Edexcel A-Level Chemistry Paper 2, June 2025: Question 1

✅ Correct Answer

C: CH₂(OH)COOH (hydroxyethanoic acid / glycolic acid)

The molecular ion peak corresponds to the intact radical cation [M]⁺•. Its m/z value equals the relative molecular mass (Mᵣ) of the compound. Molecule C has an Mᵣ of 76, not 74.

📐 Step-by-Step Mᵣ Calculations

Using accurate relative atomic masses: C = 12.0, H = 1.0, O = 16.0

🧠 Exam Technique

• Watch for the negative: The question asks which molecule would not give m/z = 74 . Circle or underline "not" immediately.
• Quick formula summing: Convert condensed structures into molecular formulas first (e.g., CHOCOOH → C₂H₂O₃) before adding masses to avoid missing hydrogen or oxygen atoms.

❌ Common Errors

• Misreading B: Students often mistake CHOCOOH for ethanoic acid (CH₃COOH, Mᵣ = 60) or miss the aldehyde carbon.
• Skipping the (OH) group in C: Overlooking the hydroxyl group leads students to count only two oxygen atoms rather than three, erroneously getting Mᵣ = 60.

✅ Correct Answer

B: 29

A peak at m/z = 29 corresponds to either:

• The ethyl cation: [CH₃CH₂]⁺ (28 + 1 = 29)
• The formyl cation: [CHO]⁺ (12 + 1 + 16 = 29)

Both are formed directly from propanal by breaking the central C–C bond. Neither fragment can be formed by a single bond cleavage from propanone.

💡 Key Knowledge: Why Other Peaks Match Both

• m/z = 15: Formed by loss of a methyl radical to yield [CH₃]⁺ . Both molecules contain methyl groups, so both give m/z = 15 .
• m/z = 43: Produced by both isomers!
  • Propanone: Cleavage of C–C gives [CH₃CO]⁺ ( m/z = 43 ).
  • Propanal: Loss of methyl radical gives [CH₂CHO]⁺ or loss of oxygen gives [C₃H₇]⁺ ( m/z = 43 ).
• m/z = 59: This is an [M + 1]⁺ peak resulting from the naturally occurring ¹³C isotope (1.1% abundance), present in both compounds.

📐 Fragmentation Breakdown

Propanal (CH₃CH₂CHO, Mᵣ = 58):

• CH₃CH₂—CHO → [CH₃CH₂]⁺ ( m/z = 29 ) + •CHO
• CH₃CH₂—CHO → CH₃CH₂• + [CHO]⁺ ( m/z = 29 )

Propanone (CH₃COCH₃, Mᵣ = 58):

• CH₃—COCH₃ → [CH₃]⁺ ( m/z = 15 ) + •COCH₃
• CH₃—COCH₃ → CH₃• + [CH₃CO]⁺ ( m/z = 43 )

Neither fragment in propanone has a mass of 29!

🧠 Examiner Guidance & Pitfalls

• The m/z = 43 Trap: Many candidates instinctively chose C (43) assuming only propanone gives an acylium ion [CH₃CO]⁺ . They forget that propanal can lose a methyl group (58 − 15 = 43) to produce an isomeric cation at m/z = 43 !
• Positive charges are mandatory: When writing fragment ions in written questions, always remember to show the positive charge (e.g. [C₂H₅]⁺ ), as uncharged species are not detected by mass spectrometers.