Edexcel A-Level Chemistry Paper 2, November 2020: Question 2
8 marks · Medium difficulty · Short Open Response
Identify formulas, systematic names, reaction products, addition polymer structures, and calculate molecular amounts for various alkenes including ethene and but-1-ene.
Practise this questionQuestion
Question text
2 This question is about alkenes.
(a) Which of these has the molecular formula C6H10?
(1)
A B
C D
(b) What is the systematic name of this alkene?
CH3
HC CH3
H2C C
CH3
(1)
A 2-methylpent-1-ene
B 3-methylpent-1-ene
C 2,3-dimethylbut-1-ene
D 2,3-dimethylbut-3-ene
(c) Two reactions of ethene are shown.
H H
C C
H H
Reaction 1 Reaction 2
HBr
Product H H
H C C OH
H H
Complete the table.
(3)
Reaction *P62669A0432*Reagent and conditionProduct
1 HBr at room temperature
H H
2 H C C OH
H H
(d) But-1-ene has the structure
H H
C C
H CH2CH3
(i) Draw the structure of the polymer formed when but-1-ene polymerises.
Include two repeat units.
(1)
*P62669A0532*
(ii) Calculate the number of molecules in 70.0g of but-1-ene.
[Avogadro constant = 6.02 × 1023 mol−1]
(2)
(Total for Question 2 = 8 marks)
Mark scheme
Show the mark scheme
How to answer it
Edexpert Study Guide: Alkenes & Reactions
What this question tests
This question assesses your core knowledge of alkene chemistry, IUPAC nomenclature rules, addition reactions of alkenes (electrophilic addition and hydration), addition polymerisation drawing conventions, and fundamental moles-to-particles calculations using the Avogadro constant.
Part (a) — Identifying Molecular Formulae from Structures
Determine which structure corresponds to C₆H₁₀ (1 mark)
✅ Correct Answer: C
Structure C (cyclohexene) has a ring structure with one double bond, giving it the general formula CₙH₂ₙ₂ , which equates to C₆H₁₀ .
❌ Common Errors & Distractors
- A: Hex-1-ene is an open-chain alkene with formula C₆H₁₂ .
- B: Contains multiple conjugated double bonds (hexa-1,3,5-triene) with formula C₆H₈ .
- D: Cyclohexa-1,4-diene has two double bonds, giving the formula C₆H₈ .
Part (b) — Systematic IUPAC Nomenclature
Name the branched alkene structure (1 mark)
✅ Correct Answer: C (2,3-dimethylbut-1-ene)
The longest continuous carbon chain containing the double bond has 4 carbons (but-ene). Numbering from the end closest to the double bond gives carbon 1 to the =CH₂ group. Methyl branches are located at carbons 2 and 3.
❌ Why other options fail
- A & B: Incorrectly assume a 5-carbon chain (pentene) by misidentifying the longest continuous carbon pathway.
- D: Fails to give the double bond the lowest possible number (numbering from the wrong end).
Part (c) — Reactions of Ethene
Complete the reaction and reagent table (3 marks)
💡 Key Knowledge & Reagents
- Reaction 1 Product: Bromoethane ( CH₃CH₂Br or displayed formula showing H-C-C-Br connectivity).
- Reaction 2 Reagents: Steam ( H₂O(g) ) and an acid catalyst (e.g., concentrated H₃PO₄ or H₂SO₄ ). Note: "water and heat" is also accepted.
🧠 Exam Technique
Make sure you explicitly state both the chemical reagent AND the catalyst for hydration reactions. Writing just "water" without specifying "steam" or an acid catalyst will often lose marks in stricter mark schemes.
Part (d)(i) — Addition Polymerisation
Draw two repeat units of poly(but-1-ene) (1 mark)
✅ Correct Answer Structure
A backbone chain of 4 carbon atoms with single bonds, showing two complete repeat units joined together, with open extension bonds passing through square brackets or simply terminating outside the ends.
Side chains: Ethyl groups ( -CH₂CH₃ or -C₂H₅ ) hanging off alternating carbon atoms in the backbone.
❌ Common Errors
- Leaving the double bond in the backbone of the polymer chain.
- Drawing only one repeat unit when the question explicitly requests two.
- Failing to include clear extension bonds through the brackets/ends.
Part (d)(ii) — Moles and Avogadro Calculations
Calculate the number of molecules in 70.0 g of but-1-ene (2 marks)
📐 Step-by-Step Calculation
- Find the molar mass of but-1-ene (C₄H₈):
M = (4 × 12.0) + (8 × 1.0) = 56.0 g mol⁻¹ - Calculate moles of but-1-ene:
Moles = Mass / Molar Mass = 70.0 / 56.0 = 1.25 mol - Calculate the number of molecules:
Molecules = Moles × Avogadro constant
= 1.25 × 6.02 × 10²³ = 7.525 × 10²³ (or 7.53 × 10²³)
❌ Calculation Traps
- Incorrect Molar Mass: Forgetting the hydrogen count or miscalculating atomic masses.
- No Working Out: A correct final answer with zero working scores full marks, but if your answer is slightly off due to rounding, intermediate working steps secure your method mark (M1)!
Topics
Organic Chemistry · Physical Chemistry · Topic 6: Organic Chemistry I · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.