O Level Chemistry Organic Chemistry Quiz

O-Level Chemistry Quiz - Organic Chemistry — Answer Key and Marking Scheme

Total Marks: 50

---

Section A: Multiple Choice (10 marks)

Question	Answer	Explanation
1	B	Alkanes are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. Option A is alkenes; C is alkynes; D is alcohols.
2	B	A homologous series is a family of organic compounds with the same functional group, similar chemical properties, and a gradual trend in physical properties. Each member differs by a –CH₂– unit.
3	B	Ethene (C₂H₄) reacts with bromine (Br₂) by addition across the C=C double bond to form C₂H₄Br₂. The double bond opens up to accommodate the bromine atoms.
4	C	Bioethanol is produced by fermentation of sugars from plants (e.g., sugarcane, corn), making it renewable. Natural gas, crude oil, and coal are fossil fuels — non-renewable resources.
5	D	Carboxylic acids contain the carboxyl functional group –COOH. Option A is alcohols; B is esters; C is aldehydes.

Marking: 1 mark per correct answer. Total = 5 × 1 = 5 marks.

---

Section B: Short Answer (20 marks)

Question 6

(a) Explain why crude oil can be separated by fractional distillation. [2 marks]

Answer: Crude oil is a mixture of hydrocarbons with different boiling points. During fractional distillation, the crude oil is heated and the components vaporise at different temperatures. As the vapours rise up the fractionating column, they cool and condense at different levels according to their boiling points. Components with lower boiling points are collected near the top, while those with higher boiling points are collected near the bottom.

Marking:

• [1] Mention that components have different boiling points.
• [1] Describe the separation process (vaporisation and condensation at different levels) OR mention that fractions are collected based on boiling point ranges.

(b) State the relationship between boiling point and number of carbon atoms. Explain. [2 marks]

Answer: As the number of carbon atoms in the hydrocarbon molecules increases, the boiling point increases. This is because larger molecules have stronger intermolecular forces of attraction (van der Waals forces) between them. More energy is required to overcome these stronger forces, resulting in a higher boiling point.

Marking:

• [1] Correct relationship stated (more carbon atoms → higher boiling point).
• [1] Explanation in terms of stronger intermolecular forces requiring more energy to overcome.

---

Question 7

(a) Draw the full structural formula of butane, showing all atoms and bonds. [2 marks]

Answer:

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

Or: CH₃–CH₂–CH₂–CH₃ (condensed structural formula acceptable if all bonds are shown).

Marking:

• [1] Correct carbon skeleton with four carbon atoms in a straight chain.
• [1] All hydrogen atoms shown with correct number (10 H atoms) and all bonds correctly drawn.

(b) Chemical test to distinguish butane and but-1-ene. [3 marks]

Answer: Add a few drops of aqueous bromine (orange/brown) to each compound and shake.

• Butane: The orange/brown colour of bromine remains (no reaction occurs).
• But-1-ene: The orange/brown colour of bromine is decolourised (turns colourless).

But-1-ene contains a C=C double bond and undergoes an addition reaction with bromine. Butane is saturated (only C–C single bonds) and does not react with bromine under these conditions.

Marking:

• [1] Correct reagent: aqueous bromine (or bromine water).
• [1] Observation for butane (no colour change / remains orange).
• [1] Observation for but-1-ene (decolourises / turns colourless) AND explanation referencing the C=C double bond / unsaturation.

---

Question 8

(a) Calculate the volume of CO₂ produced. [3 marks]

Answer: Mr of C₂H₅OH = (2 × 12) + (6 × 1) + 16 = 46

Moles of ethanol = mass / Mr = 9.2 / 46 = 0.20 mol

From the equation: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O Mole ratio: 1 mol C₂H₅OH produces 2 mol CO₂

Moles of CO₂ = 0.20 × 2 = 0.40 mol

Volume of CO₂ at r.t.p. = moles × 24 = 0.40 × 24 = 9.6 dm³

Marking:

• [1] Correct calculation of Mr and moles of ethanol (0.20 mol).
• [1] Correct use of mole ratio to find moles of CO₂ (0.40 mol).
• [1] Correct volume calculation with correct unit (9.6 dm³). Accept 9600 cm³.

(b) Environmental advantage of ethanol as fuel. [1 mark]

Answer: Ethanol is produced from renewable sources (plants/crops) by fermentation, so it is a renewable fuel / carbon-neutral (CO₂ released during combustion is offset by CO₂ absorbed during plant growth). Petrol is derived from crude oil, a non-renewable fossil fuel.

