OCR A Level Chemistry A: Alkanes (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Alkanes, section 4.1.2 of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, is the focus of this set. Every outcome of this Module 4 (Core organic chemistry) section, 4.1.2(a) to (g), is tested. Module 4 appears in two of the examined components, Unified chemistry (03) and Synthesis and analytical techniques (02). The last two questions are longer, mixed-outcome problems.

Work through the Alkanes study guide and the Alkanes revision notes first. The OCR chemistry hub lists other topics, the printable checklist tracks outcomes, and the free diagnostics give a wider check.

Any data you need is supplied in each question.

Questions

1. Ethane, CH₃CH₃, is the second member of the alkane homologous series.

(a) State what is meant by a saturated hydrocarbon. [1] (b) Describe how the orbitals overlap to form the C–C σ-bond in ethane. [1] (c) Explain why the two CH₃ groups can rotate relative to each other. [1]

2. This question is about pentane, CH₃CH₂CH₂CH₂CH₃.

(a) State the H–C–H bond angle at carbon 3. [1] (b) Explain, in terms of electron pairs, why the arrangement around carbon 3 has this angle. [2] (c) Describe how you would show the 3-D arrangement of the four bonds around carbon 1 in a drawing. [1]

3. Use these boiling points: nonane 151 °C; hexane 69 °C; 3-methylpentane 63 °C; 2,2-dimethylbutane 50 °C.

(a) Explain why nonane has a higher boiling point than hexane. [3] (b) Explain the order of boiling points of the three isomers of C₆H₁₄. [3]

4. Heptane does not react with dilute hydrochloric acid or aqueous sodium hydroxide at room temperature. Explain why alkanes such as heptane react with so few reagents. [3]

5. Pentadecane, C₁₅H₃₂, is one of the alkanes in diesel fuel.

(a) Write an equation, with whole-number coefficients, for the complete combustion of pentadecane. [1] (b) Write an equation, with whole-number coefficients, for the incomplete combustion of pentadecane to carbon monoxide and water. [1] (c) Explain why carbon monoxide from a faulty engine or heater is dangerous. [2]

6. 30.0 cm³ of a gaseous alkane, X, needs exactly 195 cm³ of oxygen for complete combustion. All gas volumes share one temperature and pressure; water is liquid.

(a) Write a general equation for the complete combustion of an alkane CₙH₂ₙ₊₂. [1] (b) Deduce the molecular formula and name of X. [2] (c) State the volume of carbon dioxide formed. [1]

7. Methane reacts with bromine when the mixture is exposed to ultraviolet radiation. One product is bromomethane, CH₃Br.

(a) Name the type of reaction and mechanism. [1] (b) Write the equation for the initiation step and name the type of bond fission. [2] (c) Write the two propagation steps that form bromomethane. [2] (d) Give one termination step whose product is ethane. [1]

8. Pentane reacts with chlorine in UV radiation.

(a) Name the three structural isomers of C₅H₁₁Cl that can form. [3] (b) Some C₅H₁₀Cl₂ also forms. Write two propagation steps to show how C₅H₁₀Cl₂ is made from C₅H₁₁Cl. [2]

9. A student reacts 3.60 g of 2,2-dimethylpropane, C(CH₃)₄, with chlorine in UV radiation. After separation, 3.73 g of 1-chloro-2,2-dimethylpropane, C₅H₁₁Cl, is obtained. Use M(C₅H₁₂) = 72.0 g mol⁻¹ and M(C₅H₁₁Cl) = 106.5 g mol⁻¹.

(a) Explain why only one monochloro isomer can form from 2,2-dimethylpropane. [1] (b) Calculate the percentage yield of 1-chloro-2,2-dimethylpropane. [3] (c) Suggest two reasons, linked to the mechanism, why the yield is below 100%. [2]

10. A portable gas heater burns propane, C₃H₈. Use M(C₃H₈) = 44.0 g mol⁻¹ and M(CO) = 28.0 g mol⁻¹.

(a) Write the equation for the complete combustion of propane. [1] (b) Write the equation for the incomplete combustion of propane to carbon monoxide and water, with whole-number coefficients. [1] (c) In a poorly ventilated room, 0.660 g of propane burns entirely by the reaction in (b). Calculate the mass of carbon monoxide formed. [3] (d) Explain why a carbon monoxide detector is recommended in rooms with such heaters. [2]

11. 2-Methylbutane, CH₃CH(CH₃)CH₂CH₃, reacts with bromine in UV radiation.

