VCE Chemistry 2021

Every question from the 2021 VCE Chemistry examination. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle. This paper was set on the previous study design, so each question also says whether it still matches the course you are studying now.

How to use this page

• Questions are from the 2021 VCE Chemistry examination, copyright Victorian Curriculum and Assessment Authority (VCAA). Each is summarised briefly here; open the official examination PDF for the full wording, diagrams, spectra and the 2021 Data Book (bound in at the back).
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2021 Chemistry external assessment report (Word document), and every Section B calculation was redone and compared with the report. Both files are listed on the VCAA Chemistry examinations page.
• Data values (heats of combustion, $E^0$ values, molar masses) are the ones in the 2021 Data Book. A few differ slightly in today's Data Book (for example, butane is now $-2877$ kJ mol⁻¹ and ethanol 29.7 kJ g⁻¹), so use the Data Book that matches the paper you are practising.
• Percentages after multiple-choice answers are the share of students who chose the correct option. In Section B, "From the report" notes summarise what the examiners said students got wrong.

Structure and timing

The paper is 120 marks in 150 minutes (plus 15 minutes reading time), with a scientific calculator and the Data Book. That is 1.25 minutes per mark.

• Section A (30 marks): 30 multiple-choice questions (Question 5 has since been withdrawn from the published paper). Aim for about 35 minutes.
• Section B (90 marks): 10 questions (9, 7, 12, 9, 8, 9, 8, 9, 11 and 8 marks). Show working, give states in equations and use sensible significant figures.

Study design check for 2026

The 2021 exam was written for the 2017 to 2023 study design. The current Units 3 and 4 study design (from 2024) keeps fuels, calorimetry, galvanic and fuel cells, electrolysis, rates and equilibrium, organic reactions and spectroscopy, but it dropped most of the old "chemistry of food" area of study and a few other details. When you practise with this paper:

• Skip (no longer in the study design): Section A Q3 (lactose intolerance) and Q15 (flashpoint); Section B 3a (essential amino acids), 3d.i (omega fatty acids), 5b (vitamin solubility), 5d (coenzymes), 9a (artificial sweeteners), 9c (HPLC operating conditions, now excluded), and the glycaemic index and carbon monoxide poisoning contexts in 10a and 10b.
• Adapt: Section A Q2 (the petrodiesel and biodiesel comparison is no longer listed, but the viscosity reasoning is), Q17 (gas calculations are now limited to SLC), Q19 (the current course names solution calorimetry, not bomb calorimetry, though $E = \text{CF} \times \Delta T$ is the same), Section B 3e (cellulose is no longer named) and 8b (Maxwell-Boltzmann curves are no longer named, though the collision-theory idea still is).
• Still fully examinable: everything else, including all the electrochemistry, equilibrium, organic pathway and spectroscopy questions.

Section A: Multiple choice

Q1

Rechargeable batteries. Answer: A - they use reversible reactions: discharge runs as a galvanic cell and recharge as an electrolytic cell that reverses it. (82%)

Q2

Biodiesel compared with petrodiesel. Answer: A - they have different viscosities: biodiesel molecules are larger and contain polar ester groups, so they attract each other more strongly and flow less easily. (92%) Study design: the explicit petrodiesel/biodiesel comparison is no longer listed, but viscosity trends from structure and bonding still are.

Q3

People who are lactose intolerant. Answer: A - without enough lactase, some of the lactose they eat is not hydrolysed (digested). Lactose is a carbohydrate, not a protein. (53%; 37% chose C.) Study design: outside (lactose intolerance is no longer listed).

Q4

Which would cause a systematic error in an ethanoic acid/NaOH titration? Answer: D - an NaOH solution weaker than its stated concentration makes every calculated result wrong in the same direction. A leaking tap, reading to 0.1 mL and inconsistent indicator drops are random errors. (62%)

Q5

This question is marked "no longer available" in the published paper and report, so there is nothing to answer.