Marking:

• [1] Any valid environmental advantage: renewable, carbon-neutral, reduces dependence on fossil fuels, biodegradable.

---

Question 9

(a) Draw the repeating unit of poly(ethene). [2 marks]

Answer:

    H   H
    |   |
— ( C — C ) —
    |   |
    H   H

Or showing the continuation bonds clearly on both sides of the repeating unit.

Marking:

• [1] Correct structure showing two carbon atoms in the backbone with single bonds.
• [1] Correct number of hydrogen atoms (4 H) and continuation bonds shown on both sides.

(b) Explain why poly(ethene) is non-biodegradable and state one environmental problem. [2 marks]

Answer: Poly(ethene) is non-biodegradable because it consists of long-chain hydrocarbon molecules with strong C–C and C–H covalent bonds that microorganisms (bacteria and fungi) cannot break down. There are no natural enzymes that can degrade these synthetic polymers.

Environmental problems include:

• Accumulation in landfills (takes hundreds of years to decompose).
• Pollution of oceans and waterways, harming marine life.
• Release of toxic gases when burned.
• Visual pollution/litter.

Marking:

• [1] Explanation: strong C–C/C–H bonds cannot be broken down by microorganisms / no natural enzymes can degrade it.
• [1] Any valid environmental problem stated.

---

Question 10

(a) Balanced equation for ethanoic acid + sodium hydroxide. [2 marks]

Answer: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

Marking:

• [1] Correct formulae for reactants and products (CH₃COOH, NaOH, CH₃COONa, H₂O).
• [1] Correctly balanced with state symbols. Accept CH₃CO₂H for ethanoic acid.

(b) Explain "weak acid". [1 mark]

Answer: A weak acid is one that only partially ionises/dissociates in water to form H⁺ ions. In ethanoic acid, only a small fraction of the acid molecules release H⁺ ions; most molecules remain undissociated. The ionisation is reversible and reaches an equilibrium: CH₃COOH ⇌ CH₃COO⁻ + H⁺.

Marking:

• [1] Partial ionisation/dissociation in water OR only a small fraction of molecules produce H⁺ ions.

---

Section C: Structured Questions (20 marks)

Question 11

(a) Name Compound A and type of reaction. [2 marks]

Answer: Compound A is ethanol. The reaction is addition (specifically, hydration of ethene / addition of steam across the C=C double bond).

Marking:

• [1] Correct name: ethanol.
• [1] Correct reaction type: addition (accept hydration).

(b) Name Compound B and balanced equation. [2 marks]

Answer: Compound B is ethane.
Equation: C₂H₄(g) + H₂(g) → C₂H₆(g)

Marking:

• [1] Correct name: ethane.
• [1] Correct balanced equation with state symbols (state symbols not essential if not specified, but good practice).

(c) Name Compound D and type of reaction. [2 marks]

Answer: Compound D is ethyl ethanoate. The reaction is esterification (or condensation).

Marking:

• [1] Correct name: ethyl ethanoate.
• [1] Correct reaction type: esterification / condensation.

(d) Commercial use of compounds like Compound D. [1 mark]

Answer: Compounds like ethyl ethanoate (esters) are used as:

• Flavourings in food
• Fragrances in perfumes
• Solvents (e.g., in nail polish remover, glues)

Marking:

• [1] Any valid commercial use of esters.

---

Question 12

(a) Explain why hex-1-ene decolourises aqueous bromine but hexane does not. [3 marks]

Answer: Hex-1-ene contains a carbon-carbon double bond (C=C), which makes it unsaturated. It undergoes an addition reaction with bromine, where the bromine molecule adds across the double bond, causing the orange colour to disappear. Hexane is a saturated hydrocarbon with only carbon-carbon single bonds (C–C), so it does not react with bromine under these conditions; the bromine remains orange.

Marking:

• [1] Hex-1-ene has C=C double bond / is unsaturated; hexane has only C–C single bonds / is saturated.
• [1] Reaction type named: addition reaction.
• [1] Explanation that bromine adds across the double bond, removing the colour; hexane cannot undergo addition.

(b) Name the gas produced when ethanol reacts with sodium. [1 mark]

Answer: Hydrogen gas (H₂).

Marking:

• [1] Hydrogen.

