QCAA Chemistry Properties and structure of organic materials

15 sample questions with marking guides and sample answers

Q39
2025
SCSA
14 marks
Q39
14 marks

Freon-11 is a colourless chlorofluorocarbon that boils at 23.77 °C. Prior to the knowledge of the ozone-depleting potential of chlorofluorocarbons (CFCs) and other possible harmful effects on the environment, it was used as a refrigerant.

The following data was used to determine that Freon-11 is trichlorofluoromethane, with a molecular formula of CCl3FCCl_3F.

A Freon-11 sample of 4.121 g was combusted in excess oxygen. All the carbon in the compound was converted to carbon dioxide and in a separate process, all its chlorine was converted into hydrochloric acid. The carbon dioxide produced had a mass of 1.320 g and the hydrochloric acid formed, required 85.70 mL of 1.050 mol L1^{-1} of ammonia solution for complete neutralisation.
Another sample of the Freon-11 with a mass of 3.721 g occupied a volume of 0.6068 L at a pressure of 120.00 kPa and temperature of 50.6 °C.

Using the same data, use calculations and reasoning to demonstrate that this is the correct molecular formula.

Reveal Answer

Carbon
n(C)=n(CO2)=1.320/44.01=0.02999n(C) = n(CO_2) = 1.320/44.01 = 0.02999 mol
m(C)=0.02299×12.01=0.3602m(C) = 0.02299 \times 12.01 = 0.3602 g

Chlorine
n(Cl)=n(HCl)=n(NH3)=1.050×0.08570=0.08999n(Cl) = n(HCl) = n(NH_3) = 1.050 \times 0.08570 = 0.08999 mol
m(Cl)=0.08999×35.45=3.1900m(Cl) = 0.08999 \times 35.45 = 3.1900 g

Fluorine
m(F)=4.121(0.3602+3.1900)=0.5708m(F) = 4.121 - (0.3602 + 3.1900) = 0.5708 g
n(F)=0.5708/19.00=0.03004n(F) = 0.5708 / 19.00 = 0.03004 mol

Mole Ratio
C: 0.02999
Cl: 0.08999
F: 0.03004

Simplify
Divide by 0.02999
C: 1
Cl: 3.00
F: 1.002

Empirical Formula
CCl3FCCl_3F

Molecular Formula
Empirical formula mass (EFM) = 137.36 amu/g mol1^{-1}
n=PV/RT=(120.0×0.6068)/(8.314×323.75)=0.027052476n = PV/RT = (120.0 \times 0.6068)/(8.314 \times 323.75) = 0.027052476 mol
Molecular formula mass (MFM) = m/n=3.721/0.0276503204=137.54m/n = 3.721/0.0276503204 = 137.54 g mol1^{-1}
Molecular formula = MFM/EFM ×\times Empirical formula = 137.54/137.36×CCl3F=1.0013×CCl3F137.54/137.36 \times CCl_3F = 1.0013 \times CCl_3F

Molecular formula = Empirical formula = CCl3FCCl_3F

Marking Criteria
DescriptorMarks

n(C) calculation

1

m(C) calculation

1

n(Cl) calculation

1

m(Cl) calculation

1

m(F) calculation

1

n(F) calculation

1

Mole Ratio setup

1

Simplify ratio

1

Empirical Formula

1

Empirical formula mass (EFM)

1

n = PV/RT calculation

1

Molecular formula mass (MFM) calculation

1

Molecular formula calculation

1

Molecular formula = Empirical formula statement

1
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.

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.

Q1
2023
QCAA
Paper 2
2 marks
Q1
2 marks

Polylactic acid (PLA) and low-density polyethylene (LDPE) are both used to produce plastic wrapping film.

PlasticCompositionDensity (g/cm3^3)Tensile stress (MPa)Elongation (%)Degradation rate
PLAplant-based1.24606slow
LDPEpetrochemical-based0.9212148none

Analyse the data to discuss one advantage and one disadvantage of using PLA rather than LDPE to produce plastic wrapping film.

Advantage:

Disadvantage:

Reveal Answer

Advantage: An advantage is that PLA is plant based, therefore it uses (renewable) natural resources while LDPE is produced from non-renewable fossil fuels.
Disadvantage: PLA has less % elongation than LDPE, therefore PLA would stretch less.

Marking Criteria
DescriptorMarks

identifies an advantage of using PLA using data

1

identifies a disadvantage of using PLA using data

1
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.

Q22
2024
VCAA
1 mark
Q22
1 mark

Use the following information to answer the question.

A triglyceride is reacted with methanol, CH3OH\text{CH}_3\text{OH}, in the presence of concentrated KOH(aq)\text{KOH}(\text{aq}). The products of this reaction are glycerol and Compound J.

