Edexcel A-Level Chemistry Paper 3, June 2018: Question 4

13 marks · Hard difficulty · Short Open Response

Deduce structural formulae, explain splitting patterns, and discuss the use of TMS in NMR spectroscopy for isomers of C6H12O2.

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Question

Question text

4 This question is about the use of NMR spectroscopy to distinguish between isomers

of C6H12O2.

(a) Tetramethylsilane (TMS) is a compound used as a standard when recording both

1H and 13C NMR spectra.

(i) Give the structural formula of TMS.

(1)

(ii) TMS is an inert and non-toxic compound. State two other reasons why TMS is

suitable for use as a standard when recording NMR spectra.

(2)

(b) (i) Draw the structural formulae of the two esters with formula C6H12O2 that

each have only two peaks, both singlets, in their high resolution proton NMR

spectra. The relative peak areas are 3:1 for both esters.

(2)

(ii) The high resolution proton NMR spectrum of another isomer of C6H12O2 is shown.

54 3 2 1 0

δ / ppm

The ratios of the number of protons for the five sets peaks in the spectrum are

given in the table.

δ / ppm 3.8 3.5 2.6 2.2 1.2 11

Ratio of the *P52304A01132*

22 2 3 3

number of protons

Show that all these data are consistent with the displayed formula shown.

Refer to the five chemical shifts and explain two of the splitting patterns.

H H H H O H

H C C O C C C C H

H H H H H

(5)

… *P52304A01232*

peaks in their carbon-13 NMR spectra.

Draw the structural formula of two of these isomers.

(2)

(ii) Draw the skeletal formula of a cyclic diol isomer of C6H12O2 that has only two

peaks in its carbon-13 NMR spectrum.

(1)

(Total for Question 4 = 13 marks)

Mark scheme

Show the mark scheme

NH + Allow CH CH(OH)CN + 2H O + HCl →

43 2

or CH3CH(OH)COOH + NH4Cl

CH3CH(OH)CN + 2H2O → CH3CH(OH)COOH + NH3 (1)

Question

Acceptable Answers Additional Guidance Mark

Number

2(b)(i) Condensation (polymerisation) Ignore esterification or (1)

addition-elimination

Do not award addition

Question

Acceptable Answers Additional Guidance Mark

Number

2(b)(ii) Repeat unit circled on diagram as follows: Allow any repeat unit (1)

e.g

Do not award circle containing more

than one repeat unit

(Total for Question 2 = 9 marks)

Question

Acceptable Answers Additional Guidance Mark

Number

3(a) 0.816 / 8.16 x 10−1 (g) (1)

Question

Acceptable Answers Additional Guidance Mark

Number

3(b) calculation of moles of CO2 Example of calculation: (1)

(moles CO2 = 225 =) 0.009375

24000

Allow 9.375 x 10−3 / 9.38 x 10−3 / 9.4 x 10−3

Ignore SF except 1SF

Question

Acceptable Answers Additional Guidance Mark

Number

3(c) Example of calculation: (4)

moles of MCO3 (1) Moles of MCO3 = moles CO2 = 0.009375 (mol)

method for calculation of molar mass of MCO3 (1) Molar mass of MCO3 = 0.816

0.009375

(= 87.04 ( g mol−1))

M2 subsumes mark for M1

molar mass final answer to 1, 2 or 3 SF (1) = 87.0 / 87 / 90 (g mol−1)

NOTE

M3 mark subsumes mark for M2 and M1

consequential identification of Group 2 metal by (87.0 – 60) = 27 AND Mg / Magnesium /

name or formula (1) MgCO3

Allow TE on answers to parts (a) and (b), with

Metal consequential on calculated molar mass

NOTE Alternative method can score 3 MAX but M must be a Group 2 element

Calculation of moles of CO 2- (1) Moles CO 2- = 0.009375

(Calculation of mass of CO 2-) (Mass of CO 2- = 0.009375 x 60 = 0.5625 g)

Deduction of mass of M by subtraction (1) Mass of M = 0.2535 g

Calculation of Ar of M to 1, 2 or 3 SF AND Identification of Ar = 0.2535/0.009375

group 2 metal (1) = 27.0 / 27 / 30 (g mol-1)

AND Mg / Magnesium / MgCO3

Question

Acceptable Answers Additional Guidance Mark

Number

3(d)(i) An explanation that makes reference to the following (2)

points:

the bung was not replaced quickly enough (1) Allow bung not fitting tightly resulting in leaks

