Edexcel A-Level Chemistry Paper 2, June 2018: Question 6
16 marks · Hard difficulty · Extended Writing
Devise an experiment to compare hydrolysis rates of halogenoalkanes, explain trends, define chirality for butan-2-ol enantiomers, and compare SN1 and SN2 mechanisms.
Practise this questionQuestion
Question text
6 This is a question about the hydrolysis of halogenoalkanes.
(a) Devise an experiment, giving outline details only, that would enable the relative
rates of hydrolysis of halogenoalkanes to be compared.
(5)
(b) Explain the trend in the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and
1-iodobutane.
(2)
(c) The product of the hydrolysis of 2-bromobutane is butan-2-ol. Both molecules are chiral.
12 State what is meant by the term chiral, using three-dimensional diagrams of the
enantiomers of butan-2-ol to illustrate your answer.*P52293A01224*
(3)
*(d) Compare and contrast the mechanism of hydrolysis, using aqueous potassium hydroxide,
of the primary halogenoalkane, RCH2X, with that of the tertiary halogenoalkane, R3CX.
Include diagrams of any intermediate or transition state.
Curly arrows are not required.
(6)
… *P52293A01324*
… *P52293A01424*
(Total for Question 6 = 16 marks)
Mark scheme
Show the mark scheme
Question
Acceptable Answer Additional guidance Mark
Number
6(a) An answer that gives reference to the following (5)
(M1) use of ethanol (as a solvent) (1) Allow “alcohol”
(M2) use of silver nitrate (solution) (1) Do not award ammoniacal silver
nitrate
Ignore use of nitric acid
(M3) equal amounts used of each halogenoalkane (1) Allow equal volumes/equal stated
volumes
(M4) measure the time taken for precipitate to form (1) Allow “time for cross to disappear”
Do not award for a colour to form.
M4 dependent on M2 or near miss.
(M5) use a water bath (to control a raised temperature)
(1)
If hydroxide (ions) used for
hydrolysis then measuring the
reaction is too quick, so no M4. The
solution would need to be acidified
before the addition of silver nitrate
if M2 is to be awarded.
If hydrochloric acid is used, then
only M1, M3 and M5 can be scored
Question
Acceptable Answer Additional guidance Mark
Number
6(b) An explanation that makes reference to the following Accept reverse arguments (2)
Incorrect trend scores (0)
the reaction rate is in the order
1-chlorobutane<1-bromobutane<1-iodobutane
(1)
because the C-Cl bond is stronger than the C-Br Allow ‘the C-Cl bond is the strongest’
bond which is stronger than the C-I bond (1) Ignore any reasoning given
Do not award if reference is made to
the bonding of the halide (ion)
Question
Acceptable Answer Additional guidance Mark
Number
6(c) Diagram must be 3-dimensional, i.e. (3)
include ‘wedges’.
Allow Br instead of OH
Ignore attachment of –OH, CH3 and
C2H5 groups
Forms (two) isomers which are non-superimposable (1) Standalone mark
Allow
a chiral carbon has four different
groups attached (so they are non-
superimposable)
Do not award has four different
‘molecules’ attached
Question
Acceptable Answer Additional Guidance Mark
Number
6(d) This question assesses the student’s ability to show a Guidance on how the mark scheme (6)
coherent and logically structured answer with linkages and should be applied:
fully sustained reasoning. The mark for indicative content should
be added to the mark for lines of
Marks are awarded for indicative content and for how the reasoning. For example, a response with
answer is structured and shows lines of reasoning. four indicative marking points that is
partially structured with some linkages
The following table shows how the marks should be and lines of reasoning scores 4 marks (3
awarded for indicative content. marks for indicative content and 1 mark
Number of indicative Number of marks awarded for partial structure and some linkages
marking points seen in for indicative marking points and lines of reasoning).
answer If there were no linkages between the
64 points, then the same indicative
5-4 3 marking points would yield an overall
3-2 2 score of 3 marks (3 marks for indicative
content and zero marks for linkages).
The following table shows how the marks should be
awarded for structure and lines of reasoning
Number of marks awarded
for structure of answer and
sustained lines of reasoning
Answer shows a coherent 2 In general it would be expected that 5
logical structure with linkages or 6 indicative points would get 2
and fully sustained lines of reasoning marks, and 3 or 4 indicative
reasoning demonstrated points would get 1 mark for reasoning,
throughout and 0, 1 or 2 indicative points would
Answer is partially structured 1 score zero marks for reasoning.
with some linkages and lines of
reasoning
Answer has no linkages 0 If there is any incorrect chemistry, deduct
between points and is mark(s) from the reasoning. If no
unstructured reasoning mark(s) awarded do not deduct
mark(s).
