VCAA Chemistry How are organic compounds analysed and used?

15 sample questions with marking guides and sample answers

Q23
2020
VCAA
1 mark
Q23
1 mark

Use the following information to answer the question.

A solution of citric acid, C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3, was analysed by titration.
25.0 mL aliquots of the C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3 solution were titrated against a standardised solution of 0.0250 M sodium hydroxide, NaOH. Phenolphthalein indicator was used and the average titre was found to be 24.0 mL.

Based on the titration, the concentration of C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3 in the solution was

A

8.0×10−3 M8.0 \times 10^{-3} \text{ M}

B

8.7×10−3 M8.7 \times 10^{-3} \text{ M}

C

2.6×10−2 M2.6 \times 10^{-2} \text{ M}

D

7.2×10−2 M7.2 \times 10^{-2} \text{ M}

Reveal Answer
A

8.0×10−3 M8.0 \times 10^{-3} \text{ M}

Correct Answer

Citric acid is triprotic, reacting with NaOH in a 1:3 ratio. The moles of NaOH used is 0.0250×0.0240=6.0×10−40.0250 \times 0.0240 = 6.0 \times 10^{-4} mol, so the moles of acid is 2.0×10−42.0 \times 10^{-4} mol. Dividing by the aliquot volume (0.0250 L) gives 8.0×10−38.0 \times 10^{-3} M.

B

8.7×10−3 M8.7 \times 10^{-3} \text{ M}

This incorrect answer is obtained by mistakenly swapping the volumes of the acid and base in the concentration calculation.

C

2.6×10−2 M2.6 \times 10^{-2} \text{ M}

This error occurs if the volumes of the acid and base are swapped and the 1:3 stoichiometric ratio between citric acid and NaOH is ignored.

D

7.2×10−2 M7.2 \times 10^{-2} \text{ M}

This incorrect result comes from multiplying the moles of NaOH by 3 instead of dividing by 3 to find the moles of the triprotic citric acid.

Q30
2021
VCAA
1 mark
Q30
1 mark

The 1H^1\text{H} NMR spectrum of an organic compound has three unique sets of peaks: a single peak, seven peaks (septet) and two peaks (doublet).

The compound is

A

3-methyl butanoic acid.

B

2-methyl propanoic acid.

C

2-chloro-2-methylpropane.

D

1,2-dichloro-2-methylpropane.

Reveal Answer
A

3-methyl butanoic acid.

3-methylbutanoic acid has four unique sets of protons, which would produce a singlet (OH), a doublet (CH3 groups), a multiplet/nonet (CH), and another doublet (CH2).

B

2-methyl propanoic acid.

Correct Answer

2-methylpropanoic acid has three unique sets of protons: the carboxylic acid proton (singlet), the CH proton split by six methyl protons (septet), and the six equivalent methyl protons split by the CH proton (doublet).

C

2-chloro-2-methylpropane.

2-chloro-2-methylpropane has only one unique set of protons (nine equivalent methyl protons), which would produce a single peak (singlet) in its 1H^1\text{H} NMR spectrum.

D

1,2-dichloro-2-methylpropane.

1,2-dichloro-2-methylpropane has two unique sets of protons that are not adjacent to any other protons, which would result in a spectrum with two singlets.

Q12
2022
VCAA
1 mark
Q12
1 mark

Enzymes are commonly not effective in acidic conditions because acids

A

change the charges on the enzymes.

B

react with the enzymes to form zwitterions.

C

esterify the enzymes into smaller molecules.

D

react with the carboxyl groups on the enzymes' amino acid residues.

Reveal Answer
A

change the charges on the enzymes.

Correct Answer

Acidic conditions increase the concentration of H+H^+ ions, which protonate amino acid side chains. This alters the enzyme's overall charge distribution, disrupting the ionic bonds that maintain its functional 3D structure.

B

react with the enzymes to form zwitterions.

Zwitterions are molecules with both positive and negative charges that net to zero, typically occurring at an amino acid's isoelectric point. In highly acidic conditions, amino acids become positively charged cations, not zwitterions.

