Secondary 4 Pure Chemistry Organic Chemistry Quiz

Secondary 4 Pure Chemistry Quiz - Organic Chemistry — Answer Key

Total Marks: 40

Section A: Multiple Choice (10 marks)

1 mark each = 10 marks

Section B: Structured Questions (20 marks)

Question 11 (7 marks)

(a) Ethane. [1 mark]

(b) Full structural formula of ethene (C₂H₄):

    H   H
     \ /
      C=C
     / \
    H   H

[1 mark for correct structure showing all atoms and the double bond]

(c) Acidified potassium dichromate(VI) / acidified potassium manganate(VII) / oxygen with a catalyst. [1 mark for any correct oxidising agent]

(d)(i) Esterification / condensation reaction. [1 mark]

(d)(ii) Ethyl ethanoate. [1 mark]

(d)(iii) CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O [2 marks: 1 for correct reactants and products, 1 for correct balancing and reversible arrow]

Question 12 (5 marks)

(a) Crude oil contains hydrocarbons with different boiling points / different molecular sizes. In the fractionating column, the temperature decreases going up. Larger hydrocarbons with higher boiling points condense lower down; smaller hydrocarbons with lower boiling points rise higher before condensing. The fractions are collected at different levels. [2 marks: 1 for mentioning different boiling points, 1 for explaining separation based on condensation at different levels]

(b)(i) To produce shorter-chain hydrocarbons which are in higher demand (e.g., petrol) / to produce alkenes for the petrochemical industry / because demand for shorter-chain hydrocarbons exceeds supply from fractional distillation alone. [1 mark for any valid reason]

(b)(ii) C₁₀H₂₂ → C₂H₄ + C₈H₁₈ [2 marks: 1 for correct product formula (C₈H₁₈), 1 for balanced equation. Accept other valid cracking products that balance, e.g., C₁₀H₂₂ → 2C₂H₄ + C₆H₁₄, etc.]

Question 13 (8 marks)

(a) Addition polymerisation. [1 mark]

(b) Poly(ethene) is an addition polymer with long chains held together by weak intermolecular forces / van der Waals' forces. When heated, these weak forces are overcome and the chains can slide past each other, allowing the polymer to melt. Nylon is a condensation polymer with strong covalent bonds (amide linkages) between polymer chains. When heated strongly, the covalent bonds break and the polymer decomposes/chars rather than melting. [3 marks: 1 for describing poly(ethene) structure and weak intermolecular forces, 1 for describing nylon's cross-linking/amide bonds, 1 for linking to melting vs decomposition]

(c)(i) Water / H₂O. [1 mark]

(c)(ii) Repeat unit of nylon (from given monomers):

    H   H   H   H   H   H   O   H   H   H   H   O
    |   |   |   |   |   |   ||  |   |   |   |   ||
...-N-(CH₂)₆-N-C-(CH₂)₄-C-...
                    |

Or more simply showing the amide linkage: –NH–(CH₂)₆–NH–CO–(CH₂)₄–CO– [2 marks: 1 for correct connectivity, 1 for showing the amide (–CONH–) linkage clearly]

(c)(iii) Ropes / clothing / parachutes / fishing lines / toothbrush bristles. Nylon is strong, lightweight, and resistant to stretching/abrasion. [1 mark for a valid use with a linked property explanation]

Question 14 (10 marks)

(a) Liquid X is the alkene. Alkenes contain a C=C double bond which undergoes addition reaction with bromine, decolourising the orange bromine water. Liquid X is the only one that decolourised bromine water immediately. [2 marks: 1 for identifying X, 1 for explaining the decolourisation due to the C=C double bond]

(b) Liquid Y is the carboxylic acid. Carboxylic acids react with sodium carbonate to produce carbon dioxide gas (effervescence). Liquid Y is the only one that produced effervescence with sodium carbonate. [2 marks: 1 for identifying Y, 1 for explaining the reaction with carbonate producing CO₂]

(c)(i) Hydroxyl group / –OH (alcohol functional group). [1 mark]

(c)(ii) C₃H₈O as an alcohol that can be oxidised: propan-1-ol or propan-2-ol.

• Propan-1-ol: CH₃–CH₂–CH₂–OH (primary alcohol, oxidised to aldehyde then carboxylic acid)
• Propan-2-ol: CH₃–CH(OH)–CH₃ (secondary alcohol, oxidised to ketone)

Either is acceptable. [2 marks: 1 for correct structural formula, 1 for correct name matching the structure drawn]

(d) Liquid W belongs to the alkane homologous series. It does not react with bromine water (no C=C bond), does not react with sodium carbonate (no –COOH group), does not react with acidified potassium dichromate(VI) (not easily oxidised, so not an alcohol), and does not react with sodium metal (not an alcohol or carboxylic acid). Its molecular formula C₅H₁₂ fits the general formula CₙH₂ₙ₊₂ for alkanes. [2 marks: 1 for identifying alkane, 1 for reasoning based on lack of reactivity and/or molecular formula]

(e) Bromine water is toxic/corrosive and produces harmful vapour. The student should wear gloves and safety goggles, and carry out the test in a fume cupboard or well-ventilated area. [1 mark for stating a valid precaution with reason]

Section C: Extended Questions (10 marks)

Question 15 (10 marks)

(a) Saturated means the hydrocarbon contains only single bonds between carbon atoms / contains the maximum number of hydrogen atoms per carbon atom. [1 mark]

(b)(i) Substitution reaction / free radical substitution. [1 mark]

(b)(ii) CH₄ + Cl₂ → CH₃Cl + HCl [1 mark: must be balanced, UV light condition not required in equation but accept if written above arrow]

(c) Isomer 1: butane

    H   H   H   H
    |   |   |   |
H — C — C — C — C — H
    |   |   |   |
    H   H   H   H

Name: Butane

Isomer 2: 2-methylpropane (or methylpropane)

    H   H   H
    |   |   |
H — C — C — C — H
    |   |   |
    H   |   H
        C — H
       / \
      H   H

Name: 2-methylpropane / methylpropane

[4 marks: 1 for each correct structure, 1 for each correct name. Accept condensed structural formulae if all bonds are shown.]

