OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 18

1 mark · Easy difficulty · Multiple Choice

Identify the statement that explains why 1-iodopropane hydrolyses faster than 1-bromopropane.

Practise this question

Question

Question 18: Multiple-choice question stating that the hydrolysis of 1-iodopropane by aqueous potassium hydroxide takes place at a faster rate than 1-bromopropane. It asks which statement explains the different rates of reaction. The options are: A - Iodine is less reactive than bromine; B - The C–I bond enthalpy is less than the C–Br bond enthalpy; C - The C–I bond is less polar than a C–Br bond; D - The C–I bond length is shorter than the C–Br bond length.
Question text

18 The hydrolysis of 1-iodopropane by aqueous potassium hydroxide takes place at a faster rate

than 1-bromopropane.

Which statement explains the different rates of reaction?

A Iodine is less reactive than bromine.

B The C–I bond enthalpy is less than the C–Br bond enthalpy.

C The C–I bond is less polar than a C–Br bond.

D The C–I bond length is shorter than the C–Br bond length.

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme table row showing the correct answer for question 18 is B, awarded 1 mark.

18 B 1

How to answer it

Rates of Hydrolysis in Haloalkanes

📋 What this question tests

This question assesses your understanding of the factors governing the rate of nucleophilic substitution (hydrolysis) in haloalkanes, specifically contrasting the influence of carbon–halogen bond enthalpy against bond polarity down Group 7.

Question 18

Multiple Choice: Factors Affecting Rate of Haloalkane Hydrolysis

✅ Correct Answer

B: The C–I bond enthalpy is less than the C–Br bond enthalpy.

Award: [1 Mark] for selecting option B.

💡 Key Knowledge

  • Bond Enthalpy vs. Polarity: Even though the C–F and C–Cl bonds are the most polar due to higher electronegativity, bond enthalpy is the overriding factor that determines the rate of reaction.
  • Trend down Group 7: Atomic radius increases down the group, so shared electrons are further from the halogen nucleus. This makes the bond longer and weaker:
    C–F > C–Cl > C–Br > C–I (decreasing bond strength).
  • Reaction Rate: The lower the bond enthalpy, the easier the carbon–halogen bond is broken, leading to a lower activation energy and a faster rate of hydrolysis:
    1-iodopropane > 1-bromopropane > 1-chloropropane .

🧠 Exam Technique & Option Elimination

  • Option A: While elemental iodine is less reactive as an oxidising agent than bromine, elemental halogen reactivity is irrelevant to the breaking of covalent bonds in an organic haloalkane molecule.
  • Option B (Correct): Weaker bonds break faster. The C–I bond requires less energy to break than the C–Br bond, directly explaining the faster rate.
  • Option C: While it is factually true that C–I is less polar than C–Br, greater polarity would actually attract nucleophiles (like OH⁻) more strongly. Thus, polarity fails to explain why iodoalkanes react faster.
  • Option D: Iodine has a larger atomic radius than bromine (more electron shells), making the C–I bond longer, not shorter.

❌ Common Student Errors

  • The Polarity Trap (Choosing C): Many students assume that since nucleophilic attack begins with an attraction to the δ+ carbon atom, the most polar bond must react fastest. Remember: bond enthalpy always outweighs bond polarity when comparing rates of haloalkane hydrolysis.
  • Confusing Length with Strength: Believing shorter bonds are present in iodoalkanes (Option D). Larger atoms form longer, weaker bonds.

Topics

Module 4: Core organic chemistry · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.