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

8.0×103 M8.0 \times 10^{-3} \text{ M}

B

8.7×103 M8.7 \times 10^{-3} \text{ M}

C

2.6×102 M2.6 \times 10^{-2} \text{ M}

D

7.2×102 M7.2 \times 10^{-2} \text{ M}

Reveal Answer
A

8.0×103 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=6.0×1040.0250 \times 0.0240 = 6.0 \times 10^{-4} mol, so the moles of acid is 2.0×1042.0 \times 10^{-4} mol. Dividing by the aliquot volume (0.0250 L) gives 8.0×1038.0 \times 10^{-3} M.

B

8.7×103 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

2.6×102 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

7.2×102 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.

Q6
2025
SCSA
1 mark
Q6
1 mark

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

A

1.0 mol L1^{-1} Fe(NO3)3Fe(NO_3)_3 solution

B

1.0 mol L1^{-1} KCl solution

C

CH3CH2CH2CH2CHOCH_3CH_2CH_2CH_2CHO

D

CH3CH2CHCHCH3CH_3CH_2CHCHCH_3

Reveal Answer
A

1.0 mol L1^{-1} Fe(NO3)3Fe(NO_3)_3 solution

This is a solution of Iron(III) nitrate. Neither Fe3+Fe^{3+} nor NO3NO_3^- ions react with bromine water, so the solution would not be decolourised.

B

1.0 mol L1^{-1} KCl solution

Chloride ions (ClCl^-) 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

Correct 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

Correct 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.

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.

Q24
2020
VCAA
1 mark
Q24
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.

Which one of the following would have resulted in a concentration that is higher than the actual concentration?

A

The pipette was rinsed with NaOH solution.

B

The pipette was rinsed with C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3 solution.

C

The conical flask was rinsed with NaOH solution.

D

The conical flask was rinsed with C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3 solution.

Reveal Answer
A

The pipette was rinsed with NaOH solution.

Rinsing the pipette with NaOH would neutralize some of the citric acid before it is transferred to the flask. This decreases the required NaOH titre, resulting in a lower calculated concentration.

B

The pipette was rinsed with C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3 solution.

Rinsing the pipette with the solution it will transfer (citric acid) is the standard correct procedure and ensures the concentration remains accurate.

C

The conical flask was rinsed with NaOH solution.

Rinsing the conical flask with NaOH adds extra base to the flask, which neutralizes some of the citric acid before titration begins. This decreases the required NaOH titre, leading to a lower calculated concentration.

D

The conical flask was rinsed with C3H5O(COOH)3\text{C}_3\text{H}_5\text{O(COOH)}_3 solution.

Correct Answer

Rinsing the conical flask with citric acid leaves residual acid in the flask, meaning there is more acid present than just the 25.0 mL aliquot. This requires a larger volume of NaOH to neutralize, leading to a higher calculated concentration.

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.

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).

Q6
2024
QCAA
Paper 1
1 mark
Q6
1 mark

The equivalence point of an acid–base titration occurs when the

A

pH equals the pKaK_a.

B

pH stops changing.

C

indicator changes colour.

D

titrant completely neutralises the analyte.

Reveal Answer
A

pH equals the pKaK_a.

The pH equals the pKaK_a at the half-equivalence point, where exactly half of the weak acid has been converted to its conjugate base.

B

pH stops changing.

The pH changes most rapidly at the equivalence point rather than stopping; it only levels off significantly after excess titrant is added.

C

indicator changes colour.

The point where the indicator changes colour is called the end point, which is an experimental approximation that should ideally be close to the equivalence point.

D

titrant completely neutralises the analyte.

Correct Answer

The equivalence point is defined as the stoichiometric point where the moles of added titrant exactly equal the moles required to completely react with (neutralise) the analyte.

Q26
2024
VCAA
1 mark
Q26
1 mark

Use the following information to answer the question.

A chemist runs a mixture of hexane, hexan-1-ol and hexan-2-one through a high-performance liquid chromatography (HPLC) column using a polar mobile phase and a non-polar stationary phase.

The chemist wants to determine the concentration of hexane in the mixture.

Which one of the following will provide information to allow the hexane concentration to be accurately calculated?

