Edexcel International A Level Chemistry: Alkenes (YCH11) – 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.

This set covers Topic 5: Alkenes of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 5.1 to 5.8. Topic 5 belongs to Unit 1: Structure, Bonding and Introduction to Organic Chemistry, an IAS unit. Learn the content first from the alkenes study guide and the alkenes revision notes, and track your progress on the printable checklist. More resources are on the course hub.

Data for this set: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5, Ca = 40.1. Molar volume of a gas = 24.0 dm³ mol⁻¹ at room temperature and pressure.

Questions

1. This question is about the structure of alkenes.

(a) State the general formula of open-chain alkenes with one C=C bond. [1] (b) State the general formula of cycloalkenes with one ring and one C=C bond. [1] (c) Describe how the σ bond and the π bond in a C=C double bond are formed. [2]

2. Ethene reacts with hydrogen bromide at room temperature.

(a) Describe, with the help of a labelled sketch, the mechanism of this reaction. Show curly arrows, dipoles, charges and the intermediate. [4] (b) Give the name of the product. [1]

3. 2-chlorobut-2-ene has the structure CH₃CCl=CHCH₃.

(a) Explain why 2-chlorobut-2-ene shows geometric isomerism. [2] (b) Explain why the cis–trans naming system is not suitable for this compound. [1] (c) Describe the Z isomer and explain how you assigned it. [2]

4. 1.75 g of an unbranched, open-chain alkene X reacts completely with 600 cm³ of hydrogen at room temperature and pressure. One molecule of X contains one C=C bond.

(a) Calculate the molar mass of X and deduce its molecular formula. [3] (b) X exists as E and Z isomers. Give the name of X. [1] (c) State the catalyst for this reaction and name the organic product. [2]

5. Cyclohexene, C₆H₁₀, is reacted separately with four reagents. For each, give the information asked for.

(a) Hydrogen bromide: the name of the organic product. [1] (b) Steam: the conditions and the name of the organic product. [2] (c) Acidified potassium manganate(VII): the observation and the name of the organic product. [2] (d) Hydrogen: the catalyst and the name of the organic product. [2]

6. Ethene is bubbled into an aqueous solution containing bromine and a high concentration of sodium chloride. The organic products include 1,2-dibromoethane and 1-bromo-2-chloroethane, but no 1,2-dichloroethane.

(a) Explain how the formation of 1-bromo-2-chloroethane supports a two-step mechanism with a carbocation intermediate. [3] (b) Explain why no 1,2-dichloroethane forms. [1] (c) Suggest the name of one other organic product, formed by reaction with water. [1]

7. 2-ethylbut-1-ene, CH₂=C(CH₂CH₃)₂, reacts with hydrogen bromide to give a major and a minor product.

(a) Using curly arrows, show how the major product forms. [4] (b) Give the name of this product. [1] (c) Explain why this product is formed in much larger amounts than the other possible product. [2]

8. Describe a simple chemical test to distinguish cyclohexane from cyclohexene. Give the reagent and the result for each compound. [3]

9. This question is about addition polymers.

(a) Draw the repeat unit of the polymer formed from propenenitrile, CH₂=CHCN, and explain why it is described as an addition polymer. [2] (b) A polymer has the repeat unit –[CH₂–C(CH₃)₂]ₙ–. Give the structure of its monomer. [1] (c) A chain of poly(propenenitrile) has Mr 3.18 × 10⁵. Calculate the number of monomer units in the chain. [2] (d) Explain why poly(alkenes) persist for a long time in landfill. [2]

10. A waste-to-energy plant burns 2.50 kg of poly(chloroethene), PVC, repeat unit –[CH₂–CHCl]–. Assume all the chlorine is released as hydrogen chloride.