Q6

Bonds broken in denaturation and in hydrolysis of a protein. Answer: D - denaturation disrupts the secondary and tertiary structure (hydrogen bonds, ionic bonds and other interactions between side chains); hydrolysis breaks the covalent peptide bonds. (61%)

Q7

Which characteristics of electrolytic and galvanic cells are correct? Answer: A - only 1 (electrolytic cathode negative, galvanic cathode positive) and 2 (non-spontaneous against spontaneous). Reduction always happens at the cathode, and it is galvanic cells that produce electricity. (77%)

Q8

A triglyceride made from C₁₇H₃₅, C₁₉H₃₉ and C₁₇H₃₃ chains. Answer: D - its long non-polar hydrocarbon chains make it insoluble in water. Only the oleic (C₁₇H₃₃) chain has a C=C, so there is one double bond, not three; that unsaturated chain can undergo oxidative rancidity; hydrolysis gives glycerol and fatty acids, not CO₂ and water. (71%)

Q9

A charge of 4.00 C in 5.00 minutes. Amount of electrons? Answer: A - $n(e^-) = \tfrac{Q}{F} = \tfrac{4.00}{96\,500} = 4.15 \times 10^{-5}$ mol. The time is not needed; B comes from treating 4.00 as a current and multiplying by 5.00 min. (62%)

Q10

Most valid test of whether polishing the Zn electrode of an Fe-Zn cell increases the current. Answer: C - measure the rate of mass change of the Fe electrode (proportional to the current) in the same cell, before and after polishing, so the only thing that changes is the polishing. Option D uses two different cells, so other variables are not controlled. (17%; 80% chose D.)

Q11

Two ¹³C peaks, three sets of ¹H peaks, last MS peak at m/z 60 and an IR peak at 3350 cm⁻¹. Answer: B - propan-2-ol: two carbon environments (the two CH₃ are equivalent), three hydrogen environments (CH₃, CH, OH), $M = 60$ and an alcohol O-H. Ethanoic acid (A) has only two hydrogen environments; the amine (C) has $M = 59$; propanone (D) has no O-H. (69%)

Q12

67.0 g of butane is burned. Mass of CO₂? Answer: D - $n(\text{C}_4\text{H}_{10}) = \tfrac{67.0}{58.0} = 1.155$ mol, so $n(\text{CO}_2) = 4 \times 1.155 = 4.62$ mol and $m = 4.62 \times 44.0 \approx 203$ g. Working from the energy ($\tfrac{3330}{2880} = 1.156$ mol butane) gives 204 g; the report accepts both routes. C forgets the 1 : 4 mole ratio. (70%)

Q13

Lead-acid battery: the cathode reaction during recharge. Answer: B - recharge reverses discharge, and reduction happens at the cathode: PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq). (62%)

Q14

The oxidising agent during discharge. Answer: B - PbO₂: lead goes from +4 in PbO₂ to +2 in PbSO₄, so PbO₂ is reduced and oxidises the Pb. (46%)

Q15

Which statement about flashpoint is correct? Answer: D - optical isomers of 3-methylhexane have identical physical properties, so equal flashpoints. Pentane's weaker dispersion forces give it a lower flashpoint than octane; structural isomers differ; a higher flashpoint means a lower fire risk. (56%) Study design: outside (flashpoint is no longer listed).

Q16

Which statement about IR spectroscopy is correct? Answer: C - comparing a sample's fingerprint region with that of the pure compound shows whether it is pure. IR changes bond vibrations, not spin states; wave number depends on bond strength and the masses of the atoms; high transmittance means low absorption. (15%; 38% chose D.)

Q17

Mass of O₂ to raise the pressure in a 200.0 L container at 22.0 °C by 250.0 kPa. Answer: C - $n = \tfrac{pV}{RT} = \tfrac{250.0 \times 200.0}{8.31 \times 295} = 20.4$ mol, so $m = 20.4 \times 32.0 = 653$ g $= 6.53 \times 10^2$ g. (67%) Study design: adapt. Gas calculations are now limited to SLC, although $pV = nRT$ is still printed in the Data Book.

Q18

Which energy profile matches the given thermochemical equations? Answer: C - it shows N₂O₄ → 2NO₂ rising by about 57 kJ, the reverse of 2NO₂ ⇌ N₂O₄ ($\Delta H = -57$ kJ mol⁻¹). A, B and D each show the opposite sign to their equation. (73%)

Q19

Bomb calorimeter: 3.95 g of muesli bar, $\Delta T = 16.7$ °C, CF $= 4780$ J °C⁻¹. Answer: B - $E = 4780 \times 16.7 = 7.98 \times 10^4$ J, and $\tfrac{7.98 \times 10^4}{3.95} = 2.02 \times 10^4$ J g⁻¹. (66%) Study design: adapt (bomb calorimetry is no longer named; the calculation is the same).