A

running a series of known concentrations of hexane through the HPLC column under the same conditions

B

running the HPLC experiment using a non-polar mobile phase and a polar stationary phase

C

using published retention times and peak sizes of standard hexane chromatographs

D

reducing the HPLC column temperature to achieve better separation of the compounds

Reveal Answer
A

running a series of known concentrations of hexane through the HPLC column under the same conditions

Correct Answer

To accurately determine concentration, a calibration curve must be constructed by running standard solutions of known concentrations under the exact same experimental conditions to compare peak areas.

B

running the HPLC experiment using a non-polar mobile phase and a polar stationary phase

Changing the polarity of the mobile and stationary phases alters the separation method (switching to normal phase chromatography) but does not provide the quantitative reference data needed to calculate concentration.

C

using published retention times and peak sizes of standard hexane chromatographs

Retention times and peak sizes are highly dependent on the specific instrument, column age, and exact experimental conditions, so published data cannot be reliably used for quantitative analysis.

D

reducing the HPLC column temperature to achieve better separation of the compounds

While reducing the temperature might improve the separation (resolution) of the peaks, it does not provide the reference standards required to calculate the actual concentration of the compound.

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
2024
VCAA
1 mark
Q24
1 mark

Evelyn titrates 10.0 mL of 0.100 M potassium permanganate, KMnO4\text{KMnO}_4, with 0.100 M oxalic acid, C2H2O4\text{C}_2\text{H}_2\text{O}_4.

The half-equation for the oxidation of oxalic acid in acidic conditions is

C2H2O4(aq)2CO2(g)+2H+(aq)+2e\text{C}_2\text{H}_2\text{O}_4(\text{aq}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{H}^+(\text{aq}) + 2\text{e}^-

What volume of C2H2O4\text{C}_2\text{H}_2\text{O}_4 should be added to reach the equivalence point?

A

4.0 mL

B

10.0 mL

C

12.0 mL

D

25.0 mL

Reveal Answer
A

4.0 mL

This incorrect volume results from using a 2:5 molar ratio of oxalic acid to permanganate, which is the inverse of the actual stoichiometric ratio.

B

10.0 mL

This assumes a 1:1 stoichiometric ratio between oxalic acid and potassium permanganate, ignoring the number of electrons transferred in the redox reaction.

C

12.0 mL

This is a miscalculation. The correct volume must be determined by balancing the electrons transferred in the oxidation and reduction half-reactions.

D

25.0 mL

Correct Answer

The reduction of MnO4\text{MnO}_4^- requires 5 electrons, while the oxidation of oxalic acid produces 2 electrons, resulting in a 5:2 molar ratio of oxalic acid to permanganate. Since the concentrations are equal, 10.0 mL×(5/2)=25.0 mL10.0 \text{ mL} \times (5/2) = 25.0 \text{ mL} of oxalic acid is needed.

Q25
2024
VCAA
1 mark
Q25
1 mark

Use the following information to answer the question.

A chemist runs a mixture of hexane, hexan-1-ol and hexan-2-one through a high-performance liquid chromatography (HPLC) column using a polar mobile phase and a non-polar stationary phase.

Which of the following shows the chemicals in order of their retention times, from lowest to highest?

A

hexane, hexan-2-one, hexan-1-ol

B

hexane, hexan-1-ol, hexan-2-one

C

hexan-2-one, hexan-1-ol, hexane

D

hexan-1-ol, hexan-2-one, hexane

Reveal Answer
A

hexane, hexan-2-one, hexan-1-ol

This order represents the highest to lowest retention time, which would be the correct order if the stationary phase were polar (normal-phase HPLC).

B

hexane, hexan-1-ol, hexan-2-one

Hexane is the most non-polar compound, meaning it will interact most strongly with the non-polar stationary phase and have the highest, not lowest, retention time.

C

hexan-2-one, hexan-1-ol, hexane

While hexane correctly has the highest retention time, hexan-1-ol is more polar than hexan-2-one due to its ability to form hydrogen bonds, so hexan-1-ol will elute before hexan-2-one.

D

hexan-1-ol, hexan-2-one, hexane

Correct Answer

In reverse-phase HPLC (non-polar stationary phase, polar mobile phase), the most polar compound has the lowest retention time. Hexan-1-ol (most polar) elutes first, followed by hexan-2-one, and the non-polar hexane elutes last.

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 (NHN-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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