OCR A-Level Chemistry AS Breadth in chemistry (01), June 2018: Question 24
10 marks · Medium difficulty · Structured Questions
Explain structural isomerism and boiling point trends of saturated hydrocarbons A, B, and C, and analyse their reactions with chlorine.
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Question text
24 This question is about saturated hydrocarbons.
(a) Compounds A, B and C are saturated hydrocarbons.
The structures and boiling points of A, B and C are shown below.
Isomer Boiling point /°C
A 36
B 28
C 9
• Use the structures to explain what is meant by the term structural isomer.
• Explain the trend in boiling points shown by A, B and C in the table.
… [5]
(b) Compounds A, B and C all react with chlorine in the presence of ultraviolet radiation to form
organic compounds with the formula C5H11Cl.
(i) Name the mechanism for this reaction.
… [1]
(ii) Complete the table to show the number of structural isomers of C5H11Cl that could be
formed from the reaction of chlorine with A and B.
A B
Number of
structural isomers …
[2]
(iii) The reaction of compound A with excess chlorine forms a compound D, which has a
molar mass of 175.5 g mol−1.
Draw a possible structure for compound D and write the equation for its formation from
compound A. Use molecular formulae in the equation.
Compound D
Equation … [2]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
24 (a) Structural isomers: 1 mark 5 For ‘structural’:
Different structural formulae ALLOW different structure
AND same molecular formula OR different displayed/ skeletal formula
DO NOT ALLOW any reference to spatial/space/3D
Same formula is not sufficient (no ‘molecular’)
Different arrangement of atoms is not sufficient
(no ‘structure’/’structural’)
Common molecular formula: 1 mark ALLOW 5 carbons and 12 hydrogens
C5H12 for all 3 hydrocarbons
ALLOW for 2 marks:
Different structural formulae
AND same molecular formula of C5H12
Boiling point and branching: Comparisons needed throughout
1 mark ORA throughout
Boiling point decreases with
more branching ALLOW comparison between any alcohols, e.g.
OR more methyl/alkyl groups/side chains A is least branched and has highest b pt
OR shorter carbon chain C is most branched and has lowest b pt
Branching and London forces: 1 mark
Could be seen anywhere within response ALLOW induced dipole(–dipole) interactions
More branching gives less (surface) contact IGNORE van der Waals’/vdw forces
AND ALLOW SA for surface area
fewer/weaker London forces
ALLOW ‘harder to overcome intermolecular forces
Energy and intermolecular forces: 1 mark ALLOW more energy to separate the molecules
Less energy to break London forces/
intermolecular forces/intermolecular bonds/ IGNORE just ‘bonds’
intermolecular/London forces required
Question Answer 13 Marks Guidance
(b) (i) Radical substitution 1 ALLOW Free radical substitution
(b) (ii) 2
A B
(b) (iii) 2
Structure of D
Structure of a trichloro isomer of A, e.g. ALLOW correct structural OR displayed
Cl Cl OR skeletal formula OR mixture of the above
(as long as unambiguous)
IGNORE molecular formula
Cl
ALLOW any trichloro isomer of A
CHECK carefully
Equation ALLOW multiples,
C5H12 + 3Cl2 C5H9Cl3 + 3HCl e.g. 2C H + 6Cl 2C H Cl + 6HCl
5 12 2 5 9 3
Molecular formulae required
NO ECF from incorrect structure of D
Total 10
How to answer it
Saturated Hydrocarbons & Isomerism Study Guide
What this question tests
This question assesses core organic chemistry principles including the definition of structural isomerism, the physical properties of alkanes (specifically how branching affects boiling points via London forces), free radical substitution mechanisms, isomer counting, and writing balanced chemical equations using molecular formulae.
Part (a): Structural Isomers & Boiling Point Trends
Understanding Isomerism and Intermolecular Forces
✅ Correct Answers
- Structural isomer definition: Same molecular formula, but different structural formulae.
- Molecular formula: C₅H₁₂ for all three isomers (A, B, and C).
- Boiling point trend: Boiling point decreases as branching increases (or as carbon chains become more branched / shorter main chains).
- Explanation: More branching leads to less surface contact between molecules, resulting in fewer/weaker London forces. Consequently, less energy is required to overcome these intermolecular forces.
💡 Key Knowledge
- Alkanes are non-polar molecules held together solely by weak induced dipole-dipole interactions (London forces).
- Straight-chain isomers (like A, pentane) have a greater surface area of contact, maximizing intermolecular attractions.
- Branched isomers (like C, 2,2-dimethylpropane) are more spherical, reducing contact area and weakening the instantaneous dipole interactions.
❌ Common Errors
- Saying "same molecular formula" without stating "different structural formula" (or vice versa).
- Using vague terms like "space", "spatial", or "3D" which describe stereoisomerism rather than structural isomerism.
- Stating "different arrangement of atoms" without qualifying it with "structural formula".
- Attributing boiling point changes to breaking "covalent bonds" rather than "intermolecular forces / London forces".
🧠 Exam Technique
For trend questions comparing molecules, ensure you make a direct comparative statement using "ORA" (Or Reverse Applies). Always link molecular shape directly to surface contact area, then to the strength of London forces, and finally to the energy required.
Part (b)(i): Reaction Mechanism
Naming the Mechanism
✅ Correct Answer
Radical substitution (or Free radical substitution)
🧠 Exam Technique
Learn reaction classifications precisely. Do not confuse substitution with addition mechanisms typical of alkenes.
Part (b)(ii): Isomer Counting
Determining Monochloro Isomers ( C₅H₁₁Cl )
✅ Correct Answers
- Isomer A (Pentane): 3 structural isomers (1-chloropentane, 2-chloropentane, 3-chloropentane).
- Isomer B (2-methylbutane): 4 structural isomers (1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 3-chloro-2-methylbutane, 1-chloro-3-methylbutane).
❌ Common Errors
Failing to systematically check all possible carbon positions for chlorine substitution, leading to missed isomers or double-counted mirror images.
Part (b)(iii): Trichloro Substitution & Equations
Advanced Halogenation and Molecular Formula Equations
✅ Correct Answers
- Structure D: A trichloro isomer of compound A ( C₅H₉Cl₃ ). Chlorine atoms can be placed on various carbon positions (e.g., 1,2,3-trichloropentane or 2,2,3-trichloropentane).
- Equation: C₅H₁₂ + 3Cl₂ → C₅H₉Cl₃ + 3HCl
📐 Working Out Molar Mass / Formula Check
- Compound A is pentane ( C₅H₁₂ ), Mr = (5 × 12.0) + (12 × 1.0) = 72.0 g mol⁻¹.
- Compound D has molar mass 175.5 g mol⁻¹. Difference = 175.5 - 72.0 = 103.5 g mol⁻¹.
- Each substituted Cl replaces an H atom (Loss of H = 1.0, Gain of Cl = 35.5, Net change per substitution = +34.5 g mol⁻¹).
- 103.5 / 34.5 = exactly 3 chlorine atoms added, confirming C₅H₉Cl₃ .
❌ Common Errors
- Using structural or displayed formulae within the chemical equation when the question explicitly requests molecular formulae.
- Incorrect stoichiometry in the equation (forgetting balancing coefficients for Cl₂ and HCl ).
🧠 Exam Technique
Always double-check what formula style is requested. If the question asks for "molecular formulae in the equation", structural drawings or expanded representations will lose marks even if balanced correctly.
Topics
Module 4: Core organic chemistry · 4.1 Basic concepts and hydrocarbons
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.