Q20

Mass of Ni plated by 320 F of charge. Answer: C - Ni²⁺ + 2e⁻ → Ni, so $n(\text{Ni}) = \tfrac{320}{2} = 160$ mol and $m = 160 \times 58.7 = 9.39 \times 10^3$ g. D forgets the 2 : 1 ratio. (52%)

Q21

Which change may reduce the Ni plated onto the Pt electrode? Answer: A - with a Cu anode instead of Ni, no Ni²⁺ is made at the anode to replace what is plated, so the Ni²⁺ runs down (and Cu²⁺, a stronger oxidant, enters the electrolyte). The other changes leave the Ni anode supplying Ni²⁺. (31%)

Q22

Volume of octane (703 g L⁻¹, 25.0% efficient engine) for 528 MJ of usable energy. Answer: D - energy needed from fuel $= \tfrac{528}{0.250} = 2112$ MJ; mass $= \tfrac{2.112 \times 10^6 \text{ kJ}}{47.9 \text{ kJ g}^{-1}} = 4.41 \times 10^4$ g; volume $= \tfrac{4.41 \times 10^4}{703} = 62.7$ L. C (15.7 L) ignores the efficiency. (37%; 34% chose C.)

Q23

Titres for three samples of an organic acid against the same KOH. Answer: B - equal 25 mL aliquots, so the largest titre (Sample 3, 21.55 mL) means the most acid. D is wrong: Sample 2's titres (19.85, 19.65, 20.45 mL) are not concordant. (49%)

Q24

Effect of a catalyst on the energy profile of an exothermic reaction. Answer: A - the energy of the products (and reactants, so $\Delta H$) is unchanged. The activation energies fall by the same amount, not the same proportion, so D is wrong. (38%; 47% chose D.)

Q25

Rate-time graph: at $t_1$ the reverse rate jumps, then both rates meet at a higher level at $t_2$. Answer: C - only a sudden addition of product raises the reverse rate while the forward rate is unchanged; the system then shifts forward, so there is more product at $t_2$ than before $t_1$. Argon has no effect, removing reactant would drop the forward rate, and a temperature change would change both rates at once. (34%)

Q26

Half-cells in decreasing voltage when paired with In³⁺/In ($-0.34$ V) as the negative electrode. Answer: C - In must be oxidised, so the partner must be higher in the series: Cu ($0.34 + 0.34 = 0.68$ V), Pb (0.21 V), Ni (0.09 V). Mn, Al and Mg are all below In. (41%)

Q27

At $t_2$ every concentration in ½H₂ + ½I₂ ⇌ HI doubles instantly. Which statement is correct? Answer: D - by the equation, HI is formed at twice the rate that H₂ is used and HI is used at twice the rate that H₂ forms, so at equilibrium the rate of production of HI is double that of H₂. The jump is a volume halving with equal numbers of gas particles on each side, so there is no shift: the amount of HI is unchanged (B is wrong) but both rates are higher (C is wrong). A catalyst (A) does not change concentrations. (13%; 45% chose B.)

Q28

A change at $t_4$ alters $K$, and the reverse rate at $t_5$ is higher than at $t_3$. Answer: A - only temperature changes $K$, and higher rates mean the temperature rose. The forward reaction is endothermic, so $K$ increases and the system absorbs energy: total chemical energy increases. (22%; 31% chose D.)

Q29

Two connected cells, Zn/Cu (Cell 1) and Cu/Ni (Cell 2). Answer: C - Cell 1 (1.10 V) is galvanic (chemical → electrical) and drives Cell 2 (0.59 V) backwards as an electrolytic cell (electrical → chemical). (48%; 36% chose A.)

Q30

¹H NMR with a singlet, a septet and a doublet. Answer: B - 2-methylpropanoic acid, (CH₃)₂CHCOOH: the six CH₃ hydrogens are a doublet (one neighbouring H), the CH is a septet (six neighbouring H), and the COOH is a singlet. (47%)

Section B

Question 1 (9 marks)

Digesters turn organic waste into biogas, which is mainly methane, CH₄.

a

Why is biogas renewable but coal seam gas is not? (1 mark)

b

1 kg of waste gives 496.0 L of biogas at SLC, 60.0% CH₄. i. Write the thermochemical equation for the complete combustion of CH₄ at SLC. (2 marks) ii. Calculate the energy the CH₄ could produce. (3 marks)

c

Biogas releasing 1.63 × 10³ kJ heats 100 kg of water from 25.0 °C. i. What is the maximum temperature reached? (2 marks) ii. Why may this not be reached? (1 mark)

Study design: all parts are in the current course.