C

esterify the enzymes into smaller molecules.

Acids do not esterify enzymes into smaller molecules. Breaking down an enzyme's protein chain into smaller molecules would involve the hydrolysis of peptide bonds, not esterification.

D

react with the carboxyl groups on the enzymes' amino acid residues.

While acidic conditions do protonate carboxylate groups into neutral carboxyl groups, this is only a partial explanation. The loss of enzyme effectiveness is due to the overall change in charges across all ionizable groups, which disrupts the active site.

Q18
2025
VCAA
1 mark
Q18
1 mark

When an excess of Molecule G was mixed with acidified potassium dichromate solution, K2_2Cr2_2O7_7(aq), the colour of the solution remained orange.

The molecular formula of Molecule G is C4_4H10_{10}O.

Which one of the following could be Molecule G?

A

2-methylbutan-2-ol

B

3-methylbutan-2-ol

C

2-methylpropan-2-ol

D

butan-2-ol

Reveal Answer
A

2-methylbutan-2-ol

Although 2-methylbutan-2-ol is a tertiary alcohol and would not be oxidized, its molecular formula is C5_5H12_{12}O, which does not match the given formula of C4_4H10_{10}O.

B

3-methylbutan-2-ol

3-methylbutan-2-ol has the incorrect molecular formula (C5_5H12_{12}O) and is a secondary alcohol, which would be oxidized and turn the dichromate solution green.

C

2-methylpropan-2-ol

Correct Answer

2-methylpropan-2-ol has the correct molecular formula (C4_4H10_{10}O) and is a tertiary alcohol. Tertiary alcohols cannot be oxidized by acidified potassium dichromate, so the solution remains orange.

D

butan-2-ol

While butan-2-ol has the correct molecular formula, it is a secondary alcohol. It would be oxidized to a ketone, causing the dichromate solution to change colour from orange to green.

Q25
2025
VCAA
1 mark
Q25
1 mark

Which one of the following sets of wave numbers for the infrared spectrum is consistent with CH3_3COCH2_2CONH2_2? Refer to item 22 of the Data Book.

A

1720–1840 cm−1^{-1} and 1660–1745 cm−1^{-1}

B

1720–1840 cm−1^{-1} and 1680–1850 cm−1^{-1}

C

1630–1680 cm−1^{-1} and 1660–1745 cm−1^{-1}

D

1630–1680 cm−1^{-1} and 1680–1850 cm−1^{-1}

Reveal Answer
A

1720–1840 cm−1^{-1} and 1660–1745 cm−1^{-1}

This option is incorrect because it does not include the characteristic absorption range for an amide C=O bond (1630–1680 cm−1^{-1}).

B

1720–1840 cm−1^{-1} and 1680–1850 cm−1^{-1}

This option is incorrect because it misses the absorption band for the amide C=O stretch, which occurs at 1630–1680 cm−1^{-1}.

C

1630–1680 cm−1^{-1} and 1660–1745 cm−1^{-1}

This option is incorrect because it lacks the correct absorption range for a ketone C=O bond, which typically absorbs between 1680–1850 cm−1^{-1}.

D

1630–1680 cm−1^{-1} and 1680–1850 cm−1^{-1}

Correct Answer

This option is correct because the molecule contains both an amide group and a ketone group. The amide C=O bond absorbs at 1630–1680 cm−1^{-1}, and the ketone C=O bond absorbs at 1680–1850 cm−1^{-1}.

Q12
2023
QCAA
Paper 1
1 mark
Q12
1 mark

Enzymes are classified as

A

carbohydrates.

B

proteins.

C

starches.

D

lipids.

Reveal Answer
A

carbohydrates.

Carbohydrates are organic compounds like sugars and fibers used primarily for energy and structure, not for catalyzing reactions.

B

proteins.

Correct Answer

Enzymes are biological catalysts composed of amino acid chains folded into specific shapes, classifying them as proteins.

C

starches.

