OCR A Level Chemistry A: Alkenes (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.

These questions practise section 4.1.3, Alkenes, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Together they test every learning outcome from 4.1.3(a) to (l) in Module 4, Core organic chemistry. H432 is the full A level. Questions 6, 8 and 11 are longer, mixed problems.

Before you start, work through the 4.1.3 study guide and the matching revision notes. See also the OCR chemistry hub, the printable checklist and the free diagnostics.

Data for this set: C = 12.0, H = 1.0, O = 16.0, F = 19.0, Cl = 35.5, Ca = 40.1. Molar gas volume = 24.0 dm³ mol⁻¹ at RTP.

Questions

1. (a) Define the term electrophile. [1] (b) State what happens to the bonding electrons when a covalent bond undergoes heterolytic fission. [1]

2. This question is about the C=C bond in alkenes.

(a) Describe how the σ-bond and the π-bond in C=C are formed. [2] (b) Explain why there is no rotation about a C=C bond. [1] (c) State the bond angle around each carbon of a C=C bond and explain this value. [2]

3. (a) State whether each compound has E/Z isomers. (i) 2-methylbut-2-ene, (CH₃)₂C=CHCH₃ (ii) 3-(chloromethyl)pent-2-ene, CH₃CH=C(CH₂Cl)CH₂CH₃ (iii) hept-3-ene, CH₃CH₂CH=CHCH₂CH₂CH₃ [3] (b) In one isomer of (ii), the CH₃ on C2 and the CH₂CH₃ are on the same side. Use CIP priority rules to decide whether this is the E or the Z isomer. Explain your answer. [3] (c) Explain why this compound cannot be described using cis–trans names. [1]

4. Octa-2,5-diene has the structural formula CH₃CH=CHCH₂CH=CHCH₂CH₃. Deduce the number of stereoisomers it has. Explain your answer. [3]

5. (a) Explain, in terms of bonding, why alkenes are much more reactive than alkanes. [2] (b) Use these average bond enthalpies, in kJ mol⁻¹, to estimate the enthalpy change when HBr adds to an alkene: C=C 612, H–Br 366, C–C 347, C–H 413, C–Br 290. [3]

6. 2,7-dimethyloct-4-ene, (CH₃)₂CHCH₂CH=CHCH₂CH(CH₃)₂, reacts with each reagent below. For each, give any catalyst needed and name the organic product.

(a) hydrogen [2] (b) chlorine [2] (c) steam [2] (d) hydrogen bromide [2]

7. (a) Describe how bromine water could show whether a liquid hydrocarbon, Z, is unsaturated. Give the result you would see in each case. [2] (b) A different hydrocarbon, C₈H₁₂, contains one ring. A 1.62 g sample reacts completely with 720 cm³ of hydrogen, measured at RTP. Deduce the number of C=C bonds in each molecule. [3] (c) Suggest why the test in (a) cannot distinguish hex-1-ene from hex-2-ene. [1]

8. 3-methylhex-2-ene, CH₃CH=C(CH₃)CH₂CH₂CH₃, is treated with hydrogen chloride.

(a) Describe the mechanism for this reaction, stating where each curly arrow starts and finishes and the intermediate formed. [4] (b) Name the main product. Use the intermediates to explain your choice. [3] (c) Name the minor product. [1]

9. Butan-2-ol can be made by reacting but-2-ene with steam.

(a) Write an equation for this reaction using structural formulae and name a suitable catalyst. [2] (b) 14.0 g of but-2-ene gives 12.6 g of butan-2-ol. Calculate the percentage yield. [3] (c) Explain why this reaction gives only one alcohol. [1]

10. (a) 3,3,3-trifluoropropene is CH₂=CHCF₃. Draw the repeat unit of its addition polymer. [1] (b) One molecule of this polymer has Mr = 1.44 × 10⁵. How many monomer molecules joined to make it? [2] (c) A section of a different polymer is –CF₂–CH(CH₃)–CF₂–CH(CH₃)–. Give the repeat unit and name the monomer. [2]