Show worked solution

a. [1 mark]. Biogas is made continually by bacteria from organic waste, so it is replaced within a short time. Coal seam gas formed over millions of years and is used far faster than it can be replaced. (Both fuels must be discussed.)

b. i. [2 marks]. At SLC, water is a liquid:

$$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \qquad \Delta H = -890 \text{ kJ mol}^{-1}$$

ii. [3 marks].

$$n(\text{biogas}) = \frac{496.0}{24.8} = 20.0 \text{ mol}, \qquad n(\text{CH}_4) = 0.600 \times 20.0 = 12.0 \text{ mol}$$

$$E = 12.0 \times 890 = 1.07 \times 10^4 \text{ kJ} \;(10.7 \text{ MJ}).$$

c. i. [2 marks]. $q = mc\Delta T$ with $m = 100$ kg $= 1.00 \times 10^5$ g:

$$\Delta T = \frac{1.63 \times 10^6 \text{ J}}{4.18 \times 1.00 \times 10^5} = 3.90 \text{ °C}, \qquad T_{\max} = 25.0 + 3.90 = 28.9 \text{ °C}.$$

ii. [1 mark]. Not all the energy reaches the water: some is lost to the surroundings (air, the tank and pipes) as heat.

From the report. In b.i (30% full marks), H₂O(g) was wrong at SLC and $\Delta H$ had to be negative and match the equation. In b.ii (37% full marks), many tried to use the 1 kg of waste instead of the biogas volume. Part c.i was assessed for significant figures (three were needed), and some wrongly brought the density of water into it.

Question 2 (7 marks)

A rechargeable Mg-Na hybrid cell has a magnesium electrode, a sodium ion electrode and a hybrid organic/salt electrolyte, X. During recharge: 2NaX + Mg²⁺ → Mg + 2Na⁺ + 2X.

a. i

Give the polarity of the Mg electrode while the cell is discharging. (1 mark) ii. Write the half-equation at the Mg electrode during discharge. (1 mark)

b

Give two potential safety hazards of using this cell in a pacemaker. (2 marks)

c

Explain how Mg can be produced from MgCl₂ in an electrolytic cell. (3 marks)

Study design: all parts are in the current course (secondary cells and commercial electrolysis).

Show worked solution

a. i. [1 mark]

Negative (−). Recharge reduces Mg²⁺ to Mg, so discharge does the reverse: Mg is oxidised, making it the anode, which is negative in a galvanic cell.

ii. [1 mark]

Mg → Mg²⁺ + 2e⁻ (a one-way arrow; states were not required).

b. [2 marks]

Any two hazards linked to this cell, for example:

• the electrolyte or cell contents could leak and be toxic or corrosive in the body;
• magnesium is a reactive metal and could react with body fluids;
• leaking Mg²⁺ or Na⁺ could upset the body's natural ion balance;
• the cell could overheat.

c. [3 marks].

• The MgCl₂ must be molten (heated until it melts). In aqueous solution, water ($-0.83$ V) is a stronger oxidant than Mg²⁺ ($-2.37$ V), so hydrogen, not magnesium, would form at the cathode.
• A DC power supply forces the non-spontaneous reaction, at least $1.36 - (-2.37) = 3.73$ V under standard conditions, using inert electrodes.
• Cathode (negative): Mg²⁺(l) + 2e⁻ → Mg(l); anode (positive): 2Cl⁻(l) → Cl₂(g) + 2e⁻. The Mg and Cl₂ must be kept apart so they do not react again.

From the report. Part b (9% full marks) needed hazards specific to this Mg-Na cell, not general ones. In c (7% full marks), many described the electrolysis of an aqueous solution and missed the need for molten MgCl₂.

Question 3 (12 marks)

Quinoa contains all nine essential amino acids.

a

What is an essential amino acid? (1 mark)

b

For the dipeptide shown: i. Circle the peptide bond. (1 mark) ii. Name the amino acid with the non-polar Z group. (1 mark)

c

Explain why eating papaya with a protein meal can aid protein digestion. (2 marks)

d

Quinoa fat contains oleic and linoleic acids. i. What type of omega fatty acid is linoleic acid? (1 mark) ii. Which has the lower melting point? Justify. (3 marks)

e

Give two similarities and one difference between the structures of amylose and cellulose. (3 marks)

Study design: a and d.i are outside the current course (essential amino acids and omega fatty acids are no longer listed). In e, cellulose is no longer named, although condensation polymers of glucose (starch) are. Parts b, c and d.ii are still in.

