OxfordAQA A-Level Chemistry: Alkenes (9620) – 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.

Every question here tests section 3.3.4 Alkenes in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, across 3.3.4.1 to 3.3.4.4: bonding and reactivity, addition reactions, addition polymers and epoxyethane. No part is International A-level only. Alkene questions belong to Unit 2 (Organic 1 and Physical 1), and the synoptic Unit 5 can include them.

If stuck, revisit the alkenes study guide or the alkenes revision notes. The OxfordAQA A-Level Chemistry hub and the printable checklist list the whole course, and the free 10-minute diagnostics show where to focus. For the mole work in questions 5 and 11, see amount of substance. Relative atomic masses: O = 16.0, N = 14.0, C = 12.0, H = 1.0.

Questions

1. Which alkene gives only one organic product when it reacts with hydrogen bromide?

• A hept-1-ene
• B 2-methylpent-2-ene
• C 3,4-dimethylhex-3-ene
• D hept-2-ene

[1]

2. Two liquid hydrocarbons, X and Y, both have the molecular formula C₇H₁₄. X decolourises bromine water. Y does not.

(a) State what the result shows about the structure of X. [1] (b) Suggest a name for Y. [1] (c) Explain, in terms of bonding, why X reacts with bromine. [2]

3. Cold, concentrated sulfuric acid is added to non-1-ene, CH₃(CH₂)₆CH=CH₂.

(a) Which species acts as the electrophile? [1] (b) Using curly arrows, show how the major addition product forms. [4]

4. 2-methylhex-2-ene, (CH₃)₂C=CHCH₂CH₂CH₃, reacts with hydrogen bromide.

(a) Draw the two carbocations that can form and classify each one. [2] (b) Give the names of the main product and of the product formed in smaller amount. [2] (c) Explain why one product is formed in a much larger amount. [2]

5. A 2.00 g mixture of cyclohexane and cyclopentene, C₅H₈, reacts with exactly 25.0 cm³ of a 0.400 mol dm⁻³ solution of bromine in an inert organic solvent. Calculate the percentage by mass of cyclopentene in the mixture. [4]

6. This question is about addition polymers.

(a) Draw the repeating unit of the polymer made from 2-methylpropenenitrile, CH₂=C(CH₃)CN, and name the polymer. [2] (b) Part of another polymer is –CH₂–CH(OCOCH₃)–CH₂–CH(OCOCH₃)–. Give the repeating unit and name the monomer. [2] (c) Explain why the polymer in part (a) does not decolourise bromine water. [2]

7. Two samples of poly(propene) have average Mr values of 8.4 × 10⁴ and 3.4 × 10⁵.

(a) Predict which sample softens at the higher temperature. Explain your answer. [3] (b) An inflatable paddling pool is made from PVC containing a plasticiser. Explain how the plasticiser makes the PVC flexible. [2] (c) Give one typical use of PVC without a plasticiser. [1]

8. Epoxyethane is made from ethene.

(a) Write an equation for the reaction. [1] (b) Name the catalyst. [1] (c) State the percentage atom economy and give a reason. [1] (d) State two hazards associated with this process. [2] (e) Suggest why the temperature of the reactor must be controlled carefully. [1]

9. Epoxyethane reacts with methanol.

(a) Explain why epoxyethane is so reactive. [2] (b) Write an equation for the reaction and name the organic product. [2] (c) Outline the mechanism for the reaction. [3]

10. A non-ionic surfactant is made from dodecan-1-ol, C₁₂H₂₅OH, and epoxyethane. A sample has Mr = 538.0.

(a) Write a general equation for the formation of the surfactant from one molecule of dodecan-1-ol and n molecules of epoxyethane. [1] (b) Calculate n. [2] (c) Suggest why this molecule can act as a surfactant. [2] (d) Give one economic reason why such surfactants matter and one environmental concern about them. [2]

11. A plant converts 7.00 tonnes of ethene into epoxyethane with an 80.0% yield. All the epoxyethane then reacts with water.