(a) Deduce the number of different monobromo isomers that can form. Explain your answer. [2] (b) Give the two propagation steps leading to 2-bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃. [2] (c) A by-product with molar mass 142.0 g mol⁻¹ (Ar: C 12.0, H 1.0) is detected. Identify it, and write an equation to show how it forms. [2] (d) Evaluate radical bromination as a way to make pure 1-bromo-2-methylbutane. [2]

Answers

1. (a) It contains carbon and hydrogen only, and has only single carbon–carbon bonds [1] (b) The orbitals overlap directly between the two carbon nuclei, on the line joining them [1] (c) Rotation about the bond axis does not change the direct overlap, so the σ-bond stays intact [1] Examiner insight: In (b), “directly between the bonding nuclei” is the key idea; any mention of sideways overlap describes a π-bond instead.

2. (a) 109.5° [1] (b) Carbon 3 has four bonding pairs and no lone pairs [1]; they repel and get as far apart as possible: tetrahedral [1] (c) Draw two bonds as plain lines in the page, a bold wedge for the bond pointing out at the viewer and a hashed wedge for the bond pointing back [1] Examiner insight: Part (b) needs two separate ideas, the count of bonding pairs and the repulsion; “tetrahedral because it has four bonds” gives only one of them.

3. (a) Nonane has more electrons and a longer chain, so more surface contact between molecules [1]; this gives stronger London forces (induced dipole–dipole interactions) [1]; more energy is needed to overcome these intermolecular forces, so the boiling point is higher [1] (b) The isomers have the same number of electrons [1]; more branching makes molecules more compact, with fewer points of contact: hexane > 3-methylpentane > 2,2-dimethylbutane [1]; London forces weaken, so less energy is needed and boiling point falls in that order [1] Examiner insight: Name London forces and say they act between molecules; an answer about C–C bonds breaking earns nothing for the reasoning.

4. The C–C and C–H bonds have high bond enthalpies, so a lot of energy is needed to break them [1]. Carbon and hydrogen have very similar electronegativities, so the σ-bonds have very low polarity [1]. There are no δ+ or δ− sites to attract the H⁺ ions (electrophiles) or OH⁻ ions (nucleophiles) [1] Examiner insight: Bond enthalpy and polarity are two separate ideas; an answer built only on “strong bonds” covers just one of them, however well it is written.

5. (a) C₁₅H₃₂ + 23O₂ → 15CO₂ + 16H₂O (or doubled) [1] (b) 2C₁₅H₃₂ + 31O₂ → 30CO + 32H₂O [1] (c) CO is toxic: it binds to haemoglobin and reduces oxygen transport [1]. It is colourless and odourless, so it goes unnoticed [1] Examiner insight: In (b) one molecule of fuel needs 15½ O₂, which does not meet the whole-number instruction; double every coefficient, then recount the oxygen atoms on both sides.

6. (a) CₙH₂ₙ₊₂ + ((3n + 1)/2)O₂ → nCO₂ + (n + 1)H₂O [1] (b) Volume ratio O₂ : X = 195 ÷ 30.0 = 6.5, so (3n + 1)/2 = 6.5 and n = 4 [1]; X is C₄H₁₀, butane [1] (c) 4 × 30.0 = 120 cm³ [1] Examiner insight: At the same temperature and pressure, gas volumes are in the same ratio as moles; show the ratio 6.5 so the deduction is visible.

7. (a) Substitution by a radical mechanism: radical substitution [1] (b) Br₂ → 2Br• [1]; homolytic fission [1] (c) CH₄ + Br• → CH₃• + HBr [1]; CH₃• + Br₂ → CH₃Br + Br• [1] (d) CH₃• + CH₃• → C₂H₆ [1] Examiner insight: The specification requires a dot on every radical; CH₄ + Br → CH₃ + HBr does not show which species are radicals.

