Edexcel A-Level Chemistry Paper 3, June 2024: Question 6

9 marks · Medium difficulty · Calculations

Calculate the relative molecular mass of an ester using titration and stoichiometry data, and interpret its high-resolution proton NMR spectrum by explaining splitting patterns and predicting chemical shifts, relative peak areas, and splitting patterns.

Practise this question

Question

Question 6 features an ester Q with molecular formula C8H16O2. Part (a) asks to show that the relative molecular mass of Q is 144 using titration data where 2.07 g of Q yielded a carboxylic acid at a 78% yield, which produced 269 cm3 of carbon dioxide upon reaction with sodium hydrogencarbonate. Part (b) provides the structure of ester Q, showing proton environment A circled on the hydrogen attached to the carbonyl carbon, and asks to explain the splitting pattern for peak A, label remaining proton environments, and predict chemical shifts, peak areas, and splitting patterns for those environments.
Question text

6 An ester Q has the molecular formula C8H16O2.

(a) When hydrolysed, 2.07g of Q formed a carboxylic acid with a 78% yield.

Once separated, the carboxylic acid was neutralised using excess

sodium hydrogencarbonate solution.

NaHCO + RCOOH → RCOO–Na+ + H O + CO

32 2

In this reaction, 269 cm3 of carbon dioxide gas was produced at room temperature

and pressure.

Show that these data confirm that the relative molecular mass of Q is 144.

(3)

(b) The high resolution proton NMR spectrum of ester Q was obtained.

The structure of the ester Q is shown.

CH3

O H2C

H2

C C C CH3

H O

H2C

proton environment A CH3

(i) Explain the expected splitting pattern for the peak due to proton

environment A circled in the structure.

(2)

(ii) Label the structure to indicate the remaining equivalent proton environments

in ester Q.

(1)

(iii) Predict the chemical shifts, relative peak areas and splitting patterns in the

high resolution proton NMR spectrum, due to the proton environments you

have labelled.

Do not consider proton environment A.

(3)

… *P74455A01232*

(Total for Question 6 = 9 marks)

Mark scheme

Show the mark scheme The mark scheme provides step-by-step calculation methods for part (a) to determine moles of CO2, moles of ester Q at 100% yield, and its Mr of 144. For part (b)(i), it awards marks for identifying peak A as a singlet due to no adjacent hydrogens. For part (b)(ii), it shows the labelling of the other proton environments. For part (b)(iii), it provides a table of expected chemical shifts, splitting patterns (quartet and triplet), and relative peak areas (6 and 9) for the remaining environments.

Question

Answer Additional Guidance Mark

Number

6(a) Example of calculation (3)

(method 1)

• (M1) calculation of moles of carbon dioxide (1) n (CO ) = 269 ÷ 24000 = 0.011208 / 1.1208 × 10–2 (mol)

n (ester) = 0.011208 × (100 ÷ 78) = 0.014370 / 1.4370 × 10–2 (mol)

• (M2) calculation of moles of Q based on 100%

yield (1)

(method 2)

(1) V (CO ) = 269 × (100 ÷ 78) = 344.87 (cm3)

• (M1) calculation of volume at 100% yield 2

(1) n (CO ) = 344.87 ÷ 24000 = 0.01437 / 1.437 × 10–2 (mol) = n (ester)

• (M2) calculation of moles at 100% yield 2

(method 1 and method 2)

• (M3) show how to calculate Mr of Q (1) Mr of ester = 2.07 ÷ 0.014370 = 144.05 / 144

or

Justification 2.07 ÷ 144 = 1.4375 × 10–2 which is ca. 1.4370 × 10–2

Accept calculation variations which involve the 78% conversion,

the molar volume of gas and the starting mass of 2.07

Ignore SF except 1 SF in M1 and M2

Ignore intermediate units even if incorrect

Note: Use of the formula pV=nRT with T =298 and P =1 × 105

gives a RMM=148 and scores (3)

Question

Answer Additional Guidance Mark

Number

6(b)(i) An explanation that makes reference to the following points: Ignore any comments about chemical shifts (2)

• (peak due to A) is a singlet (1) Allow no splitting

• as there are no adjacent carbon atoms with hydrogen atoms/ Use of n+1 rule

as the carbon is (only) bonded to oxygen atoms (1)

Do not award if it is clear that the methanoate

carbon is being referred to as the adjacent carbon

Question

Answer Additional Guidance Mark

Number

6(b)(ii) An answer that makes reference to the following points: (1)

Accept any clear means of labelling of the two different

C hydrogen environments

Ignore labelling of just one CH3 and one CH2 unless it is clearly

stated that the other groups (of each sort) are equivalent

B

Question

Answer Additional Guidance Mark

Number

6(b)(iii) (3)

• chemical shifts (1) Chemical shift (∂) Splitting Relative peak

/ ppm pattern of peak area

• splitting patterns (1) 6

B 0 − 1.9 quartet

Allow 2

• relative peak areas (1) 9

C 0 − 1.9 triplet

Allow 3

Allow any single chemical shift value or range within the MS range

Allow four splits for quartet and three splits for triplet

Ignore reference to proton environment A

Additional proton environments max 1 for chemical shifts

If no other mark awarded then allow (1) for either B or C given correctly

for chemical shift and splitting and peak area (row)

(Total for Question 6 = 9 marks)

How to answer it

Ester Hydrolysis and High-Resolution Proton NMR Spectroscopy

What this question tests

This multi-step question assesses core organic chemistry competencies: stoichiometry calculations involving percentage yield and molar gas volume ( 24000 cm³ mol⁻¹ ), interpretation of high-resolution proton NMR splitting patterns via the n+1 rule, identification of chemical environments of protons, and prediction of chemical shifts, relative peak areas, and splitting patterns.