(a) Calculate the mass of epoxyethane formed. [3] (b) Calculate the mass of ethane-1,2-diol formed, assuming full conversion. [2] (c) Explain why a large excess of water is used. [1] (d) Give two uses of ethane-1,2-diol. [2]

Answers

1. C [1]. It is symmetrical, so both possible carbocations are identical. [1] Examiner insight: Check the groups on each carbon of the double bond, not the whole molecule; B and D look balanced but their two C=C carbons carry different groups.

2. (a) X contains a C=C double bond (it is unsaturated) [1] (b) A cycloalkane, e.g. methylcyclohexane or cycloheptane [1] (c) The C=C bond is a centre of high electron density [1]. It induces a dipole in Br₂, and the δ+ Br accepts a pair of electrons from the double bond (acts as an electrophile) [1]. [4] Examiner insight: In (b) a straight-chain alkane cannot be right, because C₇H₁₄ does not fit CₙH₂ₙ₊₂; check any suggested formula against the one given.

3. (a) A hydrogen atom of H₂SO₄ (Hδ+) [1] (b) Curly arrow from the double bond to an H atom in H₂SO₄, plus an arrow moving the O–H bonding pair onto O [1]; secondary carbocation CH₃(CH₂)₆C⁺HCH₃ [1]; the hydrogensulfate ion’s negative oxygen uses a lone pair to bond to C⁺ (arrow shown) [1]; product CH₃(CH₂)₆CH(OSO₂OH)CH₃ [1]. [5] Examiner insight: Because the major product is asked for, the positive charge must be on C2; a mechanism through the primary carbocation answers a different question.

4. (a) (CH₃)₂C⁺CH₂CH₂CH₂CH₃, tertiary [1]; (CH₃)₂CHC⁺HCH₂CH₂CH₃, secondary [1] (b) Major: 2-bromo-2-methylhexane [1]. Minor: 3-bromo-2-methylhexane [1] (c) The tertiary carbocation has three electron-releasing alkyl groups that spread the positive charge, so it is more stable than the secondary one [1]. It is formed faster, so most of the HBr adds by this route [1]. [6] Examiner insight: An answer about the stability of the products, not the carbocations, misses the point; name the intermediates in your explanation.

5. n(Br₂) = 0.0250 × 0.400 = 0.0100 mol [1]. Only cyclopentene reacts (1 : 1), so n(C₅H₈) = 0.0100 mol [1]. Mr(C₅H₈) = 68.0, mass = 0.680 g [1]. Percentage = 0.680 ÷ 2.00 × 100 = 34.0% [1]. [4] Examiner insight: State that cyclohexane does not react; otherwise the 1 : 1 link is unexplained.

6. (a) Repeating unit –[CH₂–C(CH₃)(CN)]ₙ–, single C–C, both CH₃ and CN on the same carbon, extension bonds and n [1]; poly(2-methylpropenenitrile) [1] (b) Repeating unit –[CH₂–CH(OCOCH₃)]ₙ– [1]; monomer ethenyl ethanoate, CH₂=CHOCOCH₃ [1] (c) The polymer is saturated: there is no C=C bond left, so there is no centre of high electron density for bromine to attack [1]. Its backbone is made of strong C–C and C–H bonds of low polarity [1]. [6] Examiner insight: In (a) the two substituents must stay on the same carbon; splitting them across the two main-chain carbons gives a different polymer.

7. (a) The sample with Mr 3.4 × 10⁵ [1]. Its chains are longer, with more electrons and more contact between neighbouring chains [1], so the total induced dipole–dipole (van der Waals) attraction is larger and more energy is needed to let the chains move apart [1]. (b) Plasticiser molecules sit between the chains and push them further apart [1]. With the chains spaced out, the attraction between them is weaker and they can move over one another [1]. (c) Any one: water or sewage pipes, doors, bank cards [1]. [6] Examiner insight: In (a) avoid saying covalent bonds break on softening; the change is about forces between chains, and mixing the two up contradicts the answer.