Show worked solution

a. [1 mark]

An amino acid that the body cannot make, so it must come from the diet. (Both ideas are needed.)

b. i. [1 mark]

The amide link between the two amino acid residues: the C(=O)-N(H) group (circling the C-N bond, or the C and N atoms, was also accepted).

ii. [1 mark]

Isoleucine. Its Z group, -CH(CH₃)CH₂CH₃, is a hydrocarbon. (The other residue has a -(CH₂)₄NH₂ side chain: lysine, which is polar.)

c. [2 marks]

Papaya contains an enzyme (a protease) that catalyses the hydrolysis of the peptide bonds in proteins, breaking them into smaller peptides and amino acids that can be absorbed.

d. i. [1 mark]

Omega-6. (From the Data Book, linoleic acid is CH₃(CH₂)₄(CH=CHCH₂)₂(CH₂)₆COOH: the first C=C starts at the sixth carbon from the CH₃ end.)

ii. [3 marks]

Linoleic acid.

• Linoleic acid has two C=C double bonds per molecule; oleic acid has one.
• Each cis double bond puts a kink in the chain, so linoleic acid molecules pack less closely together.
• The dispersion forces between linoleic acid molecules are therefore weaker, and less energy is needed to melt it.

e. [3 marks].

• Similarities (any two): both are condensation polymers of glucose; both have glucose units joined by glycosidic (ether) links; both are unbranched (linear) chains; both have the same number of -OH groups per unit.
• Difference: the glucose units are oriented differently. In amylose (α-glucose) all the units face the same way, giving a coiled chain; in cellulose (β-glucose) alternate units are flipped, giving straight chains that hydrogen bond to each other.

From the report. In b.i, many circled other groups while trying to find the side chains. In c, the key idea was enzyme-catalysed hydrolysis; saying acid denatures the protein missed the point. In d.ii (36% full marks), answers had to compare the number of C=C bonds specifically; "number of double bonds" alone, or naming linolenic acid, lost marks. In e, the structures of both polymers are in the Data Book.

Question 4 (9 marks)

a. What is a fuel cell? (2 marks)

An ethanol fuel cell with an acidic electrolyte produces CO₂. The diagram shows the anode, a membrane and the cathode.

b. i

Name the species that crosses the membrane. (1 mark) ii. Show its direction of flow. (1 mark)

c

Write the anode half-equation. (1 mark)

d

At 25 °C and 100% efficiency, how much electrical energy could 1.0 g of ethanol produce? (1 mark)

e

Identify two aspects of electrode design that improve fuel cell efficiency. (2 marks)

f

How can the environmental impact of this fuel cell be minimised? (1 mark)

Study design: all parts are in the current course.

Show worked solution

a. [2 marks]

A galvanic cell that converts chemical energy directly into electrical energy, with a continuous supply of reactants (fuel and oxidant) fed in from outside.

b. i. [1 mark]

H⁺ ions (protons, or H₃O⁺).

ii. [1 mark]

From the anode to the cathode (left to right in the diagram): H⁺ is made at the anode and used at the cathode.

c. [1 mark]

$$\text{C}_2\text{H}_5\text{OH}(l) + 3\text{H}_2\text{O}(l) \rightarrow 2\text{CO}_2(g) + 12\text{H}^+(aq) + 12e^-$$

(The cathode reaction is O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l).)

d. [1 mark]

At 100% efficiency all the chemical energy becomes electrical energy. From the Data Book, ethanol releases 29.6 kJ g⁻¹, so 1.0 g gives $29.6 \approx 30$ kJ.

e. [2 marks]

Use porous electrodes with a large surface area, and electrodes that are, or contain, a catalyst.

f. [1 mark]

For example, use bioethanol from a renewable source, so the CO₂ released is balanced by the CO₂ absorbed as the crops grow. (Recycling electrode materials, managing the acidic electrolyte at end of life, or capturing the CO₂ were also accepted.)

From the report. Only 33% named H⁺ in b.i, and the arrow in b.ii had to be inside the box. Part c was answered correctly by 24%: students should be able to write half-equations for alkanol fuels, not only copy from the electrochemical series, and equilibrium arrows were not accepted. In d, many had the right number with the wrong unit.

Question 5 (8 marks)

One 124 g serving of sweet potato has 2.0 g protein, 3.0 g fat and 18.7 g carbohydrate, plus small amounts of vitamins C and D.

a

Calculate the energy in one serving. (1 mark)

b

Explain how boiling in water may affect its vitamin D level. (3 marks)

c

Vitamin C is titrated with iodine: C₆H₈O₆ + I₂ → 2HI + C₆H₆O₆. The average concordant titre of 0.0500 M I₂ is 21.81 mL. Calculate the mass of vitamin C. (2 marks)

d

Explain the role of coenzymes in metabolic reactions. (2 marks)

Study design: a and c are in the current course (food energy and redox titrations). Parts b (vitamin solubility) and d (coenzymes) are outside it.

