OCR A-Level Chemistry AS Breadth in chemistry (01), June 2024: Question 24

14 marks · Medium difficulty · Structured Questions

Analyze boiling points of alkane isomers, complete the mechanism for free radical substitution of ethane with bromine, and predict/explain bond angles and shapes in propene.

Practise this question

Question

A three-part chemistry question about hydrocarbons. Part (a) features a table showing three isomers of C6H14 (A, B, C) with their skeletal formulae and boiling points (69, 63, and 58 °C respectively), followed by a prompt to state and explain the trend. Part (b) asks to complete a table for the mechanism of the reaction between C2H6 and Br2 including initiation, propagation, and termination stages. Part (c) shows the displayed formula of propene, C3H6, labeling carbon atoms 1 and 2, and asks to predict bond angles and shapes of these carbon atoms with an explanation.
Question text

24 This question is about hydrocarbons.

(a) The skeletal formulae and boiling points of three isomers of C6H14 are shown in the table below.

Molecular Boiling point /

Isomer Skeletal formula

formula °°C

A C6H14 69

B C6H14 63

C C6H14 58

State and explain the trend in the boiling points shown in the table.

Refer to the isomers A, B and C in your answer.

… [4]

(b) The hydrocarbon C2H6 reacts with bromine, Br2, to form C2H5Br under suitable conditions.

Complete the table below to show the mechanism for the three stages of the reaction of C2H6

with Br2 to form C2H5Br.

The equation for one of the possible reactions for termination has been completed.

In your equations, use molecular formulae and ‘dots’ (•) with any radicals.

Initiation

Conditions …

Equation …

Propagation

1 …

2 …

Termination 1 Br• + Br• Br2

2 …

3 …

[5]

(c) Propene, C3H6, has different bond angles and shapes around the carbon atoms.

The displayed formula of a propene molecule is shown below.

H H

H C C

H C H

H

Predict the bond angles and the names of the shapes around the C atoms 1 and 2 above, and

explain why the bond angles and shapes are different.

Carbon atom Bond angle Name of shape

Explanation: …

… [5]

Mark scheme

Show the mark scheme The mark scheme details points for part (a) covering trend, branching/surface contact, London forces/intermolecular forces, and energy required. Part (b) provides the conditions and equations for initiation, propagation, and termination steps using radical dots.

Question Answer Marks Guidance

24 (a) CHECK FOR RESPONSES ON TABLE ANNOTATE WITH TICKS AND CROSSES

Comparisons needed throughout

ORA throughout

Trend

Boiling point decreases with more branching OR fewer ALLOW comparison between 2 alkanes, e.g.

methyl/alkyl groups/side chains ✓ C has greatest branching AND lowest boiling point

A has no branching AND highest boiling point

IGNORE Chain length

Branching and surface contact

Could be seen anywhere within response Surface area alone is not sufficient

Branching linked to the amount of (surface) contact / must have idea of contact.

interaction/overlap (between molecules) ✓

DO NOT ALLOW responses comparing different

numbers of electrons (as all have the same number).

Type and strength of intermolecular force

Could be seen anywhere within response ALLOW more branching results in fewer London forces

Branching/ boiling points/contact linked to strength of ORA

London forces OR induced dipole(–dipole) interactions

OR extent of surface contact ✓ IGNORE van der Waals’/vdW forces

OR IDID OR IDD

Energy and intermolecular forces

Linked to energy seen anywhere ALLOW more energy to break/overcome

More energy to break intermolecular forces with less London forces

branching ✓ OR induced dipole(–dipole) interactions

OR vdW forces

IGNORE just ‘bonds’

intermolecular or type of forces required IGNORE harder to overcome/break intermolecular

forces (no reference to energy)

(b) CORRECT DOTS REQUIRED FOR ALL MARKS 5 ALLOW any combination of skeletal OR structural OR

19 displayed formula as long as unambiguous

DO NOT ALLOW charged formulae

IGNORE position of dots within a formula

Initiation

ultraviolet / UV DO NOT ALLOW if reagents also present, e.g..steam

AND

Br2 → 2Br• OR Br2 → Br• + Br•

OR Br–Br → 2Br•, etc ✓

Propagation

1 C2H6 + Br• → C2H5• + HBr ✓ ALLOW •CCH5 for C2H5•

2 C2H5• + Br2 → C2H5Br + Br• ✓

Termination

In either order:

How to answer it

Hydrocarbons Study Guide: Isomerism, Mechanisms & Shapes

What this question tests

This question assesses core organic chemistry concepts including fractional boiling points of alkane isomers, free radical substitution mechanisms (initiation, propagation, termination) with radical dots, and VSEPR theory predicting bond angles and shapes around saturated versus unsaturated carbon atoms.

