AQA A-Level Chemistry Paper 2, 2022: Question 6

6 marks · Medium difficulty · Practical Techniques & Data Analysis

Structure Determination of Compound X This question involves deducing the structure of compound X (C₂H₄O) using its infrared spectrum, ¹H NMR data, and ¹³C NMR spectrum. The IR spectrum indicates a C=O bond, while the NMR data reveal distinct chemical environments corresponding to a methyl group, a CH group, and a carbonyl group.

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Question text

06 This question is about compound X with the empirical formula C2H4O

Figure 2 shows the infrared spectrum of X.

Figure 3 shows the 13C NMR spectrum of X.

The 1H NMR spectrum of X shows four peaks with different chemical shift values.

Table 3 gives data for these peaks.

Figure 2

Figure 3

Table 3

Chemical shift / ppm 3.9 3.7 2.1 1.2

Splitting pattern quartet singlet singlet doublet

Integration value 1 1 3 3

Show how information from Figure 2, Figure 3 and Table 3 can be used to deduce

the structure of compound X.

[6 marks]

Mark scheme

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Question Answers Additional Comments/Guidelines Mark

This question is marked using Levels of Response. Refer to the Mark Indicative Chemistry content

Scheme Instructions for Examiners for guidance. Stage 1: infrared

Level 3 All stages are covered and each stage is generally 1a) (broad peak) at 3400 cm–1 (any value from 3230-

5-6 marks correct and virtually complete. 3550) indicates OH in alcohols

1b) peak at 1720 cm–1 (any value from 1680-1750)

Answer is communicated coherently and shows a logical indicates C=O

progression from Stage 1 to Stages 2 and 3 Stage 2: 1H nmr

2a) peak at 3.9 ppm integration 1 so 1 H-C-O AND

Covers at least 1 point for stage 1, 3 for stage 2 and 3 quartet so adjacent to CH (stated or shown)

for stage 3. 2b) peak at 3.7 ppm integration 1 so HO-C (stated or

shown)

Level 2 All stages are covered but stage(s) may be incomplete or 2c) peak at 2.1 ppm integration 3 so H C-C=O AND

3-4 marks may contain inaccuracies singlet so no adjacent H (stated or shown)

Covers at least 1 point for stage 1 stage 2 and stage 3. 2d) peak at 1.2 ppm integration 3 so H3C- AND doublet

OR so adjacent to CH (stated or shown) 6

2e) sum of integration values = 8 Hence C4H8O2

06.1 two stages are covered and are generally correct and 13

Stage 3: C nmr (3 x AO1,

virtually complete. 3a) peak at 210 ppm C=O aldehydes or ketones

_ 3 x AO3)

Covers at least 1 point for stage 1, and 3 for stage 2 or 3b) peak at 75 ppm C O (alcohols, ethers or esters)

stage 3 OR 3 for stage 2 and 3 for stage 3 3c) peak at 25 ppm

Answer is communicated mainly coherently and shows a

logical progression from Stage 1 to Stages 2 and 3.

3d) peak at 20 ppm

Level 1 Two stages are covered but stage(s) may be incomplete

1-2 marks or may contain inaccuracies OR only one stage is

covered but is generally correct and virtually complete. 3e) structure

Answer includes isolated statements but these are not

presented in a logical order.

0 mark Insufficient correct chemistry to gain a mark

How to answer it

Analysis of Compound X

This guide explains the structure determination of compound X using infrared (IR) and NMR spectroscopy, alongside common student errors to help improve understanding.

Infrared Spectrum

Key observations:

Broad peak at 3400 cm^-1 indicates an OH group in an alcohol.
Sharp peak at 1720 cm^-1 suggests a C=O group (likely aldehyde or ketone).

Common Error: Failing to link the broad peak at 3400 cm^-1 to an alcohol group and confusing it with a carboxylic acid.

Tip: Use the data booklet to differentiate between alcohol and carboxylic acid peaks; carboxylic acids show broader OH peaks overlapping with the C=O peak.

1H NMR Spectrum

Key observations:

Quartet at 3.9 ppm (integration = 1): Indicates an H-C-O group adjacent to a CH₃.
Singlet at 3.7 ppm (integration = 1): Suggests a HO-C group.
Singlet at 2.1 ppm (integration = 3): Indicates an isolated CH₃ group bonded to a C=O.
Doublet at 1.2 ppm (integration = 3): Indicates a CH₃ group adjacent to a CH.

Common Error: Misinterpreting the quartet as being adjacent to a CH₂ instead of a CH₃, leading to incorrect structural deductions.

Tip: Remember the splitting rule: n+1 peaks indicate n equivalent protons on adjacent carbons.

13C NMR Spectrum

Key observations:

Peak at 210 ppm: Indicates a C=O group in an aldehyde or ketone.
Peak at 75 ppm: Suggests a C-O group in an alcohol.
Peak at 25 ppm: Suggests a CH₃ group adjacent to C=O.

Common Error: Ignoring the high chemical shift at 210 ppm and failing to identify it as a C=O group.

Tip: Peaks above 150 ppm are usually associated with unsaturated carbons or carbonyl groups.

Conclusion

Using all the data:

Empirical formula = C₂H₄O.
Structure deduced as CH₃-C(=O)-CH₂-OH (2-hydroxypropanone).

Common Error: Combining signals incorrectly, leading to structures with the wrong functional groups.

Tip: Ensure all peaks are accounted for and match the integration values with the molecular formula.

Topics

Organic Chemistry · 3.3.6 Organic Analysis · 3.3.15 Nuclear Magnetic Resonance Spectroscopy

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.