AQA A-Level Chemistry Paper 2, 2023: Question 5

11 marks · Easy difficulty · State/Explain/Describe

Spectroscopic Analysis of Compound Z This question involves identifying a bond from IR absorption, determining carbon environments using NMR data to deduce structural features of Z, and explaining difficulties in interpreting spectra without additional data, concluding with deducing the overall structure of Z.

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AQA A-Level Chemistry Paper 2, 2023: Question 5
Question text

05 This question is about compound Z, with molecular formula C7H12O3

Figure 7 shows the infrared spectrum of Z.

Figure 7

05.1 Identify the bond that causes the absorption at 1725 cm–1

[1 mark]

Figure 8 shows the 13C NMR spectrum of Z.

Figure 8

05.2 How many different carbon environments are there in a molecule of Z?

[1 mark]

56 7 8

Tick ( ) one

05.3 State the type of carbon environment that causes the peak at ẟ = 174 ppm

Use Table C in the Data Booklet to help you answer this question.

[1 mark]

05.4 Table 2 shows data from the 1H NMR spectrum for compound Z.

*12* Table 2

Chemical shift ẟ / ppm 4.10 2.60 2.56 2.19 1.26

Integration ratio 2 2 2 3 3

Splitting pattern quartet triplet triplet singlet triplet

Explain what can be deduced from the splitting patterns and chemical shift values

for the peaks at ẟ = 4.10 ppm and at ẟ = 1.26 ppm

Deduce the part of the structure of Z that causes the peaks at

ẟ = 4.10 ppm and ẟ = 1.26 ppm

Use Table B in the Data Booklet to help you answer this question.

[5 marks]

Peak at ẟ = 4.10 ppm

Peak at ẟ = 1.26 ppm

Part of structure

05.5 Deduce the part of the structure of Z that causes the peak at ẟ = 2.19 ppm

[1 mark]

Part of structure

Figure 9 shows the 1H NMR spectrum of compound Z.

Figure 9

05.6 Suggest why it would be difficult to determine the structure of Z using the

spectrum in Figure 9, without the information in Table 2 on page 13.

[1 mark]

05.7 Deduce the structure of Z.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2023: Question 5

Question Answers Additional Comments/Guidelines Mark

C=O 1

05.1

(AO1)

Tick in the box for 7 ONLY 1

05.2

(AO1)

Ignore acids

05.3

(AO1)

M1 (Quartet) because neighbouring C has 3H

Ignore use of integration

M2 (At ẟ = 4.1 ppm) because connected to single bonded O of ester

or

05.4

(5 x AO2)

M3 (Triplet) because neighbouring C has 2H

M4 (At ẟ = 1.26 ppm) because R2CH2 or RCH3

M5

05.5

(AO2)

Cannot deduce splitting patterns of peaks (at about ẟ = 2.60) Allow

Or Peaks at ẟ = 2.60 and ẟ = 2.56 ppm overlap 1

05.6

No integration values OR (AO3)

spectrum at ẟ = 2.60 is second order 23

05.7

(AO2)

How to answer it

Spectroscopy Step-by-Step Commentary

Question (a): Identifying the Bond

Step 1: Analyse the IR Spectrum

The absorption at 1725 cm-1 corresponds to a C=O (carbonyl bond) .

Why: This wavenumber is characteristic of stretching in carbonyl groups, found in aldehydes, ketones, esters, and carboxylic acids.

Answer: C=O bond

Question (b): Carbon Environments

Step 1: Count the Peaks in the 13C NMR Spectrum

The 13C NMR spectrum shows 6 distinct peaks, representing 6 unique carbon environments.

Answer: Tick 6

Tip: Each peak corresponds to a unique carbon environment. Symmetry in the molecule reduces the number of peaks.

Question (c): Peak at δ = 174 ppm

Step 1: Match Chemical Shift to Carbon Environment

The chemical shift at 174 ppm indicates a carbonyl carbon in an ester .

Why: Carbonyl carbons in esters resonate between 160–180 ppm.

Answer: Ester carbonyl carbon (C=O)

Question (d): Peaks in the 1H NMR Spectrum

Step 1: Analyse Peaks at δ = 4.10 ppm and 1.26 ppm

Peak at δ = 4.10 ppm

  • Splitting: Quartet
  • Integration: 2 protons
  • Conclusion: This represents a CH2 group bonded to an oxygen atom (O-CH2-).

The quartet indicates coupling with 3 protons (a neighbouring CH3 group).

Peak at δ = 1.26 ppm

  • Splitting: Triplet
  • Integration: 3 protons
  • Conclusion: This represents a CH3 group bonded to a CH2 group.

Summary: Together, the quartet and triplet peaks suggest the presence of an ethyl group (-CH2CH3).

Question (e): Peak at δ = 2.19 ppm

Step 1: Identify the Carbon Environment

The peak at 2.19 ppm corresponds to a CH2 group adjacent to a carbonyl group (C=O).

Answer: CH2 group next to C=O

Question (f): Difficulties Using Figure 3 Alone

Step 1: Understand the Issue

Without the splitting patterns and integration ratios provided in part (d), it is difficult to:

  • Determine the number of protons contributing to each peak.
  • Identify splitting patterns (e.g., singlet, doublet, triplet).
  • Match peaks to specific groups in the molecule.

Conclusion: The spectrum alone lacks sufficient information to deduce the full structure.

Question (g): Deducing the Structure of Z

Step 1: Combine All Data

  • IR spectrum: C=O bond suggests a carbonyl group.
  • 13C NMR: 6 carbon environments include an ester carbonyl carbon (174 ppm).
  • 1H NMR: Peaks indicate an ethyl group and a CH2 next to C=O.

Step 2: Identify the Structure

The molecule is ethyl propanoate:

CH3-CH2-O-C(O)-CH3

Topics

Organic Chemistry · 3.3.6 Organic Analysis

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.