OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 11

✅ Correct Answer: A

A is 2-methylpropanoic acid: (CH₃)₂CHCOOH

• Its primary cleavages yield:
  • [COOH]⁺ at m/z = 45
  • [(CH₃)₂CH]⁺ at m/z = 43
  • [CH₃]⁺ at m/z = 15
• Because it contains an isopropyl group rather than a straight ethyl chain, and a carboxylic acid group rather than an aldehyde, it cannot form an ion of m/z = 29 through straightforward fragmentation.

💡 Key Knowledge: The Two Key m/z = 29 Ions

At A-Level, a peak at m/z = 29 almost always corresponds to one of two stable fragment cations:

• Ethyl cation: [C₂H₅]⁺
  Mass = (2 × 12.0) + (5 × 1.0) = 29
  Produced by straight alkyl chains containing at least an ethyl group (–CH₂CH₃).
• Formyl / Acylium cation: [CHO]⁺
  Mass = 12.0 + 1.0 + 16.0 = 29
  Commonly produced by aldehydes (–CHO) via cleavage of the C–C bond adjacent to the carbonyl.

Molecule A: (CH₃)₂CHCOOH

2-methylpropanoic acid

Can it form m/z = 29? NO

The alkyl group is branched at C2 (isopropyl). Breaking any single C–C bond yields either a methyl radical/ion (mass 15) or an isopropyl ion/radical (mass 43). There is no pre-existing ethyl group or formyl group.

Molecule B: (CH₃)₂CHCH₂CHO

3-methylbutanal

Can it form m/z = 29? YES

Molecule B is an aldehyde. Cleavage of the C–C bond adjacent to the carbonyl group readily eliminates an alkyl radical to leave a [CHO]⁺ acylium fragment ion at m/z = 29.

Molecule C: CH₃CH₂COCH(CH₃)₂

2-methylpentan-3-one

Can it form m/z = 29? YES

The left side of the ketone is an ethyl group (–CH₂CH₃). α-cleavage next to the C=O group easily generates a [C₂H₅]⁺ ethyl carbocation at m/z = 29.

Molecule D: CH₃CH₂CH₂OCO₂H

Propyl hydrogen carbonate

Can it form m/z = 29? YES

Molecule D contains an unbranched propyl group (CH₃–CH₂–CH₂–). Fragmentation along the alkyl chain readily produces a [C₂H₅]⁺ ethyl cation at m/z = 29.

🧠 Exam Technique: "NOT" Strategy

• Circle the word "NOT": Negative questions are a common trap under timed conditions.
• Identify common fragment formulas first: When you see m/z = 29, immediately jot down [C₂H₅]⁺ and [CHO]⁺ .
• Inspect functional groups:
  • Does it have –CHO? → Yes (gives 29).
  • Does it have –CH₂CH₃? → Yes (gives 29).

❌ Common Errors to Avoid

• Confusing propyl and isopropyl: Both contain 3 carbons, but an unbranched propyl chain (–CH₂CH₂CH₃) can break to form ethyl (m/z = 29), whereas an isopropyl group (–CH(CH₃)₂) cleaves into methyl (m/z = 15) and isopropyl (m/z = 43).
• Forgetting the aldehyde fragment: Many students only remember [C₂H₅]⁺ as 29 and incorrectly select B, forgetting that [CHO]⁺ is also 29 (12 + 1 + 16 = 29).
• Omitting positive charges: In written questions, fragment species MUST have a positive charge (e.g., C₂H₅⁺, not C₂H₅). Uncharged radicals are not detected by the mass spectrometer.