8. (a) 2CH₂=CH₂ + O₂ → 2C₂H₄O [1] (b) Silver [1] (c) 100%, because epoxyethane is the only product [1] (d) Any two: epoxyethane is extremely flammable; it forms explosive mixtures with air; it is toxic; it is a carcinogen [1] [1] (e) The competing complete combustion of ethene is far more exothermic, so the temperature could rise out of control (and yield would fall) [1]. [6] Examiner insight: “Dangerous chemicals” alone is too vague for (d); each hazard needs to name a specific property.

9. (a) The ring bond angles are about 60°, far below 109.5°, so the ring is strained and the C–O bonds are weak [1]. The C–O bonds are polar, so the ring carbons are δ+ and attract nucleophiles [1]. (b) C₂H₄O + CH₃OH → CH₃OCH₂CH₂OH [1]; 2-methoxyethanol [1] (c) Arrow from a lone pair on the methanol oxygen to a ring carbon [1]; arrow from that C–O ring bond to the ring oxygen, giving CH₃O⁺(H)CH₂CH₂O⁻ [1]; proton transferred from O⁺–H to O⁻ to give the product [1]. [7] Examiner insight: The attacking species is the nucleophile, so the first arrow starts at methanol’s lone pair; reversing it, as if the ring attacked methanol, loses the mechanism.

10. (a) C₁₂H₂₅OH + nC₂H₄O → C₁₂H₂₅(OCH₂CH₂)ₙOH [1] (b) Mr(C₁₂H₂₅OH) = 186.0 [1]; n = (538.0 − 186.0) ÷ 44.0 = 8 [1] (c) The C₁₂H₂₅ chain is non-polar and mixes with grease or oil [1]; the ether chain ending in OH is polar and is attracted to water [1] (d) Economic: used on a very large scale in detergents and cleaning products [1]. Environmental: can harm aquatic life if it reaches waterways, so it needs to break down readily [1]. [7] Examiner insight: In (b) the whole Mr of each epoxyethane unit adds on, because the reaction is addition; subtracting water as if it were a condensation gives the wrong n.

11. (a) n(C₂H₄) = 7.00 × 10⁶ ÷ 28.0 = 2.50 × 10⁵ mol [1]. Theoretical mass of C₂H₄O = 2.50 × 10⁵ × 44.0 = 1.10 × 10⁷ g = 11.0 tonnes [1]. At 80.0%: 8.80 tonnes [1] (b) n(C₂H₄O) = 8.80 × 10⁶ ÷ 44.0 = 2.00 × 10⁵ mol [1]; mass of C₂H₆O₂ = 2.00 × 10⁵ × 62.0 = 1.24 × 10⁷ g = 12.4 tonnes [1] (c) The diol still has OH groups and could react with more epoxyethane; excess water limits this [1] (d) Any two: antifreeze, engine coolant, making polyester (PET) [1] [1]. [8] Examiner insight: Convert tonnes to grams (× 10⁶) before dividing by Mr, or work in tonnes throughout; mixing units gives answers a million times out.

Where marks are usually lost

• Opening arrow begun at an alkene carbon rather than at the double bond itself.
• Partial charges missing from HBr, or Br₂ shown without its induced dipole.
• Hydrogensulfate shown bonding to carbon through sulfur.
• Major product explained by product stability rather than carbocation stability.
• Repeating unit with a C=C, missing extension bonds, or substituents moved to the wrong carbon.
• Polymer unreactivity blamed on intermolecular forces.
• Epoxyethane mechanism missing the final proton transfer.
• Unit slips between tonnes and grams in industrial calculations.

Next steps

• Alkenes revision notes
• Alkenes study guide
• Shapes, polarity and intermolecular forces
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.4 Alkenes (International AS), points 3.3.4.1 to 3.3.4.4.