Show worked solution

a. [1 mark]. Using the Data Book values (17, 37 and 16 kJ g⁻¹):

$$E = 2.0(17) + 3.0(37) + 18.7(16) = 34 + 111 + 299 = 444 \text{ kJ} \approx 4.4 \times 10^2 \text{ kJ}.$$

b. [3 marks]. Boiling should have little effect on the vitamin D level. Vitamin D is mostly non-polar (a large hydrocarbon structure with one -OH), so it is fat-soluble, not water-soluble. It will not dissolve into the polar cooking water, so it stays in the sweet potato.

c. [2 marks]. The mole ratio is 1 : 1.

$$n(\text{I}_2) = 0.0500 \times 0.02181 = 1.09 \times 10^{-3} \text{ mol} = n(\text{C}_6\text{H}_8\text{O}_6)$$

$$M(\text{C}_6\text{H}_8\text{O}_6) = 176.0 \text{ g mol}^{-1}, \qquad m = 1.09 \times 10^{-3} \times 176.0 = 0.192 \text{ g}.$$

d. [2 marks]. Coenzymes are small organic molecules (often made from vitamins) that bind to an enzyme, changing the active site so the substrate can bind. They also act as carriers, transferring electrons, atoms or groups of atoms between reactions.

From the report. In a, the common error was the unit (kJ g⁻¹ or kJ mol⁻¹ instead of kJ). Part b was weak (6% full marks): "denaturation" of the vitamin is not relevant, because vitamin D has no secondary or tertiary structure.

Question 6 (9 marks)

A reaction pathway: 1-bromopentane → pentan-1-ol → Compound S.

a

Draw an isomer of 1-bromopentane that contains a chiral carbon and circle it. (2 marks)

b. i

Write a balanced equation for making pentan-1-ol from 1-bromopentane and a sodium salt. (2 marks) ii. Calculate the atom economy. (3 marks)

c

Pentan-1-ol is fully oxidised to Compound S. Name S. (1 mark)

d

Draw the skeletal formula of pentanamide. (1 mark)

Study design: all parts are in the current course.

Show worked solution

a. [2 marks]. Any of these three isomers (the chiral carbon, bonded to four different groups, is the one to circle):

• 2-bromopentane, CH₃CHBrCH₂CH₂CH₃: C2 carries H, Br, CH₃ and CH₂CH₂CH₃;
• 2-bromo-3-methylbutane, CH₃CHBrCH(CH₃)₂: C2 carries H, Br, CH₃ and CH(CH₃)₂;
• 1-bromo-2-methylbutane, BrCH₂CH(CH₃)CH₂CH₃: C2 carries H, CH₃, CH₂Br and CH₂CH₃.

b. i. [2 marks]. Substitution with sodium hydroxide:

$$\text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH} + \text{NaBr}$$

ii. [3 marks].

$$\%\text{ atom economy} = \frac{M(\text{C}_5\text{H}_{11}\text{OH})}{M(\text{C}_5\text{H}_{11}\text{Br}) + M(\text{NaOH})} \times 100 = \frac{88.0}{150.9 + 40.0} \times 100 = 46.1\%.$$

c. [1 mark]. Pentanoic acid (CH₃CH₂CH₂CH₂COOH), using, for example, acidified K₂Cr₂O₇ or KMnO₄.

d. [1 mark]. A four-carbon zigzag ending in the amide carbon, which has a C=O and an NH₂:

From the report. In a, many drew 1-bromopentane itself. In b.i the organic formulas and a balanced equation including NaBr were both needed. Part d was answered correctly by only 22%, who struggled to show the amide group in skeletal form.

Question 7 (8 marks)

An unknown is one of five compounds: P (propan-2-yl ethanoate), Q (3-methylbutanoic acid), R (3-hydroxy-3-methylbutan-2-one), S (2-methoxy-2-methylpropanal) and T (3-hydroxypentan-2-one). Its ¹³C NMR spectrum has four peaks (one above 200 ppm), its IR spectrum has a strong broad band near 3450 cm⁻¹ and a strong sharp peak near 1700 cm⁻¹, and its mass spectrum has a base peak near m/z 59 and a peak at 43.

a

Which compound is eliminated by the number of ¹³C peaks? (1 mark)

b

Which compounds does the IR spectrum eliminate? Justify with data. (3 marks)

c. i

Give the formula of the species at m/z = 43. (1 mark) ii. Define m/z. (1 mark) iii. Explain why one molecule gives multiple peaks. (2 marks)

Study design: all parts are in the current course.

Show worked solution

a. [1 mark]. T. Its five carbons are all in different environments (five peaks), but the spectrum shows four. P, Q, R and S each have four carbon environments.

b. [3 marks]. P, Q and S are eliminated.

