Edexcel A-Level Chemistry Paper 3, June 2022: Question 6

11 marks · Hard difficulty · Open Response

Determine the empirical formula of ester Q from combustion data and deduce its full structure using high-resolution proton NMR data.

Practise this question

Question

Question 6 about an ester Q with molecular formula C8H16O2. Part (a) asks to show that the empirical formula is C4H8O given masses of carbon dioxide and water produced from combustion. Part (b) provides high-resolution 1H NMR data in a table with chemical shift, splitting pattern, and relative peak area, and asks to complete a partial structure of Q and justify the answer.
Question text

6 An ester Q has the molecular formula C8H16O2.

(a) When burned in excess oxygen, 1.879g of Q formed 4.594g of carbon dioxide

and 1.879g of water.

Show that the empirical formula of Q is C4H8O.

(4)

(b) Data from the high resolution 1H (proton) NMR spectrum of the ester Q are shown

in the table.

Chemical shift (δ) / ppm Splitting pattern of peak Relative peak area

2.50 singlet 3

1.56 quartet 4

1.43 singlet 3

0.92 triplet 6

Part of the structure of*P67095A01436*Qis shown.

Complete the structure of Q.

Justify your answer by linking the proton environments in your structure to the

relative peak areas and the splitting pattern of the peaks.

(7)

O

C

H3C O

(Total for Question 6 = 11 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 6, detailing the 4 marks for empirical formula calculation in part (a) and the 7 marks for NMR peak assignments and complete structure determination in part (b).

5–4 3

3–2 2

The following table shows how the marks should be awarded for

structure and lines of reasoning.

Number of marks

awarded for In general it would be expected that 5 or 6

structure of answer indicative points would get 2 reasoning marks, and

and sustained line of 3 or 4 indicative points would get 1 mark for

reasoning reasoning, and 0, 1 or 2 indicative points would

Answer shows a coherent and logical score zero marks for reasoning.

structure with linkages and fully

sustained lines of reasoning

demonstrated throughout.

Answer is partially structured with 1

some linkages and lines of reasoning. General points to note

Answer has no linkages between 0 If there is any incorrect chemistry, deduct mark(s)

points and is unstructured. from the reasoning. If no reasoning mark(s)

awarded do not deduct mark(s).

Comment:

Look for the indicative marking points first, then consider the

mark for structure of answer and sustained line of reasoning

Indicative content If names and formulae are given, both must be

correct

• IP1 Type of reaction

Both reactions are (examples of) electrophilic substitution Do not award addition-elimination for substitution

• IP2 Products Allow these products shown as structures in

Benzene forms bromobenzene and equations, even if equations are not fully correct

phenol forms 2,4,6-tribromophenol Allow any feasible dibromophenol / tribromophenol

Ignore dibromobenzene / tribromobenzene

• IP3 Comparison of reactivity

Benzene is less reactive (than phenol) / phenol is more

reactive (than benzene)

• IP4 Conditions Allow Fe / FeCl3 / AlBr3 / AlCl3 / Lewis Acid

Benzene requires (a catalyst of) FeBr3 catalyst

and Allow Friedel-Crafts catalyst / halogen carrier

phenol does not require a catalyst / can react with just Can be shown in equation

bromine water Allow phenol reacts at room temperature

Ignore reference to heat / mechanism

Allow IP4 if stated that only benzene requires a

catalyst

• IP5 Lone pair Allow lone pair on oxygen is donated into the ring

(Phenol is more susceptible to electrophilic attack) because Allow OH for oxygen

the lone pair on the oxygen (atom in phenol) delocalises into

the ring / π system

• IP6 Electron density Allow activates the ring

Increasing the electron density of the ring / π system Do not award increases the electronegativity /

charge density of the ring

Penalise omission of ‘the ring / π system’ once only

in IP5 and 6

(Total for Question 5 = 6 marks)

Question

Answer Additional Guidance Mark

Number

6(a) Example of calculation (4)

C H O

• calculation of masses of carbon and hydrogen mass = 4.594 x 1.879 x 2/18 1.879 –

(1) 12/44 = 0.209(g) (1.253+0.209)

= 1.253(g) =0.417 (g)

• calculation of mass of oxygen (1) Moles = = 1.253 = 0.209 = 0.417

12 1 16

• calculation of moles of carbon, hydrogen and = 0.1044 = 0.209 0.0261

oxygen (1) Ratio 4 8 1

(Formula = C4H8O)

• calculation of ratio (and matched to empirical Note – no mark for C4H8O as this is given in the question so no

formula) (1) TE

Ignore SF except 1 SF in M1 and M2

Allow alternative methods, for example:

mol CO2 = 4.954/44 = 0.1044 (mol)

and

mol H2O = 1.879/18 = 0.1044 (mol) (1)

mol C8H16O2 = 1.879/144 = 0.013 (mol) (1)

ratio C8H16O2 : CO2 : H2O – conditional on moles of C8H16O2

= 1 : 8 : 8 or 0.5 : 4 : 4 (1)

this matches the balanced equation

C8H16O2 + 11O2 → 8CO2 + 8H2O /

C4H8O + 11/2O2 → 4CO2 + 4H2O (1)