11. A council must decide what to do with waste plastics, including PVC, poly(chloroethene).

(a) Explain two benefits for sustainability of processing waste polymers instead of sending them to landfill. [4] (b) When PVC burns, each repeat unit, –[CH₂–CHCl]–, releases one HCl molecule. Calculate the mass of calcium oxide needed to neutralise the HCl from burning 250 g of PVC. CaO + 2HCl → CaCl₂ + H₂O [3] (c) Explain how photodegradable and biodegradable polymers each benefit the environment. [2]

Answers

1. (a) An electron pair acceptor [1]. (b) Both electrons go to one of the atoms, forming a cation and an anion [1]. Examiner insight: “Something that attacks a double bond” describes behaviour, not the definition; the phrase “electron pair acceptor” is what earns the mark.

2. (a) σ-bond: direct, head-on overlap of orbitals on the line joining the carbons [1]. π-bond: p-orbitals on neighbouring carbons overlap sideways, putting electron density above and below the bonding carbons [1]. (b) Rotation would break the sideways p-orbital overlap of the π-bond [1]. (c) About 120° [1]; three regions of electron density around each carbon repel equally as far apart as possible [1]. Examiner insight: Hybridisation is not required, so an answer based on sp² orbitals alone, without describing p-orbital overlap, risks missing the π-bond mark.

3. (a) (i) no, C2 carries two CH₃ groups [1]; (ii) yes [1]; (iii) yes [1]. (b) C2: C has a higher atomic number than H, so CH₃ has priority [1]. C3: both groups start with C; the CH₂Cl carbon carries (Cl, H, H) and ethyl’s (C, H, H), so CH₂Cl has priority [1]. CH₃ is on the ethyl side, opposite CH₂Cl: E [1]. (c) Cis–trans naming needs two of the groups on each C=C carbon to be the same; here the groups are CH₃, H and CH₂Cl, CH₂CH₃, with none in common [1]. Examiner insight: A bare “E” without both priority comparisons is unlikely to gain full credit; state the atomic numbers or the atoms compared at each carbon.

4. Each C=C has two different groups on each carbon, so each can be E or Z [1]. The molecule is not symmetrical (CH₃ at one end, CH₂CH₃ at the other), so (2E,5Z) and (2Z,5E) are different compounds [1]. Stereoisomers: (2E,5E), (2E,5Z), (2Z,5E), (2Z,5Z), so 4 [1]. Examiner insight: Check for symmetry before you count; doubling for each C=C only works when the two ends of the molecule differ.

5. (a) The π-bond has a relatively low bond enthalpy, so it breaks more easily than a σ-bond [1]. Its electron density lies above and below the plane, exposed to attack by electrophiles [1]. (b) Bonds broken: C=C + H–Br = 612 + 366 = 978 kJ mol⁻¹ [1]. Bonds formed: C–C + C–H + C–Br = 347 + 413 + 290 = 1050 kJ mol⁻¹ [1]. ΔH = 978 − 1050 = −72 kJ mol⁻¹ [1]. Examiner insight: Only the bonds that change need counting; include the sign, because a positive value here would describe an endothermic addition.

6. (a) Nickel catalyst (with heat) [1]; 2,7-dimethyloctane [1]. (b) No catalyst needed [1]; 4,5-dichloro-2,7-dimethyloctane [1]. (c) Acid catalyst such as phosphoric acid, H₃PO₄ [1]; 2,7-dimethyloctan-4-ol [1]. (d) No catalyst needed [1]; 4-bromo-2,7-dimethyloctane [1]. Examiner insight: Each addition here gives one product because the alkene is symmetrical; check that locants in each name are as low as possible.

7. (a) Shake a few drops of Z with bromine water [1]. If Z is unsaturated the orange colour disappears (colourless); if saturated it stays orange [1]. (b) Mr C₈H₁₂ = 108.0, so amount = 1.62 ÷ 108.0 = 0.0150 mol [1]. Amount of H₂ = 0.720 ÷ 24.0 = 0.0300 mol [1]. Ratio H₂ : hydrocarbon = 2 : 1, so two C=C bonds per molecule [1]. (c) Both contain a C=C bond, so both decolourise bromine water [1]. Examiner insight: “Goes clear” describes transparency, not colour; the expected wording is a change from orange to colourless.

