AQA A-Level Chemistry Paper 2, June 2025: Question 7

12 marks · Hard difficulty · Long Answer

Deduce the molecular formula and possible structures of hydrocarbon P using high-resolution mass data and spectroscopy, and determine the structure of compound Q using 13C and 1H NMR spectra.

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Question

Question 07.1 gives high-resolution mass data for compound P (Mr = 84.09396, C = 12.00000, H = 1.00783) and states it has only one singlet in its 1H NMR spectrum; students are asked to deduce its molecular formula, suggest two possible structures, and explain how to identify it via infrared spectroscopy above 1500 cm⁻¹. Question 07.2 gives the molecular formula C5H8O2 for compound Q along with its 13C NMR spectrum (showing 5 peaks: two carbonyl peaks around 195–200 ppm and three alkyl peaks around 10–30 ppm) and its 1H NMR spectrum (showing a triplet at 1.1 ppm, a singlet at 2.4 ppm, and a quartet at 2.7 ppm). Students are asked to explain what can be deduced about the structure and give the final structure of Q.
Question text

07 This question is about spectroscopy.

07.1 Compound P is a hydrocarbon with only one peak (a singlet) in its 1H NMR spectrum.

P has a precise Mr of 84.09396

Carbon has a precise Ar of 12.00000

Hydrogen has a precise Ar of 1.00783

Use this information to deduce the molecular formula of P. Show your working.

Suggest two possible structures for P.

Explain how P can be identified using its infrared spectrum above 1500 cm–1

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

[5 marks]

Molecular formula

Structure 1

Structure 2

Explanation

07.2 Compound Q has the molecular formula C5H8O2

Figure 7 shows the 13C NMR spectrum of compound Q.

Figure 8 shows the 1H NMR spectrum of compound Q.

*21* Figure 7

Figure 8

Explain what you can deduce about the structure of compound Q.

In your answer you should refer to

• the peaks in the 13C NMR spectrum

• the chemical shift values for the peaks in the 1H NMR spectrum

• the splitting pattern for the peaks in the 1H NMR spectrum.

Give the structure of compound Q.

[7 marks]

Mark scheme

Show the mark scheme The mark scheme for 07.1 awards 1 mark for showing 84.09396 = (6 × 12.00000) + (12 × 1.00783) giving C6H12, 2 marks for structures (cyclohexane and 2,3-dimethylbut-2-ene), and 2 marks for IR identification (peak at 1620–1680 cm⁻¹ for C=C in alkene, absent in cycloalkane). For 07.2, a 3-level response scheme out of 7 marks assesses Stage 1 (13C NMR: 5 environments, 2 carbonyls at 190–220 ppm, 3 alkyl carbons), Stage 2 (1H NMR chemical shifts: R-CH3/R2CH2 at 0.7–1.4 ppm, R-CO-CH at 2.1–2.8 ppm), Stage 3 (splitting: triplet next to CH2, singlet with no adjacent H, quartet next to CH3), and gives the structure as pentane-2,3-dione (CH3CH2COCOCH3).

Question Answers Additional comments/Guidelines Mark

M1 84.09396 = (6 × 12.00000) + (12 × 1.00783)

Hence C6H12

M2 and M3 Structures e.g.

07.1

(2 x AO1,

3 x AO2)

M4 Look for a peak at 1620–1680 cm–1

M5 Alkene has this peak for C=C or cycloalkane doesn’t

– A-LEVEL CHEMISTRY – 7405/2 –

Question Answers Additional Comments/Guidelines Mark

This question is marked using Levels of Response. Refer to the Mark Stage 1 13C NMR

Scheme Instructions for Examiners for guidance.