The molecular formula of Compound J is C19H30O2\text{C}_{19}\text{H}_{30}\text{O}_2.

What is the molecular formula of the triglyceride?

A

C54H82O3\text{C}_{54}\text{H}_{82}\text{O}_3

B

C54H82O6\text{C}_{54}\text{H}_{82}\text{O}_6

C

C57H84O3\text{C}_{57}\text{H}_{84}\text{O}_3

D

C57H86O6\text{C}_{57}\text{H}_{86}\text{O}_6

Reveal Answer
A

C54H82O3\text{C}_{54}\text{H}_{82}\text{O}_3

This formula is incorrect because a triglyceride must contain 6 oxygen atoms, not 3, and the carbon and hydrogen counts do not balance the transesterification reaction.

B

C54H82O6\text{C}_{54}\text{H}_{82}\text{O}_6

This formula is incorrect because it fails to account for the 3 carbon atoms from the glycerol backbone, resulting in 54 carbons instead of 57.

C

C57H84O3\text{C}_{57}\text{H}_{84}\text{O}_3

This formula is incorrect because a triglyceride contains 3 ester groups, meaning it must have 6 oxygen atoms, not 3.

D

C57H86O6\text{C}_{57}\text{H}_{86}\text{O}_6

Correct Answer

This is correct. In the transesterification reaction, 1 Triglyceride + 3 Methanol (CH4O\text{CH}_4\text{O}) \rightarrow 1 Glycerol (C3H8O3\text{C}_3\text{H}_8\text{O}_3) + 3 Compound J (C19H30O2\text{C}_{19}\text{H}_{30}\text{O}_2). Balancing the atoms yields C57H86O6\text{C}_{57}\text{H}_{86}\text{O}_6.

Q17
2025
SCSA
1 mark
Q17
1 mark

Which of the following is the IUPAC name for an isomer of propyl butanoate?

A

methyl pentanoate

B

heptanoic acid

C

6-methylpentanoic acid

D

3-propylbutanoic acid

Reveal Answer
A

methyl pentanoate

Methyl pentanoate contains 6 carbons (11 from methyl + 55 from pentanoate), whereas propyl butanoate contains 7 carbons (33 from propyl + 44 from butanoate). Isomers must have the same molecular formula.

B

heptanoic acid

Correct Answer

Propyl butanoate is an ester with the formula C7H14O2C_7H_{14}O_2 (3+4=73+4=7 carbons). Heptanoic acid is a carboxylic acid with the same formula C7H14O2C_7H_{14}O_2. Esters and carboxylic acids with the same number of carbon atoms are functional isomers.

C

6-methylpentanoic acid

This name is invalid because a pentanoic acid chain only has 5 carbons, so a substituent cannot be at position 6. Furthermore, a methyl pentanoic acid derivative would only have 6 carbons total, not the required 7.

D

3-propylbutanoic acid

Although a structure constructed from this name would have 7 carbons, '3-propylbutanoic acid' is not a valid IUPAC name. The propyl group would extend the longest carbon chain, making the correct parent name hexanoic acid (specifically 3-methylhexanoic acid).

Q18
2024
VCAA
1 mark
Q18
1 mark

Each of the following compounds has a molar mass of 88 g mol188 \text{ g mol}^{-1}.

Which one has the highest boiling point?

A

CH3CH2OCOCH3\text{CH}_3\text{CH}_2\text{OCOCH}_3

B

CH3(CH2)2COOH\text{CH}_3(\text{CH}_2)_2\text{COOH}

C

CH3(CH2)3CH2OH\text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH}

D

CH3NH(CH2)2NHCH3\text{CH}_3\text{NH}(\text{CH}_2)_2\text{NHCH}_3

Reveal Answer
A

CH3CH2OCOCH3\text{CH}_3\text{CH}_2\text{OCOCH}_3

Esters cannot form intermolecular hydrogen bonds, resulting in weaker dipole-dipole interactions and a lower boiling point compared to alcohols and carboxylic acids.

B

CH3(CH2)2COOH\text{CH}_3(\text{CH}_2)_2\text{COOH}

Correct Answer

Carboxylic acids can form strong intermolecular hydrogen bonds, often forming stable dimers. This significantly increases their intermolecular forces and gives them the highest boiling point among compounds of similar molar mass.

C

CH3(CH2)3CH2OH\text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH}

While alcohols can form hydrogen bonds, they do not form stable dimers like carboxylic acids do, resulting in slightly weaker overall intermolecular forces and a lower boiling point than option B.

D

CH3NH(CH2)2NHCH3\text{CH}_3\text{NH}(\text{CH}_2)_2\text{NHCH}_3

Amines can form hydrogen bonds, but because nitrogen is less electronegative than oxygen, the N-H bond is less polar than the O-H bond. This makes their hydrogen bonds weaker than those in alcohols and carboxylic acids.

