Edexcel A-Level Chemistry AS Paper 2, June 2018: Question 6

10 marks · Medium difficulty · Short Open Response

Identify the reagent and conditions to convert but-2-ene-1,4-diol into butane-1,4-diol, describe its reaction classification, name another product made by this type of reaction, and describe how to prepare a standard solution of butanedioic acid from a solid.

Practise this question

Question

An organic synthesis question showing the conversion of but-2-ene-1,4-diol to butane-1,4-diol using reagent B. Part (a)(i) asks to identify reagent B and conditions. Part (a)(ii) is a multiple-choice question on the reaction type (hydrolysis, oxidation, reduction, substitution). Part (a)(iii) asks to name another commercially important product. Part (b) asks to describe how to prepare 250 cm^3 of a butanedioic acid solution of approximately 0.0500 mol dm^-3 from the solid given its molar mass.
Question text

6 This question is about the synthesis and reactions of butane-1,4-diol.

(a) Butane-1,4-diol can be synthesised from but-2-ene-1,4-diol, by reaction with a

reagent, B.

O H O H O H O H

CH2 CH2 + B → CH2 CH2

CH CH CH2 CH2

but-2-ene-1,4-diol butane-1,4-diol

(i) Identify reagent B and state suitable conditions for this reaction.

(2)

(ii) This reaction is best described as

(1)

A hydrolysis

B oxidation

C reduction

D substitution

(iii) Name one other commercially important product that can be manufactured

by this type of reaction with the alkene group.

(1)

*(b) Butane-1,4-diol can be oxidised to form butanedioic acid. The molecular formula

of butanedioic acid is C4H6O4 and it is a solid at room temperature.

Describe how you would make 250 cm3 of a solution of butanedioic acid with an

accurately known concentration of approximately 0.0500 mol dm−3.

Butanedioic acid is sufficiently soluble in water to achieve this concentration.

[Molar mass of butanedioic acid = 118 g mol−1]

(6)

… *P51460RA01524*

… *P51460RA01624*

(Total for Question 6 = 10 marks)

Mark scheme

Show the mark scheme The mark scheme provides the accepted answers for question 6. Part (a)(i) accepts hydrogen and a nickel catalyst. Part (a)(ii) gives C (reduction). Part (a)(iii) accepts margarine. Part (b) uses a 6-mark levels-of-response rubric detailing calculations for mass, weighing by difference or direct weighing, dissolving in distilled water, transferring to a volumetric flask with washings, and making up to the mark.

Question Acceptable Answer Additional Guidance Mark

Number

6(a)(i) Reagent: mark independently

B is hydrogen / H2 (gas) (1)

Condition:

nickel/ Ni (catalyst) (1) allow any other suitable transition metal

catalysts eg Pt, Pd

ignore additional information relating to the

support for the catalyst

ignore references to heating/pressure/UV (2)

Question Acceptable Answer Mark

Number

6(a)(ii) The only correct answer is C

A is not correct because water is not involved

B is not correct because there is no increase in number of oxygen atoms

D is not correct because no substitution has taken place (1)

Question Acceptable Answer Additional Guidance Mark

Number

6(a)(iii) margarine allow liquid coal

allow butter substitute

do not award just butter (1)

Question Acceptable Answer Additional Guidance Mark

Number

*6(b) This question assesses a student’s ability to show a Guidance on how the mark scheme should be

coherent and logically structured answer with linkages applied:

and fully-sustained reasoning.

Marks are awarded for indicative content and for how The mark for indicative content should be added

to the mark for lines of reasoning. For example,

the answer is structured and shows lines of reasoning.

an answer with five indicative marking points

The following table shows how the marks should be that is partially structured with some linkages

awarded for indicative content. and lines of reasoning, scores 4 marks (3 marks

Number of indicative Number of marks awarded for indicative content and 1 mark for partial

marking points seen in for indicative marking structure and some linkages and lines of

answer points reasoning).

5-4 3 If there are no linkages between points, the

3-2 2 same five indicative marking points would yield

11 an overall score of 3 marks (3 marks for

00 indicative content and no marks for linkages).

The following table shows how the marks should be In general it would be expected that 5 or 6

awarded for structure and lines of reasoning. indicative points would get 2 reasoning marks,

Number of marks awarded and 3 or 4 indicative points would get 1 mark for

for structure and sustained reasoning, and 0, 1 or 2 indicative points would

lines of reasoning score zero marks for reasoning.

