OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 18

1 mark · Medium difficulty · Multiple Choice

Identify the statement that explains why 1-chlorobutane is hydrolysed at a slower rate than 1-bromobutane with aqueous sodium hydroxide.

Practise this question

Question

Multiple choice question 18. The question states that when heated with NaOH(aq), 1-chlorobutane is hydrolysed at a slower rate than 1-bromobutane and asks which statement explains the different rates. Four options A, B, C, and D are provided comparing carbon-bromine and carbon-chlorine bond enthalpies and polarities. There is a box for the answer and a mark allocation of [1] at the bottom right.
Question text

18 When heated with NaOH(aq), 1-chlorobutane is hydrolysed at a slower rate than 1-bromobutane.

Which statement explains the different rates?

A The C–Br bond enthalpy is greater than the C–Cl bond enthalpy.

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

C The C–Br bond is less polar than the C–Cl bond.

D The C–Br bond is more polar than the C–Cl bond.

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 18 is B, awarding 1 mark.

18 B 1 AO1.1

How to answer it

Halogenoalkane Hydrolysis Rates

What this question tests:
This question assesses your understanding of nucleophilic substitution reactions in halogenoalkanes, specifically how carbon-halogen bond enthalpies dictate the rate of hydrolysis (reactivity) down Group 7.
Question Multiple Choice Analysis

Core Concept: Bond Enthalpy vs. Bond Polarity

✅ Correct Answer: B

The C–Br bond enthalpy is less than the C–Cl bond enthalpy.

Because the C–Br bond is weaker (requires less energy to break), it breaks much faster during nucleophilic substitution with NaOH(aq), leading to a faster rate of hydrolysis compared to 1-chlorobutane.

💡 Key Knowledge

  • Bond Enthalpy Trend: Down Group 7 (from F to I), atomic radius increases, overlapping orbitals are less effective, and carbon-halogen bonds become longer and weaker (lower bond enthalpy).
  • Reactivity Order: iodoalkanes > bromoalkanes > chloroalkanes > fluoroalkanes.

🧠 Exam Technique

Be careful not to confuse bond enthalpy with bond polarity! While C–Cl is more polar than C–Br (due to fluorine/chlorine having higher electronegativities), bond enthalpy is the dominant factor determining the rate of bond breakage and overall chemical reactivity in halogenoalkanes.

❌ Common Errors

Students frequently fall into the trap of selecting options based on bond polarity (choosing D or C). Remember that bond polarity would suggest C–Cl reacts faster, but experimental evidence proves the exact opposite because bond breaking is the rate-determining factor.

Mark Scheme Award: 1 mark for selecting option B (AO1.1 Knowledge and Understanding).

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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.