounds
Question 27 asked which laboratory test could confirm that a sample was pure geranial and not pure linalool.
The correct test was acidified dichromate.
Geranial is an aldehyde and is readily oxidised by acidified dichromate, causing the solution to change from orange to green. Linalool is a tertiary alcohol, so it cannot be readily oxidised by acidified dichromate and the solution remains orange.
This question was especially useful because several incorrect tests sounded plausible.
Bromine solution would not distinguish the compounds because both contain carbon-carbon double bonds and can decolourise bromine. Sodium hydrogen carbonate would not distinguish them because neither contains a carboxyl group, so neither would produce carbon dioxide gas. Acidified permanganate was also not correct as written because the colour change direction was wrong.
The key principle is simple.
A functional group test is useful only if it gives different observations for the compounds being compared.
Impurities affected melting point range
Question 28 asked about a blended vanillin sample.
Pure vanillin has a melting point of 82–83 °C. The question stated that vanillin is often blended with significant amounts of cheaper compounds that have similar melting points.
The correct melting point range for the blend would be broader and lower than the pure value.
This is a classic purity test. Impurities disrupt the regular packing of particles in the solid lattice, causing the sample to melt over a wider range and usually at a lower temperature than the pure substance.
The report noted that because the question stated the vanillin was blended with significant amounts of impurities, a very narrow range close to the pure melting point was not viable.
This question rewarded students who understood what melting point tells us about purity.
A sharp melting point suggests purity.
A broad, depressed melting point suggests impurities.
Triglyceride unsaturation required mole reasoning
Question 29 asked about one mole of a triglyceride containing only one type of fatty acid reacting completely with six moles of iodine.
Iodine reacts across carbon-carbon double bonds. If one mole of triglyceride reacts with six moles of iodine, the triglyceride contains six carbon-carbon double bonds in total.
A triglyceride contains three fatty acid chains. Since the triglyceride contains only one type of fatty acid, each fatty acid must contain:
6 ÷ 3 = 2 carbon-carbon double bonds
The correct fatty acid was linoleic acid.
This question required students to combine organic structure with mole reasoning. The reaction with iodine gave the total number of double bonds per triglyceride. Students then had to divide across the three identical fatty acid chains.
Organic chemistry is often assessed through quantitative structural logic.
¹³C NMR required carbon environments, not carbon count
Question 30 asked students to identify the ¹³C NMR spectrum of pentan-3-one.
Pentan-3-one is a symmetrical ketone. It contains five carbons, but only three distinct carbon environments because pairs of carbons are equivalent across the symmetrical molecule.
The correct ¹³C NMR spectrum therefore had three signals. Because the molecule is a ketone, one of those signals had to appear in the carbonyl region around 205–220 ppm.
This question exposed a common spectroscopy error.
Students often count total carbons instead of distinct carbon environments.
NMR does not show one signal for every carbon atom. It shows one signal for each unique carbon environment.
Symmetry reduces the number of signals.
Boiling points required intermolecular force reasoning
Section B Question 4 asked students to explain why carboxylic acids have higher boiling points than corresponding amines.
The table showed that the first three carboxylic acids had boiling points of 101 °C, 118 °C and 141 °C, while the first three amines had much lower boiling points of −6 °C, 17 °C and 48 °C.
A strong response needed to refer to structure and intermolecular forces.
Carboxylic acids contain the carboxyl functional group and can form strong hydrogen-bonded dimers. Amines can also form hydrogen bonds, but the hydrogen bonding is generally weaker because nitrogen is less electronegative than oxygen. Stronger intermolecular forces require more energy to overcome, so carboxylic acids have higher boiling points.
This was not a “bigger molecules have higher boiling points” question.
The comparison was between corresponding members of different homologous series. The functional group and intermolecular forces were the decisive features.
IR spectroscopy required functional group evidence
The same Section B question gave an IR spectrum from one of the separated samples and asked students to state whether it was produced by an amine or a carboxylic acid.
This required students to use characteristic absorption ranges.
A carboxylic acid would show a broad O–H absorption across a wide region and a strong C=O absorption. An amine would show N–H absorptions in a different region and would not have the same carboxylic acid carbonyl pattern.
IR spectroscopy questions reward students who can link absorption bands to bonds and functional groups.
The spectrum is evidence.
Students should not identify the compound by guessing from the surrounding context. They need to point to the features in the spectrum that support the classification.
Mass spectra required isotope logic
Section B also included Compound X, which contained oxygen and chlorine. Students were given a mass spectrum and told that oxygen has only the stable isotope ¹⁶O, while chlorine has stable isotopes ³⁵Cl and ³⁷Cl.
The question asked students to identify the base peak, explain why all possible isomers of Compound X would have the same peak at m/z = 108, and identify the species responsible for another peak.
This kind of question tests several layers of organic analysis.