(c) Balanced equation for ethanol and sodium. [2 marks]

Answer: 2C₂H₅OH(l) + 2Na(s) → 2C₂H₅ONa(aq) + H₂(g)

Marking:

• [1] Correct reactants and products (sodium ethoxide and hydrogen).
• [1] Correctly balanced equation.

(d) Why ethanol reacts with sodium but hexane does not. [2 marks]

Answer: Ethanol contains a hydroxyl group (–OH) with a polar O–H bond. The hydrogen atom in this –OH group is reactive and can be displaced by sodium. Hexane contains only C–H bonds, which are non-polar and much stronger; the hydrogen atoms in hexane are not reactive enough to be displaced by sodium.

Marking:

• [1] Ethanol has an –OH group with a reactive hydrogen atom.
• [1] Hexane has only strong, non-polar C–H bonds; hydrogen atoms are not reactive with sodium.

---

Question 13

(a) Two conditions for catalytic cracking. [2 marks]

Answer:

• High temperature (approximately 500–700 °C).
• Presence of a catalyst (e.g., aluminium oxide, Al₂O₃, or silicon dioxide, SiO₂).

Marking:

• [1] High temperature.
• [1] Catalyst (aluminium oxide/silicon dioxide).

(b) Balanced equation for cracking decane to ethene. [2 marks]

Answer: C₁₀H₂₂ → C₂H₄ + C₈H₁₈

Marking:

• [1] Correct product: octane (C₈H₁₈).
• [1] Correctly balanced equation.

(c) Economic importance of cracking. [1 mark]

Answer: Cracking converts less useful, long-chain hydrocarbons (low demand) into more useful, short-chain hydrocarbons (high demand) such as petrol, alkenes (for polymers), and hydrogen. This maximises the value obtained from crude oil.

Marking:

• [1] Produces higher-demand products / matches supply with demand / produces alkenes for polymer industry.

---

Question 14

(a) Two types of compounds that form an ester. [2 marks]

Answer: A carboxylic acid and an alcohol.

Marking:

• [1] Carboxylic acid.
• [1] Alcohol.

(b) Balanced equation for methyl ethanoate formation. [2 marks]

Answer: CH₃OH + CH₃COOH ⇌ CH₃COOCH₃ + H₂O

Marking:

• [1] Correct reactants and products (methanol, ethanoic acid, methyl ethanoate, water).
• [1] Correctly balanced equation with reversible arrow.

(c) Reason for using a catalyst in esterification. [1 mark]

Answer: To speed up the rate of reaction / to increase the rate at which equilibrium is reached. (Concentrated sulfuric acid is commonly used as a catalyst.)

Marking:

• [1] Speeds up reaction / increases rate of reaction.

---

Question 15

(a) Balanced equation for complete combustion of propane. [2 marks]

Answer: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Marking:

• [1] Correct products: CO₂ and H₂O.
• [1] Correctly balanced equation.

(b) Why incomplete combustion is dangerous. [2 marks]

Answer: Incomplete combustion produces carbon monoxide (CO), a toxic gas that is colourless and odourless. When inhaled, it binds irreversibly to haemoglobin in red blood cells, preventing oxygen transport, which can lead to death. It also produces soot (carbon particles), which can cause respiratory problems and dirty surroundings.

Marking:

• [1] Produces carbon monoxide (CO).
• [1] CO is toxic / binds to haemoglobin / prevents oxygen transport (or mention soot/respiratory issues).

(c) Measure to ensure complete combustion. [1 mark]

Answer: Ensure an adequate supply of air/oxygen to the appliance (e.g., keep air vents open, regular servicing).

Marking:

• [1] Ensure sufficient oxygen/air supply.

---

Section D: Data-Based and Application Questions (10 marks)

Question 16

(a) Trend in boiling points. [1 mark]

Answer: As the number of carbon atoms (or molecular size) increases, the boiling point increases.

Marking:

• [1] Correct trend stated.

(b) Explain the trend. [2 marks]

Answer: As the molecular size (number of carbon atoms) increases, the number of electrons in the molecule increases. This leads to stronger intermolecular forces of attraction (van der Waals forces) between the molecules. More energy is required to overcome these stronger forces, resulting in higher boiling points.

Marking:

• [1] Larger molecules have more electrons / larger electron cloud.
• [1] Stronger van der Waals forces require more energy to overcome.