• The broad, strong absorption near 3200 to 3550 cm⁻¹ is an alcohol O-H. P (an ester) and S (an ether aldehyde) have no O-H group.
• Q is a carboxylic acid, which would show a very broad acid O-H band from 2500 to 3300 cm⁻¹ overlapping the C-H peaks. That band is absent.

That leaves R, which fits every spectrum: a ketone C=O near 1700 cm⁻¹ and an alcohol O-H in the IR, and four ¹³C environments, including the C=O above 200 ppm.

c. i. [1 mark]

[CH₃CO]⁺ (C₂H₃O⁺). [C₃H₇]⁺ was also accepted. The positive charge is essential.

ii. [1 mark]

The mass-to-charge ratio of the ion.

iii. [2 marks]

In the mass spectrometer the molecule is ionised to a positive molecular ion. This ion is unstable and fragments into smaller positive ions, each giving its own peak; ions containing different isotopes also appear at different m/z values.

From the report. In b (11% full marks), many eliminated P and S but did not say the O-H was an alcohol O-H, or missed that Q's acid would give a broad band from 2500 to 3500 cm⁻¹. In c.i the positive charge was often left off, and only 31% defined m/z correctly.

Question 8 (9 marks)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), $\Delta H = -197$ kJ mol⁻¹.

a

1.00 mol SO₂ and 1.00 mol O₂ are placed in an evacuated 3.00 L container at 100 °C. At equilibrium there is 20.0 g of SO₃. Calculate $K_c$ at 100 °C. (4 marks)

b

On the Maxwell-Boltzmann distribution for SO₃, draw the curve at a significantly lower temperature. (2 marks)

c

The volume is doubled. Describe the effect on [SO₂] from just before the change until equilibrium is re-established. (3 marks)

Study design: a and c are in the current course. Maxwell-Boltzmann curves (b) are not named in the current study design; collision theory and activation energy are, and the curve is still a common way to explain them.

Show worked solution

a. [4 marks]. $n(\text{SO}_3) = \dfrac{20.0}{80.1} = 0.250$ mol at equilibrium.

	SO₂	O₂	SO₃
Initial (mol)	1.00	1.00	0
Change (mol)	$-0.250$	$-0.125$	$+0.250$
Equilibrium (mol)	0.750	0.875	0.250
Concentration (M, ÷ 3.00)	0.250	0.292	0.0833

$$K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{0.0833^2}{0.250^2 \times 0.292} = 0.380 \text{ M}^{-1}.$$

b. [2 marks]. At a lower temperature the peak is higher and further left (lower kinetic energy), the tail at high energy is lower, and the area under the curve (the number of particles) is the same.

c. [3 marks].

• The instant the volume doubles, [SO₂] halves.
• Lower pressure favours the side with more gas particles (3 on the left against 2 on the right), so the equilibrium shifts left and [SO₂] gradually increases.
• At the new equilibrium, [SO₂] is higher than just after the change but still lower than before the change (the shift only partially opposes it).

From the report. In a (31% full marks), $K_c$ needed its unit, M⁻¹. In b, the new peak had to be higher and to the left, with roughly the same area. In c (16% full marks), many explained the cause of the shift instead of describing what happens to [SO₂].

Question 9 (11 marks)

Kim studies how storage temperature affects the rate at which aspartame in lemonade breaks down to DKP, using HPLC. She dilutes pure aspartame and pure DKP standards (1000 mg L⁻¹, diluted 10.00 mL to 1.000 L) to find their retention times, then stores unopened cans at 15, 25 and 35 °C and analyses a diluted sample from one can at each temperature each day for three days. In the lemonade chromatogram the two peaks overlap near 6 minutes.

a

Explain why aspartame is added to lemonade. (2 marks)

b. i

What is the dependent variable? (1 mark) ii. What extra steps are needed to use the HPLC data to measure it? (3 marks)

c. i

State a change to the HPLC operating conditions that would reduce errors. (1 mark) ii. How would this reduce the errors? (1 mark)

d

Write a conclusion from her table of results. (1 mark)

e. i

Identify a variable that was not controlled. (1 mark) ii. Explain how it affects validity. (1 mark)

Study design: a (artificial sweeteners) and c (HPLC operating conditions, now explicitly excluded) are outside the current course. Parts b, d and e are still in (calibration curves and investigation skills).