Answer shows a coherent

and logical structure with 2 If there is any incorrect chemistry, deduct

linkages and fully sustained mark(s) from the reasoning. If no reasoning

lines of reasoning mark(s) awarded do not deduct mark(s).

demonstrated throughout.

Answer is partially structured Comment: Look for the indicative marking

points first, then consider the mark for the

with some linkages and lines 1

structure of the answer and sustained line of

of reasoning.

reasoning.

Answer has no linkages

between points and is 0

unstructured.

(6)

Indicative content:

Ignore anything to do with oxidation even if

incorrect

example of calculation

calculate approximate mass of solute to be

weighed out 0.050 mol dm-3 = 0.050 x 118 g dm-3

=5.90 g dm-3

=1.47(5) g in 250 cm3

details of how to weigh out required mass

do not award just ‘weigh by difference’

transfer solute to beaker/conical flask and add

distilled/deionised water and dissolve transfer of solute directly to volumetric flask

gets IP3 and IP4 but must mention

dissolving for IP3

transfer to (250 cm3) volumetric flask

any mention of volumetric/graduated flask

scores IP4

add washings from beaker

direct transfer from weighing container to

volumetric flask must mention washing of

solute into the flask (e.g. through funnel).

make up to mark/line and shake/invert (to mix).

mix on its own is insufficient

(Total for Question 6 = 10 marks)

How to answer it

Synthesis and Reactions of butane-1,4-diol

What this question tests

This question assesses core organic synthesis reagents and conditions (specifically alkene hydrogenation), reaction classification, commercial applications of alkenes, and standard practical chemistry techniques for preparing a standard solution from a solid solute.

Part (a)(i) — Reagents and Conditions

Identifying Reagent B

✅ Correct Answer

Reagent: Hydrogen / H₂ (gas)

Condition: Nickel / Ni (catalyst)

2 marks available (1 mark per independent point; Pt or Pd also accepted).

💡 Key Knowledge

Converting an alkene into an alkane (or in this case, hydrogenating the carbon-carbon double bond while leaving the hydroxyl groups intact) requires electrophilic addition of hydrogen gas over a transition metal catalyst.

Part (a)(ii) — Reaction Classification

Classifying the Conversion

✅ Correct Answer

C — reduction

1 mark available.

❌ Common Errors & Examiner Insight

Students often incorrectly choose oxidation (B) due to confusion over functional group transformations. Remember that adding hydrogen to a molecule is defined as reduction in organic chemistry.

Part (a)(iii) — Commercial Applications

Commercial Product from Alkene Hydrogenation

✅ Correct Answer

Margarine (or liquid coal / butter substitute)

1 mark available. Note: "butter" alone is not awarded.

🧠 Exam Technique

This tests industrial context. The catalytic hydrogenation of unsaturated vegetable oils (alkenes) produces saturated fats used to manufacture margarine.

Part (b) — Extended Writing & Calculation (6 marks)

Preparing a Standard Solution of Butanedioic Acid

📐 Step-by-Step Calculation

  1. Calculate moles required:
    Moles = Concentration × Volume = 0.0500 mol dm⁻³ × (0.250 dm³) = 0.0125 mol
  2. Calculate mass required:
    Mass = Moles × Molar Mass = 0.0125 mol × 118 g mol⁻¹ = 1.475 g (or ~1.48 g)

💡 Practical Procedure (Indicative Content)

  • Calculate the approximate mass of solid needed (~1.48 g).
  • Weigh the solid accurately using a balance (weighing by difference).
  • Transfer the solid to a beaker, add distilled/deionised water, and stir to dissolve completely.
  • Transfer the solution quantitatively into a 250 cm³ volumetric flask.
  • Rinse the beaker and glass rod with distilled water and add washings to the flask.
  • Make up to the graduation mark with distilled water, then stopper and invert/shake to mix thoroughly.

🧠 Marking Structure & Communication

This is a levels-of-response question worth 6 marks total: up to 4 marks for indicative content (covering the calculation and practical steps) and up to 2 marks for the structure and logical flow of the explanation.

❌ Common Practical Pitfalls

  • Failing to mention dissolving the solid in a beaker *before* transferring to the volumetric flask.
  • Forgetting to rinse the beaker/glass rod and transfer the washings into the volumetric flask.
  • Inverting/mixing the flask inadequately at the end.

Topics

Organic Chemistry · Core Practicals · Physical Chemistry · Topic 6: Organic Chemistry I · Core Practical 2: Preparation of a standard solution from a solid acid · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.