8. (a) 1-chloropentane [1]; 2-chloropentane [1]; 3-chloropentane [1] (b) C₅H₁₁Cl + Cl• → C₅H₁₀Cl• + HCl [1]; C₅H₁₀Cl• + Cl₂ → C₅H₁₀Cl₂ + Cl• [1] Examiner insight: “4-chloropentane” and “5-chloropentane” are the same compounds as 2- and 1-chloropentane numbered from the wrong end; listing them as extra isomers is incorrect.

9. (a) All twelve hydrogen atoms are on equivalent CH₃ groups attached to the same central carbon, so substitution of any one gives the same product [1] (b) n(C₅H₁₂) = 3.60 ÷ 72.0 = 0.0500 mol, so the theoretical n(C₅H₁₁Cl) = 0.0500 mol [1]; actual n = 3.73 ÷ 106.5 = 0.0350 mol [1]; yield = 0.0350 ÷ 0.0500 × 100 = 70.0% [1] (c) Further substitution turns some product into C₅H₁₀Cl₂ and more highly substituted compounds [1]; termination steps form other products such as C₁₀H₂₂ from two C₅H₁₁• radicals [1] Examiner insight: “Loss during transfer” is not linked to the mechanism; name the side reactions.

10. (a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [1] (b) 2C₃H₈ + 7O₂ → 6CO + 8H₂O [1] (c) n(C₃H₈) = 0.660 ÷ 44.0 = 0.0150 mol [1]; n(CO) = 3 × 0.0150 = 0.0450 mol [1]; mass = 0.0450 × 28.0 = 1.26 g [1] (d) CO binds to haemoglobin and cuts oxygen transport [1]; it is colourless and odourless, so only a detector warns of it [1] Examiner insight: In (c), each propane gives three CO; leaving out that ratio gives 0.42 g, so show the mole step on its own line.

11. (a) Four [1]. The two CH₃ groups on carbon 2 are equivalent; carbon 2 (CH), carbon 3 (CH₂) and carbon 4 (CH₃) are each different, so there are four distinct sets of hydrogen atoms [1] (b) CH₃CH(CH₃)CH₂CH₃ + Br• → (CH₃)₂C•CH₂CH₃ + HBr [1]; (CH₃)₂C•CH₂CH₃ + Br₂ → (CH₃)₂CBrCH₂CH₃ + Br• [1] (c) C₁₀H₂₂, formed when two C₅H₁₁• radicals combine [1]; for example 2(CH₃)₂C•CH₂CH₃ → (CH₃)₂C(CH₂CH₃)C(CH₂CH₃)(CH₃)₂ [1] (d) It gives a mixture of four monobromo isomers, plus dibromo compounds and C₁₀H₂₂ [1]; separation is needed and the yield of the wanted isomer is low, so it is a poor route [1] Examiner insight: “Evaluate” expects a judgement; describing the mixture without concluding on the method is incomplete.

Where marks are usually lost

• Describing σ overlap as sideways, or saying single bonds cannot rotate (question 1).
• Stating 109.5° without the four bonding pairs and their repulsion (question 2).
• Talking about bonds breaking on boiling, or giving branched isomers “fewer electrons” (question 3).
• Giving only bond strength for low reactivity and forgetting polarity (question 4).
• Leaving fractional coefficients when whole numbers are asked for (questions 5 and 10).
• Missing radical dots, or writing H• rather than HBr in the first propagation step (question 7).
• Counting the same isomer twice by numbering from the other end of the chain (questions 8 and 11).
• Giving no judgement in an “evaluate” answer (question 11).

Next steps

• Reasoning chains and a self-test: Alkanes revision notes
• Worked examples for every outcome: Alkanes study guide
• Short checks across the course: all free 10-minute diagnostics
• Other H432 topics: OCR chemistry course hub
• Record secure outcomes: printable OCR checklist
• Mole calculations: Amount of substance practice
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.2 Alkanes, learning outcomes (a) to (g). Published by OCR.