Question 6 (a) — Stoichiometry and Molar Mass Calculation

Proving the Relative Molecular Mass of Q (3 Marks)

📐 Step-by-Step Calculation (Method 1)

  1. Find moles of CO₂ produced:
    n(CO₂) = 269 ÷ 24000 = 0.011208 mol (using molar volume at rtp).
  2. Adjust for 78% percentage yield to find 100% yield moles of Q:
    Since 1 mole of carboxylic acid produces 1 mole of CO₂, and ester Q gives 1 mole of carboxylic acid upon hydrolysis:
    n(Q) = 0.011208 × (100 ÷ 78) = 0.01437 mol .
  3. Calculate Mr of Q:
    Mr = mass ÷ moles = 2.07 ÷ 0.01437 = 144.05 (rounds to 144 ).

🧠 Exam Technique & Mark Breakdown

  • (M1): Awarded for correct calculation of moles of carbon dioxide using 24000 cm³ mol⁻¹ .
  • (M2): Awarded for factoring in the 78% yield correctly to calculate the 100% theoretical moles of the ester.
  • (M3): Awarded for showing that mass divided by moles equals 144.

❌ Common Student Errors

  • The Ideal Gas Equation Trap: Using pV = nRT with standard room conditions ( P = 1 × 10⁵ Pa , T = 298 K ) yields an Mr = 148 , which loses the final mark because the question explicitly specifies room temperature and pressure where 24 dm³ mol⁻¹ must be used.
  • Yield Inversion: Multiplying by 78 ÷ 100 instead of dividing, which miscalculates the original amount of reactant.
Mark Allocation: 3 marks total. Intermediate rounding errors are ignored as long as the final value clearly demonstrates an Mr of 144.
Question 6 (b)(i) — Proton NMR Splitting Patterns

Explaining the Splitting Pattern of Environment A (2 Marks)

✅ Correct Answer

  • The peak is a singlet (1 mark) .
  • Because there are no adjacent carbon atoms with hydrogen atoms, or because the carbon is only bonded to oxygen atoms / carbonyl carbon with no neighbouring hydrogens (1 mark) .

💡 Key Knowledge: The n+1 Rule

Splitting is caused by spin-spin coupling with non-equivalent protons on adjacent carbon atoms. If a carbon has n equivalent protons attached to adjacent carbons, the peak splits into n + 1 peaks.

❌ Common Errors

  • Vague references to "no adjacent hydrogens" without clarifying the absence of adjacent carbon-bonded hydrogens.
  • Incorrectly treating the oxygen atom as an adjacent proton-bearing unit.
Mark Allocation: 2 marks. Do not award the second mark if it is implied that the methanoate carbon itself counts as an adjacent carbon with hydrogens.
Question 6 (ii) — Identifying Equivalent Protons

Labelling Remaining Proton Environments (1 Mark)

✅ Correct Labelling Strategy

To gain the mark, clearly label the two remaining distinct proton environments on the provided skeletal structure:

  • Environment B: The central CH₂ group (or the group of equivalent CH₂ protons in the branched structure).
  • Environment C: The surrounding equivalent CH₃ methyl groups.

🧠 Examiner Guidance

Accept any clear method of annotating/labeling the two different sets of hydrogen environments. Ensure distinct letters (e.g., B and C) are assigned clearly to avoid ambiguity.

Mark Allocation: 1 mark. Ignore labeling of only a single CH₃ or CH₂ unless it is explicitly stated that all equivalent groups of that sort share the environment.
Question 6 (b)(iii) — NMR Spectral Data Prediction

Predicting Shifts, Splitting, and Peak Areas (3 Marks)

✅ Expected Spectral Data Table

  • Environment B ( CH₂ ):
    Chemical shift: 0 – 1.9 ppm | Splitting: quartet | Relative peak area: 6 (allow 2)
  • Environment C ( CH₃ ):
    Chemical shift: 0 – 1.9 ppm | Splitting: triplet | Relative peak areas: 9 (allow 3)

🧠 Mark Allocation Breakdown

  • Mark 1: Chemical shifts within the correct ranges (alkyl protons).
  • Mark 2: Correct splitting patterns (quartet for B due to adjacent CH₃ , triplet for C due to adjacent CH₂ ).
  • Mark 3: Correct relative peak areas matching the ratio of hydrogens (or simplest whole-number ratio).
Mark Allocation: 3 marks total (one per correct row/property category). If no other marks are awarded, a consolation mark may be given if either B or C is fully correct across chemical shift, splitting, and peak area combined.

Topics

Physical Chemistry · Organic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 17: Organic Chemistry II · Topic 19: Modern Analytical Techniques II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.