Question

Answer Additional Guidance Mark

Number

6(b) An answer that makes reference to the following points: Allow credit for annotations on table in p14 and (7)

on labelled structures

Allow adjacent protons / hydrogens for protons

on adjacent C

Penalise H+ for protons once only

Peak at 2.50 ppm

• identified as CH3CO (as relative peak area = 3 / singlet so no Allow ester group / H-C-C=O / CH3 on left of

protons on adjacent C) (1) structure given is indicated

Do not award if aldehyde / ketone mentioned

Peak at 1.56 ppm

• 2 CH2 groups as relative peak area = 4 (1) Allow 4 protons / hydrogens

• (the 2 CH2 groups / hydrogen environment) next to CH3

groups as peak is a quartet (1)

Peak at 0.92 ppm

• 2 CH3 groups as relative peak area = 6 (1) Allow 6 protons / hydrogens

• (the 2 CH3 groups / hydrogen environment) next to CH2

groups as peak is a triplet (1)

Peak at 1.43 ppm

• CH3 group with no protons on adjacent carbon atoms as Allow just CH3 identified in M6 if singlet

(relative peak area = 3 and) singlet (1) explained in M1

• structure of Q (1)

(Total for Question 6 = 11 marks)

How to answer it

Structure Determination of an Ester using Combustion and High-Resolution 1H NMR Spectroscopy

What this question tests

This 11-mark multi-part question tests your mastery of organic analysis. Part (a) assesses quantitative combustion analysis, converting masses of combustion products into empirical formulas. Part (b) evaluates advanced spectroscopic interpretation, combining chemical shift values, relative peak areas, and splitting patterns (n+1 rule) from high-resolution proton NMR data to deduce complex molecular structures.

Part (a): Combustion Analysis & Empirical Formula

When burned in excess oxygen, 1.879 g of Q formed 4.594 g of carbon dioxide and 1.879 g of water. Show that the empirical formula of Q is C₄H₈O. (4 marks)

📐 Step-by-Step Calculation

  1. Find mass of Carbon:
    Mass C = 4.594 × (12.0 / 44.0) = 1.253 g
  2. Find mass of Hydrogen:
    Mass H = 1.879 × (2.0 / 18.0) = 0.209 g
  3. Find mass of Oxygen (by difference):
    Mass O = 1.879 g (total) - (1.253 + 0.209) = 0.417 g
  4. Calculate Moles:
    Moles C = 1.253 / 12.0 = 0.1044 mol
    Moles H = 0.209 / 1.0 = 0.209 mol
    Moles O = 0.417 / 16.0 = 0.0261 mol
  5. Determine Ratio:
    Divide by smallest (0.0261):
    C : H : O = 4 : 8 : 1 → C₄H₈O

❌ Common Calculation Traps

  • Forgetting oxygen by difference: Students often calculate C and H, see they don't add up to the total sample mass, and panic. Always subtract to find the oxygen mass.
  • Molar mass errors: Using 18 instead of 18.0 for H₂O or 44 instead of 44.0 for CO₂. Keep at least 1 decimal place for molar masses.
  • Rounding too early: Rounding intermediate mole values too harshly leads to ratios that don't whole-number round cleanly.
Mark Breakdown (4 marks): 1 mark for calculating masses of C and H | 1 mark for mass of O | 1 mark for moles of C, H, and O | 1 mark for final ratio / stated empirical formula.

Part (b): High-Resolution 1H NMR Interpretation

Complete the structure of Q. Justify your answer by linking proton environments, peak areas, and splitting patterns. (7 marks)

✅ Correct Final Answer Structure

Ester Q is methyl pivalate (methyl 2,2-dimethylpropanoate):

H₃C-C(=O)-O-CH₃ (with a tert-butyl group attached to the carbonyl carbon: (CH₃)₃CCOOCH₃ )

Drawing check: Ensure the ester linkage matches the provided skeleton H₃C-C(=O)-O-CH₃ extended with the tertiary butyl structural features.

💡 Spectral Breakdown Table

  • 2.50 ppm | Singlet | Area 3: -O-CH₃ protons. No adjacent hydrogens (hence singlet).
  • 1.56 ppm | Quartet | Area 4: Two -CH₂- groups adjacent to a -CH₃ (triplet).
  • 1.43 ppm | Singlet | Area 3: Isolated -CH₃ group with no adjacent protons.
  • 0.92 ppm | Triplet | Area 6: Two -CH₃ groups adjacent to -CH₂- groups.

🧠 Exam Technique & Examiner Insights

  • Systematic matching: Always start by matching the relative peak areas to the molecular formula or fragment sizes. An area total matching numbers of hydrogens is vital.
  • Splitting (n+1 rule): Clearly state what neighboring protons cause each splitting pattern. E.g., a quartet means 3 adjacent hydrogens (n = 3).
  • Connecting fragments: Use chemical shifts to position oxygen atoms (e.g., 2.50 ppm shifts downfield due to deshielding by the adjacent electronegative oxygen atom in the ester link).
Mark Breakdown (7 marks): 1 mark for identifying the CH₃-C=O environment at 2.50 ppm | 1 mark for the two CH₂ groups at 1.56 ppm | 1 mark for noting they neighbor a CH₃ (quartet) | 1 mark for two CH₃ groups at 0.92 ppm | 1 mark for noting they neighbor a CH₂ (triplet) | 1 mark for the singlet at 1.43 ppm | 1 mark for the fully correct drawn structure of Q.

Topics

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

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.