1a (5 peaks so) there are 5 (different) C

Level 3 All stages are covered and the explanation of each stage environments

5–6 marks is correct and virtually complete

(i.e. two from stages 1, 2 and 3) 1b (2) peaks with chemical shift 190–220 ppm so

there are (2 different C=O) aldehyde / ketone

Answer communicates the whole explanation, including groups

equations, coherently and shows a logical progression

through all three stages 1c (3) peaks with chemical shift 5–50 ppm so could

be C–C alone or C–C and R–CO–C 7

07.2 Level 2 All stages are covered (NB ‘covered’ means min one (3 x AO2,

3–4 marks from each stage) but the explanation of each stage may Stage 2 1H NMR chemical shift data 4 x AO3)

be incomplete or may contain inaccuracies

28 2a peak at chemical shift 0.7 – 1.4 R–CH3 or R2CH2

OR two stages covered and the explanations are

generally correct and virtually complete 2b peak at chemical shift 2.1 – 2.6 R–CO–CH

(i.e. two from 2 stages either 1, 2 or 3)

2c peak at chemical shift 2.6 – 2.8 R–CO–CH

Answer is coherent and shows some progression

through all three stages. Some steps in each stage may

be incomplete

How to answer it

Multi-Technique Spectroscopy & Structure Elucidation

What this question tests

This question assesses your ability to combine high-resolution mass spectrometry ( Aᵣ values), molecular symmetry, infrared spectroscopy, ¹³C NMR, and ¹H NMR (chemical shifts and splitting patterns via the n + 1 rule) to determine unknown organic structures systematically.

Question 07.1 • 5 Marks

High-Resolution Mass, Symmetry & Infrared Spectroscopy

Deducing molecular formula and distinguishing symmetrical isomers

📐 Step-by-Step Calculation: Molecular Formula

  1. Identify the general class: P is a hydrocarbon containing only C and H. The approximate Mᵣ is 84.
  2. Find number of carbons: 84 / 12 = 7, but 7 × 12 leaves 0 mass for hydrogen. If 6 carbons are present: 6 × 12.00000 = 72.00000.
  3. Find number of hydrogens: Remaining mass = 84.09396 − 72.00000 = 12.09396. Number of H = 12.09396 / 1.00783 = 12.00000.
  4. Confirm with exact calculation:
    (6 × 12.00000) + (12 × 1.00783) = 72.00000 + 12.09396 = 84.09396
  5. Conclusion: Molecular formula = C₆H₁₂

✅ Correct Answers

  • Molecular formula: C₆H₁₂
  • Structure 1: Cyclohexane
    A regular planar hexagon representing cyclohexane (all 12 H atoms are equivalent, giving 1 singlet).
  • Structure 2: 2,3-dimethylbut-2-ene
    (CH₃)₂C=C(CH₃)₂
    All 12 hydrogens belong to four chemically equivalent methyl groups with no adjacent protons.
  • IR Explanation: Look for an absorption peak at 1620–1680 cm⁻¹. The alkene has this peak due to the C=C bond, whereas the cycloalkane does not.

💡 Key Knowledge

  • Degree of unsaturation: A molecular formula of C₆H₁₂ fits general formula CₙH₂ₙ , indicating either 1 ring (cycloalkane) or 1 double bond (alkene).
  • One peak in ¹H NMR: All 12 hydrogens must be in identical chemical environments with no coupling (a singlet). Only highly symmetrical structures satisfy this.
  • IR Table A ranges: An alkene C=C stretch appears characteristically in the diagnostic functional group region above 1500 cm⁻¹ at 1620–1680 cm⁻¹. C–C and C–H bonds alone do not absorb in this region.

🧠 Exam Technique

  • Show all working: Explicitly show the sum (6 × 12.00000) + (12 × 1.00783) = 84.09396 to secure Mark 1.
  • Quote the data booklet exact range: The mark scheme strictly expects the specific range 1620–1680 cm⁻¹. Vague values like "~1650 cm⁻¹" or broad ranges risk losing marks.
  • State the difference both ways: State that the alkene shows the peak AND the cycloalkane does not (or which bond causes the peak).

❌ Common Errors

  • Suggesting hex-1-ene or other unsymmetrical alkenes which produce multiple complex splitting patterns.
  • Proposing methylcyclopentane — although it is C₆H₁₂, the hydrogens are in several different chemical environments.
  • Confusing the C=C stretch (1620–1680 cm⁻¹) with the C=O stretch (1680–1750 cm⁻¹).
Mark Breakdown (5 marks):
[M1] Calculation showing (6 × 12.00000) + (12 × 1.00783) = 84.09396 and formula C₆H₁₂.
[M2] First valid symmetrical structure (e.g. cyclohexane).
[M3] Second valid symmetrical structure (e.g. 2,3-dimethylbut-2-ene).
[M4] Identification of peak at 1620–1680 cm⁻¹.
[M5] Stating the alkene has this C=C peak while the cycloalkane does not.
Question 07.2 • 7 Marks