(c) Predict state of pentane at 25 °C. [2 marks]

Answer: Pentane will be a liquid at room temperature (25 °C). The trend shows that boiling point increases with chain length. Butane (C₄H₁₀) has a boiling point of -1 °C, and pentane (C₅H₁₂) is expected to have a boiling point higher than -1 °C, likely above 25 °C. Therefore, pentane will exist as a liquid at 25 °C.

Marking:

• [1] Prediction: liquid.
• [1] Explanation: boiling point of pentane is expected to be above 25 °C based on the trend.

---

Question 17

(a) Raw material for bioethanol by fermentation. [1 mark]

Answer: Sugars / glucose (from plants such as sugarcane, corn, or sugar beet).

Marking:

• [1] Sugar/glucose/carbohydrates from plants.

(b) Word equation for fermentation. [2 marks]

Answer: Glucose → Ethanol + Carbon dioxide
(C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂)

Marking:

• [1] Correct reactants and products in words.
• [1] Correctly written as a word equation (or balanced chemical equation).

(c) One advantage and one disadvantage of bioethanol as fuel. [2 marks]

Answer:

• Advantage: Renewable (produced from crops that can be regrown); carbon-neutral (CO₂ absorbed during growth offsets CO₂ released during combustion); reduces dependence on fossil fuels.
• Disadvantage: Requires large areas of land for growing crops, which can lead to deforestation and competition with food production; energy-intensive production process; lower energy density than petrol.

Marking:

• [1] Valid advantage.
• [1] Valid disadvantage.

---

Question 18

(a) Meaning of "addition polymerisation". [2 marks]

Answer: Addition polymerisation is a process in which many small unsaturated monomer molecules (containing C=C double bonds) join together to form a long-chain polymer, without the loss of any atoms or molecules. The double bonds open up and link the monomers together.

Marking:

• [1] Monomers with C=C bonds join together.
• [1] No other product formed / no loss of atoms.

(b) Monomer for poly(propene) and its structural formula. [2 marks]

Answer: The monomer is propene.
Structural formula:

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

Or: CH₂=CH–CH₃

Marking:

• [1] Correct name: propene.
• [1] Correct structural formula showing the C=C double bond.

(c) Way to reduce environmental impact of addition polymers. [1 mark]

Answer: Recycling; developing biodegradable polymers; incineration with energy recovery; reducing usage (e.g., using reusable bags).

Marking:

• [1] Any valid method.

---

Question 19

(a) Balanced equation using structural formulae. [2 marks]

Answer:

    H   O                    H   O
    |   ||                   |   ||
H — C — C — O — H  +  H — O — C — C — H  ⇌
    |                       |
    H                       H
(Ethanoic acid)         (Ethanol)

    H   O       H   H
    |   ||      |   |
H — C — C — O — C — C — H  +  H₂O
    |           |   |
    H           H   H
(Ethyl ethanoate)

Or: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (with structural formulae shown for organic compounds).

Marking:

• [1] Correct structural formulae for reactants.
• [1] Correct structural formula for ester product and water; reversible arrow.

(b) Name of ester and characteristic smell. [2 marks]

Answer: The ester is ethyl ethanoate. It has a characteristic sweet, fruity smell (often described as pear drops or nail polish remover).

Marking:

• [1] Correct name: ethyl ethanoate.
• [1] Sweet/fruity smell.

(c) Why it is a condensation reaction. [1 mark]

Answer: It is a condensation reaction because two molecules combine to form a larger molecule (ester) with the elimination of a small molecule (water).

Marking:

• [1] Water molecule is eliminated / small molecule removed.

---

Question 20

(a) Functional groups. [3 marks]

Answer:

• (i) Propan-1-ol: Hydroxyl group (–OH) / alcohol group.
• (ii) Propanal: Aldehyde group (–CHO) / carbonyl group at end of chain.
• (iii) Propanoic acid: Carboxyl group (–COOH).

Marking:

• [1] Each correct functional group (1 mark each, total 3 marks).

(b) Organic product from oxidation of propan-1-ol. [1 mark]

Answer: Propanoic acid (or propanal, depending on conditions; but with acidified potassium dichromate(VI) and reflux, propanoic acid is the final product).

Marking:

• [1] Propanoic acid (accept propanal if distillation is specified, but standard oxidation with excess oxidising agent yields propanoic acid).

(c) Colour change during oxidation. [1 mark]

Answer: The colour changes from orange to green. (Acidified potassium dichromate(VI) is orange; it is reduced to chromium(III) ions, which are green.)

Marking:

• [1] Orange to green.

---

END OF ANSWER KEY