Question Part (a)

Boiling Point Trends in Isomers of C₆H₁₄

✅ Correct Answer

  • Boiling point decreases as branching increases (from A to C).
  • Isomer A has the longest straight chain and highest boiling point (69 °C); Isomer C is the most branched and has the lowest boiling point (58 °C).

💡 Key Knowledge

  • All isomers share the molecular formula C₆H₁₄, meaning they have the same number of electrons. Do not credit arguments based on electron counts!
  • Branching creates a more spherical, compact shape, reducing surface contact between molecules.
  • Fewer points of contact mean fewer London forces (induced dipole-dipole interactions) can form.

🧠 Exam Technique

  • Use the 4-mark checklist: (1) State the trend, (2) Link branching to surface contact, (3) Mention London forces/induced dipoles, (4) Explicitly refer to the energy required to break forces.
  • Always make explicit comparative references using isomers A, B, and C.

❌ Common Errors

  • Claiming branched isomers have "fewer electrons" or "weaker individual covalent bonds" (covalent bonds do not break during boiling!).
  • Forgetting to mention energy (e.g., "less energy needed to overcome intermolecular forces").
Question Part (b)

Free Radical Substitution Mechanism for C₂H₆ with Br₂

✅ Correct Answer

  • Initiation: Ultraviolet / UV light AND Br₂ → 2Br•
  • Propagation 1: C₂H₆ + Br• → C₂H₅• + HBr
  • Propagation 2: C₂H₅• + Br₂ → C₂H₅Br + Br•
  • Termination 2: C₂H₅• + Br• → C₂H₅Br
  • Termination 3: C₂H₅• + C₂H₅• → C₄H₁₀

💡 Key Knowledge

  • Free radical substitution occurs in three distinct phases: Initiation, Propagation, and Termination.
  • The radical dot ( • ) represents an unpaired electron and is crucial for marks.

🧠 Exam Technique

  • Check that every radical equation has a dot clearly placed on the radical species.
  • Ensure propagation steps always consume a radical and produce a radical.

❌ Common Errors

  • Omitting UV light conditions from the initiation step.
  • Using ionic species (like Br⁻ or C₂H₅⁺) instead of neutral radicals.
  • Missing radical dots entirely in propagation equations.
Question Part (c)

Bond Angles and Shapes in Propene (C₃H₆)

✅ Correct Answer

  • Carbon 1 (saturated, sp³): Bond angle 109.5°, Name of shape: Tetrahedral.
  • Carbon 2 (unsaturated, sp²): Bond angle 120°, Name of shape: Trigonal planar.
  • Explanation: Carbon 1 has 4 bonding pairs of electrons which repel equally into a tetrahedral geometry. Carbon 2 has 3 regions of electron density around it (double bond counts as 1 single effective region of repulsion) which repel to maximum separation in a plane.

💡 Key Knowledge

  • VSEPR (Valence Shell Electron Pair Repulsion) theory dictates that electron pairs repel as far apart as possible.
  • Single bonds create tetrahedral geometry (109.5°), while double bonds combined with single bonds around a carbon atom create trigonal planar geometry (120°).

🧠 Exam Technique

  • Clearly set out answers in a table format for angles and shapes to gain immediate clarity marks.
  • Explicitly link the number of electron pairs/regions around each specific carbon atom to its resultant repulsion and shape.

❌ Common Errors

  • Treating the double bond on Carbon 2 as two separate regions of electron repulsion instead of counting it as one unit.
  • Confusing tetrahedral angles (109.5°) with trigonal planar angles (120°).

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.2 Electrons, bonding and structure · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.