8. (a) In H–Cl, H is δ+ and Cl is δ− [1]. Curly arrow from the C=C bond to the H, and from the H–Cl bond to the Cl [1]. A carbocation forms, with Cl⁻ [1]. Curly arrow from a lone pair on Cl⁻ to the positive carbon [1]. (b) 3-chloro-3-methylhexane [1]. H adding to C2 puts the positive charge on C3, a tertiary carbocation; H adding to C3 gives a secondary carbocation on C2 [1]. The tertiary carbocation is more stable, so more product forms through it (Markownikoff’s rule) [1]. (c) 2-chloro-3-methylhexane [1]. Examiner insight: Name the carbocation types for both routes; stating “Markownikoff’s rule” alone, without comparing tertiary and secondary carbocations, gives little to credit.

9. (a) CH₃CH=CHCH₃ + H₂O → CH₃CH(OH)CH₂CH₃ [1]; phosphoric acid, H₃PO₄ [1]. (b) Amount of but-2-ene = 14.0 ÷ 56.0 = 0.250 mol [1]. Maximum mass of butan-2-ol = 0.250 × 74.0 = 18.5 g [1]. Yield = 12.6 ÷ 18.5 × 100 = 68.1% [1]. (c) But-2-ene is symmetrical, so adding OH to either C=C carbon gives the same alcohol [1]. Examiner insight: Set out the theoretical mass as a separate line; if the final percentage is wrong, visible working is the only thing left that can earn credit.

10. (a) –[CH₂–CH(CF₃)]–, with an extension bond on each side and CF₃ as a side group [1]. (b) Mr C₃H₃F₃ = 96.0 [1]. Monomer molecules = 1.44 × 10⁵ ÷ 96.0 = 1500 [1]. (c) Repeat unit –[CF₂–CH(CH₃)]– [1]; monomer CF₂=CHCH₃, 1,1-difluoropropene [1]. Examiner insight: A repeat unit must contain exactly two main-chain carbons from each monomer and no C=C; extension bonds should cross the brackets.

11. (a) Burning waste polymers releases their stored energy for electricity [1], saving fuel and reducing landfill [1]. Using waste as an organic feedstock for new plastics and chemicals [1] reduces the use of crude oil, a finite resource [1]. (b) Mr repeat unit = 62.5; amount = 250 ÷ 62.5 = 4.00 mol, so 4.00 mol HCl [1]. Amount of CaO = 4.00 ÷ 2 = 2.00 mol [1]. Mass = 2.00 × 56.1 = 112 g [1]. (c) Photodegradable polymers break down when they absorb light [1]; biodegradable polymers are broken down by microorganisms; both reduce persistent plastic waste [1]. Examiner insight: For “explain two benefits”, each benefit needs the action and its consequence; listing “recycling” and “burning” without saying what each saves earns little.

Where marks are usually lost

• Describing the π-bond as lying between the nuclei rather than above and below them.
• Missing that a carbon with two identical groups rules out E/Z isomerism.
• Ranking CIP groups by size instead of comparing atomic numbers atom by atom.
• Ignoring symmetry when counting stereoisomers of a diene.
• Forgetting the sign on an enthalpy change from bond enthalpies.
• Drawing the first curly arrow from a carbon atom or to the wrong end of H–X.
• Leaving out the lone pair on the halide ion in the final step.
• Naming the minor product as the major one in Markownikoff questions.
• Keeping C=C in a repeat unit or putting a side group into the main chain.

Next steps

• Recap the key facts: revision notes for 4.1.3
• Relearn a method: study guide for 4.1.3
• OCR A Level chemistry hub
• Printable OCR chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.3 Alkenes, learning outcomes (a) to (l). Published by OCR.