AQA A-Level Chemistry Paper 2, 2024: Question 6
12 marks · Medium difficulty · State/Explain/Describe
Isomer Identification Using NMR and Reactions of Carboxylic Acids
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Question text
06 Compounds V, W, X and Y are isomers with the molecular formula C5H10O2
Isomers V and W are carboxylic acids with formulas that can be written as C4H9COOH
06.1 Give an equation for the reaction of C4H9COOH with sodium hydrogencarbonate.
[1 mark]
06.2 Isomer V has an asymmetric carbon atom.
Deduce the structure of V.
[1 mark]
06.3 Isomer W has four peaks in its 1H NMR spectrum.
Deduce the structure of W.
Deduce the integration ratio for the four peaks in the 1H NMR spectrum of W.
[2 marks]
Structure
Integration ratio 17
06.4 Isomer X has three singlets with integration ratio 1:3:6 in its 1H NMR spectrum.
Deduce the structure of X.
Explain why the peaks in the 1H NMR spectrum are singlets.
[2 marks]
Structure
Explanation
06.5 Table 2 shows information about the peaks in the 1H NMR spectrum of isomer Y.
Table 2
Chemical shift δ / ppm Integration ratio Splitting pattern
3.65 2 singlet
1.19 3 singlet
Draw the parts of the structure of Y that can be deduced from each of these peaks.
Deduce the structure of Y.
State how many peaks are in the 13C NMR spectrum of Y.
[6 marks]
Part of structure from peak at δ = 3.65 ppm
Part of structure from peak at δ = 1.19 ppm
Structure of Y
Number of peaks in 13C NMR spectrum of Y
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidelines Mark
C4H9COOH + NaHCO3 ⟶ C4H9COONa + CO2 + H2O
06.1
(1 x AO1)
CH3CH2CH(CH3)COOH
06.2
(1 x AO2)
M1 M1 (CH3)2CHCH2COOH
06.3
(2 x AO2)
M2 6:1:2:1 (Any order) M2 Allow ECF for a 5 carbon carboxylic acid
– A-LEVEL CHEMISTRY – – JUNE 2024
M1 M1 (CH3)2C(OH)COCH3 or CH3OC(CH3)2CHO
06.4
(2 x AO2)
Or
M2
Adjacent C has no (non-equivalent) H attached (so no splitting/spin-
spin coupling takes place)
– A-LEVEL CHEMISTRY – –
Scores M1 Allow M1 for
and M2
Scores M3 Allow M3 for RCH3
and M4
Scores M5 This structure 6
06.5 also scores M5 (1 x AO1,
5 x AO3)
26 13C peaks = 3 M6 Allow ECF
from their M5
of C5H10O2
How to answer it
Isomers V, W, X & Y (C₅H₁₀O₂): acids & NMR detective work
What this question tests
Applying acid–hydrogencarbonate equations, spotting a chiral centre in a carboxylic acid, building structures from ¹H NMR clues (number of peaks & integrations), explaining singlets by the n+1 rule/absence of neighbouring non-equivalent H, and inferring a highly symmetrical structure for an oxygenated isomer from very simple NMR data.
Equation for C₄H₉COOH with NaHCO₃
✅ Correct answer
C₄H₉COOH + NaHCO₃ → C₄H₉COONa + CO₂ + H₂O
💡 Key knowledge
All carboxylic acids effervesce with NaHCO₃ , producing CO₂ and the sodium carboxylate.
Isomer V has an asymmetric carbon
✅ Structure of V
CH₃CH₂CH(CH₃)COOH (2-methylbutanoic acid). The starred C (CH*) is attached to four different groups → chiral.
Isomer W
✅ Structure
(CH₃)₂CHCH₂COOH
✅ Integration ratio
6 : 1 : 2 : 1 (any order for the four signals).
🧠 Exam technique
Count different proton environments from the structure first; then match groups to plausible areas (CH₃, CH, CH₂, OH).
Isomer X
✅ Structure
One acceptable answer: (CH₃)₂C(OH)COCH₃ (alternative accepted in mark scheme: CH₃OC(CH₃)₂CHO).
✅ Why singlets?
Each set of H has no non-equivalent hydrogens on adjacent carbons → no spin–spin coupling → all signals are singlets (n+1 with n = 0).
Isomer Y from minimal ¹H NMR data
💡 Parts of structure from each peak
- δ = 3.65 ppm, 2H, singlet: an –O–CH₂– unit with no neighbouring hydrogens (e.g. CH₂ flanked by O/quaternary C).
- δ = 1.19 ppm, 3H, singlet: a –C(CH₃)– methyl with no adjacent hydrogens (on a quaternary or carbonyl carbon).
✅ Deduced structure of Y
2,2-dimethyl-1,3-dioxolane (an acetal with two identical O–CH₂ groups and a quaternary carbon bearing two equivalent CH₃ groups).
This gives only two types of protons: the four O–CH₂ H (seen as 2 in the simplified ratio) and the six equivalent CH₃ H (seen as 3 in the simplified ratio). Both are singlets because there are no vicinal non-equivalent H.
✅ Number of ¹³C NMR peaks for Y
3 (one for the two equivalent O–CH₂ carbons, one for the quaternary acetal carbon, one for the two equivalent CH₃ carbons).
❌ Common errors
- Copying fragments from the data booklet inaccurately (e.g. losing an oxygen or wrong connectivity).
- Forgetting that singlet implies no adjacent non-equivalent hydrogens.
- Giving ratios that don’t match the stated integration or the molecular formula C₅H₁₀O₂.
Final takeaways
💡 Key knowledge
- Acid + hydrogencarbonate → salt + CO₂ + H₂O (effervescence test).
- Chiral carbon = C attached to four different groups.
- ¹H NMR: number of signals = number of distinct proton environments; splitting obeys n+1.
- Highly symmetrical molecules can collapse several carbons/protons into few NMR signals.
🧠 Exam technique
- Write the integration next to each group in your sketch to keep ratios consistent.
- Check final structures fit the formula C₅H₁₀O₂ and the count of signals asked for.
Topics
Organic Chemistry · 3.3.7 Optical Isomerism · 3.3.15 Nuclear Magnetic Resonance Spectroscopy
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.