Show worked solution

a. [2 marks]

Aspartame is far sweeter than sugar (sucrose), so only a tiny amount is needed for the same sweetness. The drink therefore provides much less energy (fewer kilojoules) than one sweetened with sugar.

b. i. [1 mark]

The rate of production of DKP (the change in DKP concentration over time).

ii. [3 marks]

• Prepare a series of DKP standard solutions of known concentrations.
• Run them through the HPLC under the same conditions and plot a calibration curve of peak area against concentration.
• Use the DKP peak area of each lemonade sample and the calibration curve to find its DKP concentration at each time (then allow for the 1 in 100 dilution).

c. i. [1 mark]

For example, use a longer column, change the mobile phase or stationary phase, or change the column temperature.

ii. [1 mark]

The aspartame and DKP peaks overlap. The change improves their separation (resolution), so each peak area, and hence each concentration, can be measured accurately.

d. [1 mark]

In general, the higher the storage temperature, the faster aspartame breaks down and DKP forms. (At 15 °C there is almost no change; at 35 °C aspartame falls most, with the 25 °C day-three value an outlier.)

e. i. [1 mark]

For example: the starting aspartame and DKP concentrations of each can were not measured (only one can was tested at the start); or the cans may have come from different batches or been of different ages.

ii. [1 mark]

If the cans did not start with the same concentrations, the concentrations measured later cannot be put down to storage temperature alone, so the results do not validly measure the effect of temperature on DKP production.

From the report. Part b.ii was very weak (6% full marks, 73% scored zero): using a calibration curve made under the same conditions is a core instrumental-analysis skill. In c.i (11% correct), "repeat more trials" is not a change to operating conditions.

Question 10 (8 marks)

A source says carbohydrate is an indispensable energy source for high-intensity exercise.

a. Starch is about 70% amylopectin and 30% amylose. Discuss why a starchy meal several hours before a run could help a runner, referring to the glycaemic indexes of amylopectin and amylose and to cellular respiration. (3 marks)

b. Tien breathes easily in a park but is short of breath in heavy traffic, where CO is 50 to 100 ppm. Use Le Chatelier's principle and competing equilibria involving CO to discuss this. (5 marks)

Study design: cellular respiration (a) and Le Chatelier's principle (b) are in the current course, but the glycaemic index and the carbon monoxide poisoning context are no longer listed, so treat this question as optional practice.

Show worked solution

a. [3 marks].

• Amylopectin is branched, with many exposed chain ends, so enzymes hydrolyse it to glucose quickly: it has a high GI and gives a fast rise in blood glucose. Amylose is an unbranched, tightly coiled chain that is hydrolysed slowly: it has a lower GI and releases glucose gradually.
• Cellular respiration, C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), releases the energy the muscles need.
• Eating the meal hours beforehand means the amylopectin tops up glucose (and stored glycogen) before the run, while the slowly digested amylose keeps supplying glucose for respiration during the run.

b. [5 marks]. Two equilibria compete for haemoglobin (Hb):

$$\text{Hb}(aq) + \text{O}_2(aq) \rightleftharpoons \text{HbO}_2(aq) \quad (1) \qquad \text{Hb}(aq) + \text{CO}(aq) \rightleftharpoons \text{HbCO}(aq) \quad (2)$$

• CO binds to haemoglobin far more strongly than O₂ (equilibrium 2 has a much larger $K$).
• In heavy traffic, the higher [CO] shifts equilibrium 2 to the right, using up Hb. The lower [Hb] shifts equilibrium 1 to the left, so less HbO₂ forms and less oxygen is carried to the cells: Tien is short of breath.
• In the park, [CO] is very low, so equilibrium 2 shifts left, releasing Hb from HbCO. Equilibrium 1 shifts right, more HbO₂ forms and oxygen transport recovers: breathing is easier.
• Both reactions are reversible, which is why the effect wears off once Tien moves away from the traffic.

From the report. In a, many did not mention cellular respiration, or did not link the answer back to the runner; both GIs had to be discussed. Part b was the hardest on the paper (53% scored zero, 5% full marks): the top marks needed both equilibria, their reversibility and an explanation of breathing in both places.

General advice from the 2021 report

• Know the study design and the Data Book. Several questions came straight from key knowledge, and the polymer structures needed in Q3e were in the Data Book.
• Give the unit the question asks for, or a correct unit if none is specified; many correct numbers lost marks for kJ g⁻¹ against kJ, and $K_c$ must be quoted with its unit.
• Answer every part of a descriptive question, for example both glycaemic indexes in Q10a, and describe the effect when the question asks for it (Q8c) rather than the cause.
• Write half-equations yourself for less familiar reactions such as fuel-cell oxidation of ethanol, with one-way arrows when a process goes in one direction.