Starches are complex carbohydrates (polysaccharides) used for energy storage in plants, whereas enzymes are the proteins that might break them down.

D

lipids.

Lipids include fats, oils, and waxes used for long-term energy storage and membrane structure, but they do not function as biological catalysts.

Q6
2025
SCSA
1 mark
Q6
1 mark

Which of the following will decolourise a solution of bromine water?

A

1.0 mol L−1^{-1} Fe(NO3)3Fe(NO_3)_3 solution

B

1.0 mol L−1^{-1} KCl solution

C

CH3CH2CH2CH2CHOCH_3CH_2CH_2CH_2CHO

D

CH3CH2CHCHCH3CH_3CH_2CHCHCH_3

Reveal Answer
A

1.0 mol L−1^{-1} Fe(NO3)3Fe(NO_3)_3 solution

This is a solution of Iron(III) nitrate. Neither Fe3+Fe^{3+} nor NO3−NO_3^- ions react with bromine water, so the solution would not be decolourised.

B

1.0 mol L−1^{-1} KCl solution

Chloride ions (Cl−Cl^-) cannot reduce bromine (Br2Br_2) because chlorine is a stronger oxidising agent than bromine. Therefore, no reaction occurs to remove the bromine colour.

C

CH3CH2CH2CH2CHOCH_3CH_2CH_2CH_2CHO

Accepted answer

Pentanal (C₅H₁₀O) can cause a colour change with bromine because it is an aldehyde. Aldehydes are easily oxidised to carboxylic acids, and in this reaction, bromine (Br₂), which is orange-brown, is reduced to colourless bromide ions (Br⁻). As a result, the orange colour of bromine water disappears when it reacts with pentanal.

D

CH3CH2CHCHCH3CH_3CH_2CHCHCH_3

Accepted answer

The condensed formula represents 2-pentene, which contains a carbon-carbon double bond (C=CC=C). Alkenes undergo a rapid addition reaction with bromine water, consuming the Br2Br_2 and turning the orange solution colourless.

Q25
2025
VCAA
1 mark
Q25
1 mark

During fermentation, yeast will produce other volatile polar compounds that have similar boiling points to ethanol, C2H6OC_2H_6O.

Which one of the following methods would be most suitable to separate these compounds from C2H6OC_2H_6O?

A

solvent extraction

B

simple distillation

C

fractional distillation

D

solvent extraction and distillation

Reveal Answer
A

solvent extraction

Solvent extraction relies on differences in solubility, which would likely be ineffective here since both ethanol and the other compounds are polar and may dissolve in similar solvents.

B

simple distillation

Simple distillation is only effective for separating liquids with significantly different boiling points (typically a difference of at least 25°C).

C

fractional distillation

Correct Answer

Fractional distillation is specifically designed to separate miscible volatile liquids that have very similar boiling points.

D

solvent extraction and distillation

Adding solvent extraction is unnecessary and less efficient, as fractional distillation alone is the standard and most effective method for separating liquids with similar boiling points.

Q19
2020
QCAA
Paper 1
1 mark
Q19
1 mark

Dispersion forces, hydrogen bonding, disulfide bridges and ionic bonding all contribute to the

A

primary structure of proteins.

B

secondary structure of proteins.

C

tertiary structure of proteins.

D

quaternary structure of proteins.

Reveal Answer
A

primary structure of proteins.

Primary structure refers to the linear sequence of amino acids held together specifically by covalent peptide bonds, not by the various side-chain interactions listed.

B

secondary structure of proteins.

Secondary structure, such as α\alpha-helices and β\beta-sheets, is primarily stabilized by hydrogen bonds between the backbone amide and carbonyl groups, rather than interactions between side chains.

C

tertiary structure of proteins.

Correct Answer

Tertiary structure is the overall three-dimensional shape of a polypeptide, stabilized by interactions between R groups (side chains) including dispersion forces, hydrogen bonds, ionic bonds (salt bridges), and covalent disulfide bridges.

D

quaternary structure of proteins.