Q18
2021
QCAA
Paper 1
1 mark
Q18
1 mark

Identify the major product when 2-methylbut-2-ene reacts with water under acidic conditions.

A

(CH3_3)2_2CHCOCH3_3

B

(CH3_3)2_2C(OH)CH2_2CH3_3

C

(CH3_3)2_2CHCH(OH)CH3_3

D

(CH3_3)2_2C(OH)CH(OH)CH3_3

Reveal Answer
A

(CH3_3)2_2CHCOCH3_3

This is a ketone (3-methyl-2-butanone). Acid-catalyzed hydration of alkenes produces alcohols, not carbonyl compounds; ketones are typically formed from the hydration of alkynes or oxidation of alcohols.

B

(CH3_3)2_2C(OH)CH2_2CH3_3

Correct Answer

The reaction follows Markovnikov's rule, proceeding via the most stable tertiary carbocation intermediate. The hydroxyl group (OH-OH) attaches to the more substituted carbon, yielding the tertiary alcohol 2-methyl-2-butanol.

C

(CH3_3)2_2CHCH(OH)CH3_3

This is the anti-Markovnikov product (3-methyl-2-butanol). The reaction does not favor the formation of the less stable secondary carbocation intermediate required to produce this secondary alcohol.

D

(CH3_3)2_2C(OH)CH(OH)CH3_3

This is a vicinal diol. Acid-catalyzed hydration adds water (HOHH-OH) across the double bond to form a mono-alcohol, whereas diols are formed through oxidation reactions like dihydroxylation.

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.

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.

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.

Q8
2023
NESA
1 mark
Q8
1 mark

How many structural isomers have the molecular formula C3H6F2\text{C}_3\text{H}_6\text{F}_2?

A

2

B

3

C

4

D

5

Reveal Answer
A

2

Incorrect. There are more than 2 structural isomers; this answer likely only accounts for the geminal isomers (both fluorines on the same carbon).

B

3

Incorrect. There are more than 3 structural isomers; this answer likely misses one of the possible arrangements, such as 1,3-difluoropropane.

C

4

Correct Answer

Correct. There are exactly 4 structural isomers for C3H6F2\text{C}_3\text{H}_6\text{F}_2: 1,1-difluoropropane, 2,2-difluoropropane, 1,2-difluoropropane, and 1,3-difluoropropane.

D

5

Incorrect. There are only 4 unique structural isomers; any fifth arrangement would just be a duplicate of one of the four due to the symmetry of the propane chain.

Q9
2021
QCAA
Paper 1
1 mark
Q9
1 mark

The boiling points of methane, ethane and propane increase as the lengths of the carbon chains increase because more energy is required to overcome the

A

intramolecular hydrogen bonds.

B

intermolecular hydrogen bonds.

C

intramolecular dispersion forces.

D

intermolecular dispersion forces.

Reveal Answer
A

intramolecular hydrogen bonds.

Intramolecular forces hold atoms together within a molecule; boiling involves separating molecules from one another, not breaking bonds within them. Furthermore, alkanes do not contain the electronegative atoms necessary for hydrogen bonding.

B

intermolecular hydrogen bonds.

Alkanes are nonpolar hydrocarbons that lack the hydrogen atoms bonded to highly electronegative atoms (N, O, F) required to form hydrogen bonds.

C

intramolecular dispersion forces.

Dispersion forces are intermolecular (between molecules), not intramolecular. Intramolecular forces refer to the covalent bonds holding the carbon and hydrogen atoms together.

D

intermolecular dispersion forces.

Correct Answer

Boiling requires overcoming intermolecular forces. As the carbon chain length increases, the molecule's surface area and electron count increase, leading to stronger London dispersion forces that require more energy to overcome.

Q20
2024
SCSA
1 mark
Q20
1 mark

Classify the type of reaction represented in the following equation:

CH2CH2(g)+HCl(g)CH2ClCH3(l)CH_2CH_2(g) + HCl(g) \rightarrow CH_2ClCH_3(l)

A

addition

B

oxidation

C

combustion

D

condensation

Reveal Answer
A

addition

Correct Answer

This is an addition reaction because two reactant molecules (CH2CH2CH_2CH_2 and HClHCl) combine to form a single product molecule (CH2ClCH3CH_2ClCH_3) without losing any atoms.

B

oxidation

Oxidation typically involves the loss of electrons or the addition of oxygen, whereas this reaction simply adds HClHCl across a carbon-carbon double bond.

C

combustion

Combustion requires oxygen (O2O_2) as a reactant and typically produces carbon dioxide and water, which are not present in this equation.

D

condensation

A condensation reaction joins two molecules together while eliminating a small molecule like water (H2OH_2O), but here all reactant atoms are incorporated into the single product.

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