Extended Reasoning: Multi-Spectroscopic Analysis of Compound Q

Deducing the structure of C₅H₈O₂ using ¹³C and ¹H NMR (Level of Response)

💡 Three-Stage Systematic Deduction Strategy

To secure top Level 3 marks (5–6 or 7 marks including structure), your response must cover all three stages with logical clarity:

Stage Spectral Data Deduction & Structural Evidence
Stage 1
¹³C NMR		 • 5 peaks in total
• 2 peaks at 190–220 ppm
• 3 peaks at 5–50 ppm 		• All 5 carbon atoms are in different chemical environments (no symmetry).
• 2 separate carbonyl (C=O) groups, specifically ketones/aldehydes.
• 3 aliphatic carbon environments (alkyl C–C or C adjacent to C=O: R–CO–C).
Stage 2
¹H NMR Shifts		 • Shift 0.7–1.4 ppm (approx. 1.1 ppm)
• Shift 2.1–2.6 ppm (approx. 2.4 ppm)
• Shift 2.6–2.8 ppm (approx. 2.7 ppm) 		• R–CH₃ methyl protons attached to standard alkyl carbon.
• R–CO–CH₃ methyl protons directly adjacent to a carbonyl group.
• R–CO–CH₂– methylene protons directly adjacent to a carbonyl group.
Stage 3
¹H Splitting ( n + 1 )		 • Triplet (3 peaks) at 1.1 ppm
• Singlet (1 peak) at 2.4 ppm
• Quartet (4 peaks) at 2.7 ppm 		• Triplet coupled to 2 adjacent protons (next to a –CH₂– group).
• Singlet has 0 adjacent protons (isolated –CO–CH₃ group).
• Quartet coupled to 3 adjacent protons (next to a –CH₃ group) → reveals an isolated CH₃–CH₂–CO– ethyl ketone fragment!

✅ Final Deduced Structure

Pentane-2,3-dione

CH₃–CH₂–C(=O)–C(=O)–CH₃

or structural representation:
    H   H   O   O   H
    |   |   ||  ||  |
H — C — C — C — C — C — H
    |   |           |
    H   H           H

Matches formula C₅H₈O₂: Contains an ethyl group (triplet + quartet), an isolated methyl carbonyl (singlet), and two adjacent ketone groups accounting for the two peaks at 190–220 ppm.

🧠 Level of Response Mark Scheme Criteria

  • Level 3 (5–6 marks): All 3 stages are covered, and each stage is virtually complete. Explanations are coherent with logical progression.
  • 7 marks (Top Mark): Level 3 achieved plus the correct final structure is clearly drawn/stated.
  • Level 2 (3–4 marks): Covers all three stages with minor omissions, OR two stages fully correct.
  • Level 1 (1–2 marks): Only one stage fully correct, or two stages partially covered.

❌ Common Misconceptions & Traps

  • Assuming ester or carboxylic acid: A peak at ~200 ppm in ¹³C NMR is diagnostic of aldehyde/ketone C=O. Ester/acid C=O appears lower (160–185 ppm).
  • Symmetrical dione mistake: Drawing pentane-2,4-dione (CH₃COCH₂COCH₃). This has a plane of symmetry, which would give only 3 ¹³C peaks, contradicting the observed 5 peaks.
  • Confusing quartet and triplet coupling: Remember the (n+1) rule: a triplet has 2 neighbouring protons (–CH₂–), and a quartet has 3 neighbouring protons (–CH₃).
Examiner Insight: High-scoring students structured their responses under clear subheadings: ¹³C NMR , ¹H Chemical Shifts , and Splitting Patterns . They explicitly quoted values from Table B and Table C of the Data Booklet and paired each peak with its specific structural fragment before drawing the overall molecule.

Topics

Organic Chemistry · Physical Chemistry · 3.3.15 Nuclear Magnetic Resonance Spectroscopy · 3.3.6 Organic Analysis · 3.3.8 Aldehydes and Ketones · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.