Quaternary structure refers specifically to the arrangement of multiple polypeptide subunits; while these forces hold subunits together, the question describes the forces defining the 3D fold of a single chain (tertiary structure).

Q23
2025
VCAA
1 mark
Q23
1 mark

Consider the following statements regarding melting point determination of a sample.

I If a sample starts melting at 49 °C and is not fully melted at 53 °C, then the sample is a mixture.
II If the melting point of a sample is lower than expected, then the sample is a mixture.

Which of the statements is/are correct?

A

I only

B

II only

C

both I and II

D

neither I nor II

Reveal Answer
A

I only

Correct Answer

This is correct. Statement I is true because a broad melting range (greater than 2 °C) is characteristic of an impure sample or mixture. Statement II is false because a lower-than-expected melting point could simply mean the sample is a different pure substance or there was an experimental error.

B

II only

This is incorrect because Statement I is true and Statement II is false. Pure substances have sharp melting points, so a broad range of >4 °C indicates a mixture, but a low melting point alone does not guarantee the presence of a mixture.

C

both I and II

This is incorrect because Statement II is false. While impurities do cause melting point depression, a lower melting point could also result from thermometer miscalibration or the sample being a completely different pure compound.

D

neither I nor II

This is incorrect because Statement I is true. A pure substance typically melts over a narrow range of 1-2 °C, so a sample melting over a range of more than 4 °C is definitively a mixture.

Q19
2024
VCAA
1 mark
Q19
1 mark

HPLC is used to separate a mixture containing methyl ethanoate, propanoic acid and butan-1-ol.
This technique uses a polar stationary phase and a non-polar mobile phase to achieve separation.

Which one of the following options is correct?

A

The sample is vaporised before passing through the stationary phase.

B

A calibration curve can be used to identify each component of the mixture.

C

Pressure is applied to overcome resistance of the mobile phase in the column.

D

A high retention time indicates that a component is highly attracted to the mobile phase.

Reveal Answer
A

The sample is vaporised before passing through the stationary phase.

Vaporising the sample is a specific requirement for gas chromatography (GC), but it is not used in liquid chromatography techniques like HPLC.

B

A calibration curve can be used to identify each component of the mixture.

A calibration curve is used to determine the concentration or amount of a component (quantitative analysis), whereas retention time is used to identify the component itself (qualitative analysis).

C

Pressure is applied to overcome resistance of the mobile phase in the column.

Correct Answer

In techniques like High-Performance Liquid Chromatography (HPLC), high pressure must be applied to force the liquid mobile phase through the densely packed stationary phase, which creates significant resistance to flow.

D

A high retention time indicates that a component is highly attracted to the mobile phase.

A high retention time means the component moves slowly through the column, which indicates it has a strong attraction (affinity) to the stationary phase, not the mobile phase.

Q21
2023
VCAA
1 mark
Q21
1 mark

Which one of the following statements about denaturation is correct?

Denaturation

A

can only be caused by changes in temperature or the addition of acids.

B

causes the reversible change of a protein's shape.

C

hydrolyses the primary structure of a protein.

D

alters the secondary structure of a protein.

Reveal Answer
A

can only be caused by changes in temperature or the addition of acids.

Incorrect. Denaturation can be caused by many other factors besides temperature and acids, such as bases, heavy metals, organic solvents, and mechanical agitation.

B

causes the reversible change of a protein's shape.

Incorrect. Denaturation is most commonly an irreversible process, such as cooking an egg, rather than a reliably reversible change.

C

hydrolyses the primary structure of a protein.

Incorrect. Denaturation does not break the covalent peptide bonds of the primary structure; a different process called hydrolysis is required to break down a protein's primary sequence.

D

alters the secondary structure of a protein.

Correct Answer

Correct. Denaturation disrupts the non-covalent interactions, such as hydrogen bonds, that maintain a protein's secondary, tertiary, and quaternary structures.