The base peak is the most intense peak. The molecular ion peak relates to the molecular mass. Isomers have the same molecular formula and therefore the same molecular ion mass. Chlorine isotopes produce characteristic M and M+2 patterns because ³⁵Cl and ³⁷Cl differ by two mass units.
Students need to interpret the spectrum as molecular evidence, not just as a set of peaks.
Isomers have the same molecular formula
One of the most important ideas in the Compound X question was that all possible isomers would have the same peak at m/z = 108.
That is because structural isomers have the same molecular formula but different arrangements of atoms. Since molecular formula determines molecular mass, isomers of the same compound formula have the same molecular ion peak.
This is a simple idea, but it is often missed.
A different structure does not necessarily mean a different molecular mass.
Organic analysis questions often require students to separate formula, structure and spectrum.
Spectroscopy evidence had to be combined
Compound X also included ¹³C NMR information, followed by a request for the skeletal structure and IUPAC name.
This kind of question requires synthesis.
A mass spectrum may give molecular mass and isotope evidence. IR may indicate functional groups. ¹³C NMR gives the number and type of carbon environments. The final structure must be consistent with all the evidence.
Students often lose marks when they treat each technique separately. The stronger approach is to build a profile:
- molecular formula or molecular mass
- atoms present
- functional groups
- number of carbon environments
- symmetry or lack of symmetry
- possible structural isomers
- final structure and name
Organic analysis is cumulative.
Each piece of evidence narrows the possibilities.
Organic pathways required reaction conditions
Section B Question 4a asked students to write a balanced equation for the production of C₃H₇Cl from C₃H₈, including reaction conditions.
This is a substitution reaction involving an alkane and chlorine, typically requiring ultraviolet light.
A strong equation needed correct reactants, products and conditions. For example:
C₃H₈ + Cl₂ → C₃H₇Cl + HCl
with UV light.
The same part of the question asked for two reactants required to form propan-1-amine in a single substitution reaction. Students needed to recognise that a haloalkane and ammonia can react to form a primary amine.
This shows that organic pathways require both structural knowledge and reaction conditions.
Knowing the product is not always enough.
Students must know how it is made.
Organic chemistry rewarded evidence-based choice
Across the 2025 exam, organic questions repeatedly asked students to choose the method, test, structure or conclusion that matched the evidence.
Fractional distillation was chosen because boiling points were similar. Acidified dichromate was chosen because it distinguished an aldehyde from a tertiary alcohol. A lower, broader melting point range indicated impurities. Linoleic acid was chosen because each fatty acid chain needed two C=C bonds. The ¹³C NMR spectrum of pentan-3-one needed three signals including a ketone carbonyl. Compound X needed to fit mass spectrum and NMR evidence.
This is structural reasoning.
The student has to ask:
What does this evidence show?
Which structure is consistent with it?
Which test would produce different observations?
Which technique suits the physical property?
Which reaction type matches the functional group?
That is how organic chemistry is assessed.
Why organic chemistry errors happen
Organic chemistry errors often occur when students rely on labels rather than structures.
They remember that alcohols react with acids but build the ester incorrectly. They remember that bromine tests for unsaturation but forget both compounds are unsaturated. They remember that melting point indicates purity but ignore the word “significant” in the stem. They count carbon atoms instead of carbon environments. They identify a functional group but choose a test that would not distinguish the two molecules.
The 2025 exam rewarded students who slowed down and looked at the structure.
In organic chemistry, the structure decides the answer.
What future Chemistry students should learn from 2025
The 2025 VCE Chemistry exam shows that organic chemistry preparation must focus on structural reasoning.
Students should be able to:
- assemble ester products from alcohols and carboxylic acids
- distinguish hydrolysis from condensation
- choose separation techniques based on boiling point and polarity
- interpret skeletal structures and molecular formulas
- select functional group tests that distinguish compounds
- explain melting point changes caused by impurities
- use iodine reactions to infer carbon-carbon double bonds
- interpret ¹³C NMR using carbon environments and symmetry
- explain boiling point differences using intermolecular forces
- identify functional groups from IR spectra
- interpret mass spectra using molecular ions and isotope patterns
- combine spectral evidence to propose structures
- write organic reaction equations with conditions
These skills are connected.
Organic chemistry is not a memory test of isolated reactions.
It is a reasoning system based on structure.
How ATAR STAR approaches organic chemistry
At ATAR STAR, organic chemistry is taught through structure first.
Students learn to read skeletal formulas, identify functional groups, predict reactions, justify separation methods and interpret analytical data using chemical evidence. They practise combining IR, NMR, mass spectrometry and reaction information so that their answers are consistent across all the data provided.
The 2025 Examination Report confirms why this matters. High-scoring students did not simply recognise organic compounds.
They reasoned from the structure.
That is what VCE Chemistry rewards.