OCR A-Level Chemistry AS Breadth in chemistry (01), June 2025: Question 19
1 mark · Easy difficulty · Multiple Choice
Identify which alkene is likely to produce a fragment ion at m/z = 29 in its mass spectrum.
Practise this questionQuestion
Question text
19 In a mass spectrum, which alkene is likely to have a fragment ion at m/z = 29?
A H2C=CHCH=CH2
B H2C=C(CH3)2
C H3CCH=CHCH3
D H3CCH=CHCH2CH3
Your answer [1]
Mark scheme
Show the mark scheme
19 D 1
How to answer it
Mass Spectrometry: Fragment Ion Identification
This question assesses your understanding of mass spectrometry fragmentation patterns in organic molecules. Specifically, it tests your ability to identify common fragment ions from their mass-to-charge ratio ( m/z ) and relate them to structural alkyl groups (such as the ethyl carbocation, [C₂H₅]⁺ ) within isomeric and related alkene structures.
Multiple Choice Analysis & Deduction
Identifying the Alkene Precursor for Peak m/z = 29
✅ Correct Answer
Correct Option: D ( H₃CCH=CHCH₂CH₃ , pent-2-ene)
• 1 mark for selecting D.
Note on MS snippet: The provided mark scheme excerpt shows row 20 (answer B); row 19 for this exam question awards the mark for D.
📐 Calculation of Fragment Ion Mass
- Identify the fragment with m/z = 29 :
Carbon: 2 × 12.0 = 24.0
Hydrogen: 5 × 1.0 = 5.0
Total = 29.0 - Formula of fragment: [C₂H₅]⁺ or [CH₂CH₃]⁺ (an ethyl cation).
- Check each molecule for a terminal or cleavable ethyl group ( -CH₂CH₃ ):
• A: Buta-1,3-diene ( C₄H₆ ) — No ethyl group.
• B: 2-Methylpropene — Only methyl ( -CH₃ ) groups.
• C: But-2-ene — Only methyl ( -CH₃ ) groups.
• D: Pent-2-ene — Contains a terminal ethyl group: -CH₂CH₃ .
💡 Key Knowledge: Common Fragment Ions
Memorise these standard alkyl fragment peaks for OCR AS Chemistry:
- m/z = 15 : [CH₃]⁺ (methyl cation)
- m/z = 29 : [C₂H₅]⁺ (ethyl cation)
- m/z = 43 : [C₃H₇]⁺ (propyl / isopropyl cation)
- m/z = 57 : [C₄H₉]⁺ (butyl cation)
Fragmentation equation for pent-2-ene:
[CH₃CH=CHCH₂CH₃]⁺• → [CH₂CH₃]⁺ + •CH=CHCH₃
The detected ion must carry the positive charge ( + ); uncharged radicals are not detected.
🧠 Exam Technique: Elimination Strategy
- Step 1: Immediately translate common m/z numbers into alkyl fragments. Seeing 29 should trigger ethyl ( C₂H₅⁺ ).
- Step 2: Scan the structural formulae for an intact ethyl ( -CH₂CH₃ ) group.
- Step 3: Notice that A, B, and C all have only 4 carbons and cannot form an ethyl group without breaking double bonds or deep skeletal rearrangements. D is the only 5-carbon alkene with a clear ethyl branch/chain.
❌ Common Errors & Pitfalls
- Confusing methyl and ethyl masses: Students frequently confuse m/z = 15 ( CH₃⁺ ) with m/z = 29 ( C₂H₅⁺ ). Selecting B or C often occurs because candidates think of splitting a 4-carbon chain in half.
- Symmetrical splitting assumption: In but-2-ene ( CH₃CH=CHCH₃ ), breaking the central double bond does not occur easily during simple electron impact fragmentation, and even if it did, a [CH₃CH] fragment would have a mass of 28 ( [C₂H₄]⁺• ), not 29.
- Forgetting the positive charge: In written explanations, forgetting to write the positive charge on the fragment ion ( [C₂H₅]⁺ instead of C₂H₅ ) loses marks automatically.
Topics
Module 4: Core organic chemistry · 4.2 Alcohols, haloalkanes and analysis · 4.1 Basic concepts and hydrocarbons
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.