Q24
2025
VCAA
1 mark
Q24
1 mark

10.0 mL of 0.882 M hydrogen peroxide, H2_2O2_2, solution is titrated with a 0.100 M solution of acidified potassium permanganate, KMnO4_4. The equation for the reaction is shown below.

2MnO4−(aq)+6H+(aq)+5H2O2(aq)→2Mn2+(aq)+8H2O(l)+5O2(g)2\text{MnO}_4^-\text{(aq)} + 6\text{H}^+\text{(aq)} + 5\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{Mn}^{2+}\text{(aq)} + 8\text{H}_2\text{O(l)} + 5\text{O}_2\text{(g)}

What volume of KMnO4_4(aq) is expected to react completely with the 10.0 mL of H2_2O2_2(aq)?

A

17.7 mL

B

22.1 mL

C

35.3 mL

D

88.2 mL

Reveal Answer
A

17.7 mL

This incorrect volume results from using a 1:5 stoichiometric ratio instead of the correct 2:5 ratio of MnO4−_4^- to H2_2O2_2.

B

22.1 mL

This incorrect answer results from applying the stoichiometric ratio upside down (5:2 instead of 2:5) and making a decimal place error.

C

35.3 mL

Correct Answer

First, find the moles of H2_2O2_2: 10.0 mL×0.882 M=8.82 mmol10.0 \text{ mL} \times 0.882 \text{ M} = 8.82 \text{ mmol}. Using the 2:5 stoichiometric ratio, 8.82 mmol×(2/5)=3.528 mmol8.82 \text{ mmol} \times (2/5) = 3.528 \text{ mmol} of KMnO4_4 is needed, which requires 3.528 mmol/0.100 M=35.3 mL3.528 \text{ mmol} / 0.100 \text{ M} = 35.3 \text{ mL}.

D

88.2 mL

This calculation assumes a 1:1 stoichiometric ratio between H2_2O2_2 and KMnO4_4, failing to account for the 2:5 ratio in the balanced equation.

Q16
2021
VCAA
1 mark
Q16
1 mark

Which one of the following statements about IR spectroscopy is correct?

A

IR radiation changes the spin state of electrons.

B

Bond wave number is influenced only by bond strength.

C

An IR spectrum can be used to determine the purity of a sample.

D

In an IR spectrum, high transmittance corresponds to high absorption.

Reveal Answer
A

IR radiation changes the spin state of electrons.

IR radiation causes changes in the vibrational states of molecules, not the spin states of electrons (which is associated with Electron Paramagnetic Resonance spectroscopy).

B

Bond wave number is influenced only by bond strength.

The wavenumber of a bond's vibration depends on both the bond strength (force constant) and the reduced mass of the atoms involved, as described by Hooke's Law.

C

An IR spectrum can be used to determine the purity of a sample.

Correct Answer

An IR spectrum can reveal the presence of impurities if unexpected absorption peaks appear that do not belong to the pure compound.

D

In an IR spectrum, high transmittance corresponds to high absorption.

Transmittance and absorbance are inversely related; high transmittance means that most of the light passed through the sample, indicating low absorption.

Q2
2021
QCAA
Paper 1
1 mark
Q2
1 mark

Intra-chain hydrogen bonding between peptide groups occurs in

A

primary protein structures.

B

secondary protein structures.

C

tertiary protein structures.

D

quaternary protein structures.

Reveal Answer
A

primary protein structures.

Primary structure refers to the linear sequence of amino acids connected by covalent peptide bonds, not hydrogen bonds.

B

secondary protein structures.

Correct Answer

Secondary structures, such as the α\alpha-helix, are stabilized by intra-chain hydrogen bonds formed between the carbonyl (C=OC=O) and amide (N−HN-H) groups of the peptide backbone.

C

tertiary protein structures.

Tertiary structure is the overall 3D shape determined primarily by interactions between amino acid side chains (R-groups), such as disulfide bridges, hydrophobic interactions, and ionic bonds.

D

quaternary protein structures.

Quaternary structure involves the arrangement of multiple polypeptide subunits held together by inter-chain interactions, rather than intra-chain bonding within a single backbone.

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