More than one indicative marking point
Indicative content may be made within the same comment
or explanation
(similarity)(both) are nucleophilic substitution (1) Words needed at least once provided
SN1 and SN2 are given
Hydrolysis mechanism for RCH2X/primary is SN2 via
a transition state and R3CX/tertiary is SN1 via a
carbocation/intermediate (1)
RCH X and OH− in the RDS (1) Allow “both/two species in the RDS”
R3CX only in the RDS (1) Allow correct rate equations for IP3 and
IP4
(RCH X forms a transition state with OH−)
diagram, including dotted lines and charge (1)
Allow “-“ either on the “OH” or the “X”
Ignore point of attachment of OH
Ignore dipoles within structure
(R3CX forms a carbocation / intermediate)
diagram, including charge (1)
Ignore shape
Ignore references to comparative rates
of reaction between 1o and 3o even if
incorrect
Ignore references to optical activity.
(Total for Question 6 = 16 marks)
How to answer it
Edexcel A-Level Chemistry: Hydrolysis of Halogenoalkanes
What this question tests
This multi-topic question evaluates your understanding of aliphatic chemistry mechanisms and practical techniques. It tests your ability to design a rate-comparison experiment, apply carbon-halogen bond enthalpy data to reactivity trends, draw 3D stereochemical representations of chiral enantiomers, and compare contrasting reaction mechanisms (Sn1 vs Sn2) using structural intermediates and transition states.
Devising an Experiment to Compare Rates of Hydrolysis
✅ Correct Answer / Mark Scheme
- M1: Use of ethanol (as a solvent/common solvent).
- M2: Use of silver nitrate solution.
- M3: Equal amounts (volumes/concentrations) of each halogenoalkane.
- M4: Measure the time taken for a precipitate to form.
- M5: Use a water bath to control/maintain a constant temperature.
❌ Common Errors & Traps
- Forgetting that halogenoalkanes are insoluble in water, making a mutual solvent like ethanol essential to mix them with aqueous silver nitrate.
- Using aqueous sodium/potassium hydroxide directly for rate measurement without acidification beforehand will react with the silver ions, forming a brown Ag₂O precipitate and ruining the test.
Explaining the Trend in Rates of Hydrolysis
✅ Correct Answer / Mark Scheme
- Rate order: 1-chlorobutane < 1-bromobutane < 1-iodobutane.
- Reason: The C-Cl bond is stronger than the C-Br bond, which is stronger than the C-I bond (decreasing bond enthalpies down the group).
💡 Key Knowledge
As you go down Group 7, the halogen atoms increase in atomic radius, leading to poorer overlap of orbitals with carbon. This results in longer, weaker carbon-halogen bonds that require less activation energy to break.
Defining Chirality and Drawing Enantiomers of Butan-2-ol
✅ Correct Answer / Mark Scheme
- 3D Diagrams: Two mirror-image tetrahedral structures of butan-2-ol featuring a central carbon attached to -H, -OH, -CH₃, and -C₂H₅ groups, utilizing wedged and dashed bonds.
- Definition: Forms two isomers which are non-superimposable mirror images (due to a chiral carbon with four different groups attached).
🧠 Exam Technique
When drawing 3D structures, make sure your wedges and dashes clearly indicate spatial orientation. To show enantiomers easily, draw the first structure, then draw a vertical dashed mirror line and reflect every group accurately.
Comparing Primary (Sn2) and Tertiary (Sn1) Hydrolysis Mechanisms
💡 Key Comparison Points (Indicative Content)
- Similarities: Both are nucleophilic substitution reactions.
- Primary (RCH₂X): Proceeds via an Sn2 mechanism involving a 5-coordinate transition state containing both the nucleophile and halogenoalkane in the rate-determining step (RDS).
- Tertiary (R₃CX): Proceeds via an Sn1 mechanism involving a planar carbocation intermediate, where only the halogenoalkane is in the RDS.
📐 Visualising Intermediates & Transition States
- Sn2 Transition State: Draw central carbon bonded to R and 3 other groups in a trigonal plane, with partial bonds (dotted lines) to incoming HO⁻ and outgoing X⁻ , enclosed in square brackets with a negative charge [⚲]⁻.
- Sn1 Intermediate: Draw a planar carbocation intermediate featuring the central carbon bonded to 3 groups with a positive charge ( C⁺ ).
❌ Common Errors
Students frequently confuse the terms transition state (maxima on a reaction profile, cannot be isolated) and intermediate (minima on a profile, species that exists briefly). Ensure you label square brackets and partial bonds correctly for Sn2, and display the full positive charge on the carbocation for Sn1.
Topics
Organic Chemistry · Core Practicals · Core Practical 4: Investigate the hydrolysis of halogenoalkanes · Topic 6: Organic Chemistry I · Topic 17: Organic Chemistry II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.