Ignore references to CO2 dissolving

Ignore references to other types of gas leak

(So) CO2 / gas lost (to the surroundings) (1) Allow ‘smaller volume of gas collected’ / lower

reading of gas volume

Mark points M1 and M2 independently

Question Mark

Number Acceptable Answers Additional Guidance

3(d)(ii) An answer that makes reference to the following point: Allow (1)

The acid was (already) in excess (and more acid won’t The carbonate is the limiting reactant / the

affect this) acid is not the limiting reactant

Question Mark

Number Acceptable Answers Additional Guidance

3(d)(iii) An explanation that makes reference to the following Mark points M1 and M2 independently

points: (2)

rate of reaction is faster and powder has greater surface Both parts of statement needed

area (1)

no effect on (final) volume of gas and moles of (metal) Both parts of statement needed

carbonate are unchanged Allow mass / amount for moles

or Allow reactant for metal carbonate

because the rate is faster more gas will be lost before

the bung is replaced so the (final) volume will be less

(1)

Question Mark

Number Acceptable Answers Additional Guidance

3(e)(i) Example of equation: (1)

balanced equation with state symbols MCO3(s) → MO(s) + CO2(g)

Allow a correct equation for the decomposition

of any Group 2 carbonate

Question Mark

Number Acceptable Answers Additional Guidance

3(e)(ii) Example of calculation: (3)

(mass of CO2 = 20.447 – 20.205) = 0.242

subtractions to obtain masses (1) AND

(mass of MCO3 = 20.447 – 19.996) = 0.451

moles of CO2 = 0.242

calculation of moles of CO2 (1) 44

= 0.0055(0) (mol) / 5.5(0) x 10−3 (mol)

ALLOW TE from M2 to M3

Mr of MCO3 = 0.451

calculation of molar mass of MCO3 (1) 0.0055(0)

= 82 (g mol−1)

Correct answer with or without working scores 3

Ignore SF except 1

Ignore attempts to identify the metal

Question Mark

Number Acceptable Answers Additional Guidance

3(f) An answer that makes reference to the following point: Allow calculations comparing the two (1)

percentage errors:

Student 3 used a smaller mass / less (and the uncertainty e.g.

of the balance was the same) Student 1:-

or (0.001/0.816) x 100% = 0.12% and

Student 1 used a larger mass / more (and the uncertainty Student 3:-

of the balance was the same) 0.001/0.451 x 100% = 0.22%

Question Mark

Number Acceptable Answers Additional Guidance

3(g) An explanation that makes reference to the following (2)

points:

more CO2 (would appear to be) given off (1)

(So) calculated molar mass is smaller (1) M2 dependent on M1

Less MO would appear to have been formed (1)

Calculated molar mass would be greater (1) M2 dependent on M1

(Total for Question 3 = 18 marks)

Question

Acceptable Answers Additional Guidance Mark

Number

4(a)(i) (CH3)4Si Allow partially or fully displayed formula (1)

Ignore connectivity

CH3

H3C Si CH3

CH3

Question

Acceptable Answers Additional Guidance Mark

Number

4(a)(ii) An answer that makes reference to any two of the following: (2)

single peak / all H or all C in same environment / no Allow 12 H or 4 C in the same environment

splitting pattern (1) Ignore references to inertness / non-

toxicity / cost / non-polar(ity)

(TMS) peak to the right / upfield / out of the way of other

peaks / peak doesn’t overlap with other peaks (1) Ignore chemical shift = 0

(TMS) low boiling temperature / volatile / can be easily

removed (1)

gives a strong signal so only a small amount needed (1) 12 H / 4 C are equivalent so gives a strong

signal scores 2 marks

Question

Acceptable Answers Additional Guidance Mark

Number

4(b)(i) C(CH3)3COOCH3 Allow displayed or skeletal formulae (2)

CH3

H3C C C O CH3

CH3 O

(1)

CH3COOC(CH3)3

O CH3

H3C C O C CH3

CH3

(1)

Question

Acceptable Answers Additional Guidance Mark

Number

4(b)(ii) An answer that makes reference to the following points: (5)

the chemical shift δ 2.2 identified (1) CH3C=O / methyl attached to C=O

four remaining chemical shifts identified (2) Identifies 2 or 3 chemical shifts correctly

scores 1

δ 1.2 3.5 3.8 2.6 (2.2)

two splitting patterns given and explained (2) 1 specific splitting patterns explained

scores 1

Questio

n Acceptable Answers Additional Guidance Mark

Number

4(c)(i) Any two of the following Allow displayed or skeletal formulae (2)