(a) Calculate the mass of hydrogen chloride formed. [2] (b) The exhaust gas passes through a scrubber containing calcium hydroxide. Write an equation for the reaction and calculate the minimum mass of calcium hydroxide needed. [3] (c) Give one advantage of incinerating polymer waste, other than reducing landfill. [1] (d) Give one benefit and one limitation of replacing PVC packaging with a biodegradable polymer. [2]

11. A student makes cyclohexene by heating 6.00 g of cyclohexanol, C₆H₁₁OH, with a phosphoric acid catalyst. She collects 3.69 g of pure cyclohexene.

(a) Give a balanced equation for the dehydration. [1] (b) Determine the percentage yield of cyclohexene. [3] (c) Suggest one practical reason why the yield is less than 100%. [1] (d) A drop of the product is shaken with bromine water. State what is seen. [1]

Answers

1. (a) CₙH₂ₙ [1] (b) CₙH₂ₙ₋₂ [1] (c) σ bond: head-on overlap of orbitals along the axis between the two carbon nuclei [1]; π bond: sideways overlap of p orbitals, giving electron density above and below the plane of the molecule [1]. Examiner insight: For each bond give the type of overlap and where the electron density lies; “above and below the plane” is the key π detail.

2. (a) HBr shown with Hδ+ and Brδ− [1]; curly arrow from the C=C to the H atom [1]; curly arrow from the H–Br bond onto Br, giving Br⁻ and the carbocation CH₃CH₂⁺ [1]; Br⁻ uses a lone pair to bond to C⁺, shown by a third curly arrow [1]. (b) Bromoethane [1] Examiner insight: The specification expects arrows to start from a bond or a lone pair, so an arrow drawn from the Br⁻ charge or from a carbon atom is unlikely to be credited.

3. (a) Rotation about the C=C is restricted because twisting would break the π bond overlap [1]; carbon 2 holds Cl and CH₃ while carbon 3 holds CH₃ and H, so both have two unlike groups [1]. (b) Both carbons carry a CH₃ group but the other groups differ, so “cis” could refer to the methyl groups or to a different pair; the name is ambiguous [1]. (c) Priorities: Cl over CH₃ on carbon 2 (atomic number 17 vs 6), and CH₃ over H on carbon 3 [1]; the Z isomer has Cl and the carbon-3 CH₃ on the same side, which places the two methyl groups on opposite sides [1]. Examiner insight: State the atomic-number comparison for each carbon; a correct letter with no stated priorities leaves the explanation incomplete.

4. (a) n(H₂) = 0.600 ÷ 24.0 = 0.0250 mol, so n(X) = 0.0250 mol [1]; M = 1.75 ÷ 0.0250 = 70.0 g mol⁻¹ [1]; CₙH₂ₙ: 14n = 70.0 so n = 5, C₅H₁₀ [1]. (b) Pent-2-ene (pent-1-ene has two H atoms on carbon 1 and cannot show E–Z isomerism) [1]. (c) Nickel [1]; pentane [1]. Examiner insight: Convert cm³ to dm³ before dividing by 24.0, and show each step of the working.

5. (a) Bromocyclohexane [1] (b) Phosphoric acid catalyst with high temperature and pressure [1]; cyclohexanol [1]. (c) Purple solution turns colourless [1]; cyclohexane-1,2-diol [1]. (d) Nickel catalyst (with heat) [1]; cyclohexane [1]. Examiner insight: Give the starting colour as well as the final one; “turns colourless” alone does not describe a change.

6. (a) Bromine is polarised and the δ+ bromine adds first, forming the carbocation BrCH₂CH₂⁺ [1]; the carbocation can then be attacked by any nucleophile present [1]; attack by Cl⁻ gives BrCH₂CH₂Cl, which could not form if both bromine atoms added in one step [1]. (b) Cl⁻ is a nucleophile, not an electrophile, so it cannot make the first attack on the electron-rich C=C; every product must contain at least one bromine [1]. (c) 2-bromoethanol [1] Examiner insight: Link each product to a step of the mechanism; a bare list of products explains nothing.

