Edexcel A-Level Chemistry Paper 2, November 2020: Question 6

13 marks · Medium difficulty · Practical Techniques and Data Analysis

Investigate the kinetics, mechanism, and optical activity of the nucleophilic substitution of bromoalkanes with hydroxide ions.

Practise this question

Question

A multi-part exam question about the nucleophilic substitution reaction of a bromoalkane (RBr) with hydroxide ions. It includes a table of experimental concentration-time data, a grid for plotting a graph of [RBr] against time, questions on half-lives, rate orders, mechanism with curly arrows for an SN1 pathway, and the optical inactivity of the product from 2-bromobutane.
Question text

6 A bromoalkane, RBr, reacts with aqueous hydroxide ions in a nucleophilic substitution

reaction.

RBr + OH− → R OH + Br−

This reaction is first order with respect to the bromoalkane and the rate equation is

rate = k[RBr]1[OH−]x

where x is the order of the reaction with respect to hydroxide ions.

In an experiment, a sample of the bromoalkane was added to a large excess

of aqueous sodium hydroxide and the concentration of the bromoalkane was

determined at regular time intervals.

Results

Time / s [RBr] / mol dm–3

0 0.100

30 0.065

60 0.042

90 0.028

120 0.019

150 0.014

(a) This experiment is carried out using the bromoalkane dissolved in ethanol and

the hydroxide ions dissolved in water.

Give a reason why a solution of hydroxide ions dissolved in pure ethanol should

not be used.

(1)

(b) Plot a graph of [RBr] against time.

(3)

*P62669A01832*

(c) Explain how the graph shows that the reaction is first order with respect to RBr.

Include the values of two consecutive half-lives.

You must show your working for the half-lives on the graph.

(2)

(d) The experiment was repeated using equal concentrations of RBr and varying the

concentration of hydroxide ions.

A graph was plotted of the results. 19

*P62669A01932*

Rate

/ mol dm−3 s−1

[OH−] / mol dm−3

(i) Deduce the value of x in the rate equation.

rate = k[RBr]1[OH−]x

(1)

(ii) Give the mechanism for the reaction that is consistent with the orders of reaction

with respect to R Br and hydroxide ions.

Include curly arrows and relevant lone pairs.

(3)

(e) 2-bromobutane can react with aqueous hydroxide ions by an SN1 mechanism.

Explain why the butan-2-ol produced from a single optical isomer of

2-bromobutane, using this mechanism, is not optically active.

(3)

… 20

*P62669A02032*(Total for Question 6 = 13 marks)

Mark scheme

Show the mark scheme The mark scheme providing answers and guidance for all parts of question 6, including expected text explanations for solvent choice, an example of the completed concentration-time graph, half-life calculations, rate order deduction, curly arrow mechanisms for the SN1 reaction, and optical isomerism explanations.

How to answer it

Kinetics, Reaction Mechanisms & Optical Activity

What this question tests

This comprehensive Edexcel A-Level Chemistry question tests your understanding of reaction kinetics (interpreting concentration-time graphs, constant half-lives, and zero-order reactions), organic reaction mechanisms (curly arrow notation for SN1 nucleophilic substitution), and stereochemistry (formation of racemic mixtures via planar carbocation intermediates).

Question Part (a)

Solvent Selection in Nucleophilic Substitution

✅ Correct Answer

Hydroxide ions dissolved in pure ethanol would cause an elimination reaction (forming an alkene) instead of substitution.

💡 Key Knowledge

Ethanolic OH⁻ acts predominantly as a strong base (proton acceptor), whereas aqueous OH⁻ acts as a nucleophile. Water as a solvent encourages substitution, while alcohol encourages elimination.

Mark: 1 mark for referencing an elimination reaction / formation of an alkene.
Question Part (b)

Plotting a Concentration-Time Graph

✅ Correct Answer

A smooth downward curving line plotted from the provided data set, showing concentration decreasing over time.

🧠 Exam Technique

  • Axes: Must be fully labelled as [RBr] / mol dm⁻³ (y-axis) and Time / s or T / s (x-axis). Do not omit square brackets for concentration!
  • Scale: Points must use at least half of the grid in both directions.
  • Curve: Draw a smooth curve of best fit. Never connect points point-to-point with straight lines.
Marks: 3 marks total (1 for labelled axes with units, 1 for suitable scale using >50% of grid, 1 for correct coordinate plots and smooth curve).
Question Part (c)

Deducing Order from Half-Life

✅ Correct Answer

First half-life = 50 s (from 0.100 to 0.050 mol dm⁻³). Second half-life = 50 s (from 0.050 to 0.025 mol dm⁻³). Because the consecutive half-lives are constant, the reaction is first order with respect to RBr.

📐 Working Out Half-Lives

  1. Find initial concentration (0.100) and halve it to get 0.050 mol dm⁻³. Read corresponding time off the graph (~50 s).
  2. Find half of 0.050 (0.025 mol dm⁻³) and read the time when concentration drops to this value (~100 s).
  3. Calculate time difference: 100 s - 50 s = 50 s. State clearly that half-lives are constant.

❌ Common Errors

Students lose marks here by failing to show construction lines directly on the graph grid to prove how they read their half-life values.

Marks: 2 marks (1 for values of two consecutive half-lives with working shown on graph, 1 for stating reaction is first order due to constant half-lives).
Question Part (d)

Rate Equations and SN1 Mechanisms

✅ Correct Answer for (i)

x = 0 (zero order with respect to hydroxide ions). The horizontal line on the rate vs [OH⁻] graph proves that rate is independent of [OH⁻].

💡 Key Mechanism Details for (ii)

  • Step 1: Curly arrow starting from the C–Br bond going to (or just beyond) the Br atom to form a carbocation intermediate ( R⁺ ) and Br⁻ .
  • Step 2: Lone pair explicitly shown on the oxygen atom of the OH⁻ ion, with a curly arrow going from the lone pair to the positive carbon ( R⁺ ).

❌ Common Errors in Mechanisms

Starting curly arrows in the middle of open space rather than precisely from a bond or a lone pair results in lost marks.

Marks: (i) 1 mark for zero order. (ii) 3 marks for correct curly arrow notation and ionic species in an SN1 mechanism framework.
Question Part (e)

Stereochemistry and Optical Inactivity

✅ Correct Answer

The product is optically inactive because a racemic mixture (equimolar amounts of both optical isomers/enantiomers) is formed.

🧠 Explanation Breakdown

  1. The SN1 mechanism proceeds via a carbocation intermediate.
  2. This central carbon in the carbocation is trigonal planar (flat around the reaction site).
  3. The incoming nucleophile ( OH⁻ ) has an equal probability of attack from either side (above and below the plane).
  4. This produces equal quantities of both enantiomers, whose opposite optical rotations cancel each other out completely.

❌ Common Errors

Do not simply write "the molecule is planar"—you must specify that the carbocation intermediate or reaction site/carbon center is planar.

Marks: 3 marks total (1 for stating a racemic mixture/equal amounts of enantiomers forms, 1 for stating the carbocation intermediate is planar, 1 for equal probability of attack from both sides/above and below).

Topics

Organic Chemistry · Physical Chemistry · Topic 6: Organic Chemistry I · Topic 16: Kinetics II · Topic 17: Organic Chemistry II

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.