(CH3)2CHCH(CH3)COOH /

CH3 CH3

H3C C C C OH

H H O (1)

CH3CH2C(CH3)2COOH /

(1)

(CH3)2CHCH2CH2COOH /

(1)

Question Acceptable Answers Additional Guidance Mark

Number

4(c)(ii) Do not award other types of structure (1)

HO OH

(Total for Question 4 = 13 marks)

How to answer it

Mastering NMR Spectroscopy: Isomers of C₆H₁₂O₂

📌 What this question tests

This question assesses your understanding of Nuclear Magnetic Resonance (NMR) spectroscopy principles, including the role of TMS as a reference standard, interpreting high-resolution proton (¹H) NMR splitting patterns and integration traces, deducing ester and carboxylic acid structures from molecular formulae, and relating carbon-13 (¹³C) environments to structural symmetry.

Question 4(a): Tetramethylsilane (TMS) as a Reference Standard

Parts (a)(i) and (a)(ii)

✅ Correct Answers

(a)(i): Structural formula: (CH₃)₄Si (or fully/partially displayed structural formula).
(a)(ii): Any two of the following:
- Gives a single peak (all H or C atoms in the same chemical environment).
- Peak is far to the right / upfield (δ = 0 ppm), avoiding overlap with other sample peaks.
- Inert and non-toxic (does not react with the sample).
- Low boiling point / volatile (can be easily removed from the sample afterwards).
- Gives a strong signal, so only a small amount is needed.

💡 Key Knowledge

TMS has 12 equivalent protons in identical chemical environments, producing a sharp singlet reference peak at exactly 0.0 ppm.
Its chemical shift is isolated from almost all organic protons, making calibration straightforward.

Question 4(b): Interpreting Proton NMR Spectra and Structures

Parts (b)(i) and (b)(ii)

✅ Correct Answers

(b)(i) The two esters with C₆H₁₂O₂ having only 2 peaks (both singlets, ratio 3:1):

C(CH₃)₃COOCH₃ (methyl 2,2-dimethylpropanoate)
CH₃COOC(CH₃)₃ (tert-butyl ethanoate)

(b)(ii) Proving consistency for the given ester structure:

Identify the chemical shift at δ 2.2 ppm as corresponding to the CH₃C=O (methyl protons attached to carbonyl).
Correlate all 4 remaining chemical shift regions to distinct proton environments.
Correctly explain at least one splitting pattern using the n + 1 rule.

🧠 Exam Technique & Common Errors

❌ Common Errors

Forgetting that high-resolution NMR requires explaining splitting patterns using neighbouring protons, not just stating the number of peaks.
Confusing ester orientations when drawing isomers.

Top-Level Tip: When a question asks to "show that all these data are consistent", make sure you systematically account for every single peak listed in the table, linking proton ratios to the number of hydrogens in each environment.

Question 4(c): Carbon-13 NMR and Isomerism

Parts (c)(i) and (c)(ii)

✅ Correct Answers

(c)(i) Two carboxylic acid isomers of C₆H₁₂O₂ with 5 peaks in ¹³C NMR:
(Any two of the following structural/skeletal formulae)

(CH₃)₂CHCH(CH₃)COOH (2,3-dimethylbutanoic acid)
CH₃CH₂C(CH₃)₂COOH (2,2-dimethylbutanoic acid)
(CH₃)₂CHCH₂CH₂COOH (4-methylpentanoic acid)

(c)(ii) Cyclic diol isomer of C₆H₁₂O₂ with only 2 peaks in ¹³C NMR:

Symmetrical cyclohexane-1,4-diol (skeletal formula showing a 6-membered ring with two -OH groups in opposite/para positions to ensure high molecular symmetry yielding only 2 carbon environments).

💡 Key Knowledge for ¹³C NMR

The number of peaks in a ¹³C NMR spectrum corresponds directly to the number of non-equivalent carbon environments.
High molecular symmetry drastically reduces the number of peaks (e.g., para-disubstituted rings or quaternary carbon centres).
Carboxylic acid carbons appear far downfield (δ ~ 160–180 ppm due to deshielding by two oxygens).

Total Marks available for Question 4: 13 marks. Ensure clear structural connectivity and accurate functional group placement to secure all points.

Topics

Organic Chemistry · Topic 19: Modern Analytical Techniques II · Topic 17: Organic Chemistry II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.