7. (a) HBr drawn with Hδ+–Brδ−; arrow from the C=C to H [1]; arrow from the H–Br bond onto Br [1]; tertiary carbocation CH₃C⁺(CH₂CH₃)₂ with Br⁻ shown with a lone pair [1]; arrow from the lone pair on Br⁻ to C⁺ [1]. (b) 3-bromo-3-methylpentane [1] (c) H adding to the CH₂ end gives a tertiary carbocation, while H adding to the other carbon gives a primary carbocation [1]; the tertiary carbocation is more stable because three alkyl groups release electron density towards C⁺, so it forms faster and its product dominates [1]. Examiner insight: A tidy mechanism through the primary carbocation still leads to the wrong product, so check which carbocation you drew.

8. Reagent: bromine water [1]; cyclohexene: orange-brown to colourless [1]; cyclohexane: stays orange-brown (no visible change) [1]. Examiner insight: Give a separate result for each compound; one observation covers only half of a two-compound test.

9. (a) –[CH₂–CH(CN)]ₙ– with bonds through the brackets [1]; the C=C bonds open and monomers join with no other product formed [1]. (b) CH₂=C(CH₃)₂ (2-methylpropene) [1] (c) Mr(C₃H₃N) = 36.0 + 3.0 + 14.0 = 53.0 [1]; 3.18 × 10⁵ ÷ 53.0 = 6000 [1]. (d) The chains are saturated and held together by strong C–C and C–H bonds [1]; so they are unreactive and microorganisms cannot break them down [1]. Examiner insight: A repeat unit with a remaining C=C, or without extension bonds, is a common way to lose the structure mark.

10. (a) Mr(C₂H₃Cl) = 24.0 + 3.0 + 35.5 = 62.5, so n(HCl) = 2500 ÷ 62.5 = 40.0 mol [1]; mass = 40.0 × 36.5 = 1460 g (1.46 kg) [1]. (b) Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O [1]; n(Ca(OH)₂) = 40.0 ÷ 2 = 20.0 mol [1]; Mr = 74.1, mass = 20.0 × 74.1 = 1480 g (1.48 kg) [1]. (c) The energy released can generate electricity or heat [1]. (d) Benefit: it is broken down by microorganisms, so less long-lived waste [1]; limitation: it may need composting conditions to degrade, or may cost more to produce [1]. Examiner insight: Use the 1 : 2 ratio from your equation; forgetting to halve the moles of HCl doubles the mass of calcium hydroxide.

11. (a) C₆H₁₁OH → C₆H₁₀ + H₂O [1] (b) n(cyclohexanol) = 6.00 ÷ 100.0 = 0.0600 mol [1]; theoretical mass of cyclohexene = 0.0600 × 82.0 = 4.92 g [1]; yield = 3.69 ÷ 4.92 × 100 = 75.0% [1]. (c) Some product is lost during transfer, separation or distillation (or the reaction is incomplete) [1]. (d) Orange-brown bromine water turns colourless [1]. Examiner insight: Find the theoretical yield via moles first; dividing 3.69 by 6.00 directly gives a meaningless figure.

Where marks are usually lost

• Using CₙH₂ₙ for a cycloalkene.
• Describing π overlap without saying where the electron density lies.
• Starting curly arrows from atoms or charges instead of from the C=C, a bond or a lone pair.
• Missing the δ+/δ− labels on HBr or Br₂, or the charge on the carbocation.
• Choosing the primary carbocation when a more stable one is available.
• Assigning E or Z without stating the atomic-number priorities on each carbon.
• Giving the conditions for hydration without the acid catalyst.
• Leaving a C=C in a polymer repeat unit.
• Forgetting mole ratios from the equation in scrubbing and yield calculations.
• Describing only the final colour in a colour-change test.

Next steps

• Alkenes revision notes
• Alkenes study guide
• Edexcel International A Level Chemistry hub
• Printable Edexcel Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 1, Topic 5: Alkenes, outcomes 5.1 to 5.8.