(d) As the number of carbon atoms increases, the molecular size / relative molecular mass increases. The number of electrons in the molecule increases, leading to stronger intermolecular forces of attraction (van der Waals' forces / London dispersion forces) between molecules. More energy is required to overcome these stronger forces, so the boiling point increases. [3 marks: 1 for larger molecular size/Mr, 1 for stronger intermolecular forces, 1 for more energy needed to overcome forces]

Question 16 (5 marks)

(a) C₂H₄ + H₂O → C₂H₅OH [1 mark: balanced equation, accept reversible arrow]

(b) Catalyst: Phosphoric acid (H₃PO₄) [1 mark] Temperature: 300°C [1 mark]

(c) The hydration of ethene produces a continuous stream of ethanol (continuous process), whereas fermentation is a batch process that is slower. Hydration produces pure ethanol without the need for further purification by fractional distillation, which is required for the dilute ethanol solution from fermentation. The higher rate and purity make it more economically viable despite using non-renewable feedstock. [2 marks: 1 for continuous process/faster rate, 1 for higher purity/no need for distillation]

Question 17 (5 marks)

(a)(i) Methyl ethanoate. [1 mark]

(a)(ii) Structural formula of methyl ethanoate:

        H   O   H
        |   ||  |
H — C — C — O — C — H
        |       |
        H       H

Or: CH₃COOCH₃ [2 marks: 1 for correct connectivity, 1 for showing all atoms and bonds]

(b) Any two of:

• Volatile (evaporate easily to carry the smell)
• Have pleasant/fruity smells
• Non-toxic (safe for consumption in small amounts)
• Insoluble in water (for perfumes, so they don't wash off easily) [2 marks: 1 for each valid property with brief explanation]

Question 18 (7 marks)

(a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [2 marks: 1 for correct products, 1 for correct balancing]

(b) Carbon monoxide binds irreversibly with haemoglobin in red blood cells, forming carboxyhaemoglobin. This prevents haemoglobin from carrying oxygen around the body, leading to oxygen deprivation (suffocation) which can be fatal. [2 marks: 1 for binding to haemoglobin, 1 for preventing oxygen transport]

(c) Molar mass of propane (C₃H₈) = (3 × 12) + (8 × 1) = 44 g/mol Number of moles of propane = 0.44 g ÷ 44 g/mol = 0.01 mol From equation: 1 mol C₃H₈ produces 3 mol CO₂ Moles of CO₂ produced = 0.01 × 3 = 0.03 mol Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol Mass of CO₂ = 0.03 mol × 44 g/mol = 1.32 g

[3 marks: 1 for moles of propane, 1 for moles of CO₂, 1 for final mass with correct unit]

Question 19 (6 marks)

(a)(i) Repeat unit of PVC:

    H   H
    |   |
...— C — C —...
    |   |
    H   Cl

[2 marks: 1 for correct carbon backbone with single bonds, 1 for correct atoms/groups attached]

(a)(ii) Uses: window frames, pipes, electrical cable insulation, flooring, clothing (e.g., raincoats). Explanation: PVC is strong, durable, and resistant to weathering/chemicals (for pipes and window frames) OR PVC is flexible and waterproof (for raincoats and cable insulation). [2 marks: 1 for valid use, 1 for linking property to use]

(b) Addition polymers have long non-polar carbon-carbon backbones that are not easily broken down by microorganisms. Microorganisms lack enzymes that can hydrolyse or break the strong C–C covalent bonds in the polymer chains. Therefore, they persist in the environment for many years. [2 marks: 1 for non-polar/unreactive nature, 1 for microorganisms lacking enzymes to break C–C bonds]

Question 20 (6 marks)

(a) The boiling points increase as the number of carbon atoms increases / as the molecular formula goes from CH₄ to C₄H₁₀. [1 mark]

(b) As the number of carbon atoms increases, the molecular size / relative molecular mass increases. The number of electrons in the molecule increases, leading to stronger intermolecular forces of attraction (van der Waals' forces / London dispersion forces) between molecules. More energy is required to overcome these stronger forces, so the boiling point increases. [3 marks: 1 for larger molecular size/Mr, 1 for stronger intermolecular forces, 1 for more energy needed to overcome forces]

(c) Pentane (C₅H₁₂) will be a liquid at room temperature (25°C). The trend shows that boiling point increases with chain length; butane (C₄H₁₀) boils at -0.5°C, so pentane will have a boiling point above room temperature (approximately 36°C). Therefore, it will exist as a liquid at 25°C. [2 marks: 1 for predicting liquid, 1 for reasoning based on trend/extrapolation]

END OF ANSWER KEY

Marking Notes:

• Award marks for correct chemical principles even if terminology is slightly imprecise (e.g., "double bond opens" is acceptable for "addition reaction").
• In structural formula questions, all atoms and bonds must be shown clearly. Deduct ½ mark for missing hydrogen atoms.
• For explanation questions, look for cause-and-effect reasoning, not just restating observations.
• Accept alternative correct answers where indicated (e.g., alternative oxidising agents, alternative cracking products that balance).
• Spelling errors in chemical names should not be penalised if the intended compound is clear.