OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 20
1 mark · Medium difficulty · Multiple Choice
Identify which isomeric alcohol is likely to produce fragment ions at m/z = 15, 29, and 43 in its mass spectrum.
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Question text
20 Which alcohol is likely to have fragment ions at m/z = 15, 29 and 43 in its mass spectrum?
OH
A
B OH
OH
C
OH
D
Your answer [1]
Mark scheme
Show the mark scheme
20 C 1 AO2.5
How to answer it
Mass Spectrometry: Fragment Ions of Alcohol Isomers
This question assesses your ability to deduce the structures of alkyl fragment ions formed during mass spectrometry fragmentation, relate m/z values to specific carbocation fragments ( CH₃⁺ , C₂H₅⁺ , C₃H₇⁺ ), and identify which alcohol isomer can produce all three fragments via simple single-bond cleavage of its carbon backbone.
Question 20 (Multiple Choice)
Identifying an alcohol yielding m/z = 15, 29, and 43 fragments
✅ Correct Answer: C (Pentan-2-ol)
Mark Scheme Award: 1 Mark (AO2.5)
Pentan-2-ol ( CH₃–CH₂–CH₂–CH(OH)–CH₃ ) has an unbranched propyl chain attached to the –CH(OH)–CH₃ group. Breaking single C–C bonds along this straight section directly generates all three carbocation fragments:
- m/z = 15: methyl cation, [CH₃]⁺
- m/z = 29: ethyl cation, [CH₃CH₂]⁺ (or [C₂H₅]⁺ )
- m/z = 43: propyl cation, [CH₃CH₂CH₂]⁺ (or [C₃H₇]⁺ )
💡 Key Knowledge: Common Fragment Ions
Memorise the masses of common alkyl fragments formed by homolytic/heterolytic C–C bond cleavage in the mass spectrometer:
- m/z = 15: [CH₃]⁺ (12 + 3 = 15)
- m/z = 29: [C₂H₅]⁺ (24 + 5 = 29)
- m/z = 43: [C₃H₇]⁺ (36 + 7 = 43)
- m/z = 57: [C₄H₉]⁺ (48 + 9 = 57)
Note: Fragment ions must carry a positive charge ( + ) to be detected by the mass spectrometer.
🧠 Exam Technique: Systematic Elimination
Break the four isomers into their skeletal fragments by cutting individual C–C bonds:
- To yield m/z = 29, the molecule must contain a straight ethyl group ( –CH₂CH₃ ).
- To yield m/z = 43, the molecule must contain a 3-carbon unit, either propyl ( –CH₂CH₂CH₃ ) or isopropyl ( –CH(CH₃)₂ ).
- Eliminate options that lack either an ethyl group or a 3-carbon alkyl group!
📐 Detailed Isomer Analysis
| Option | Structure & Name | Can it form m/z = 29 ( C₂H₅⁺ )? | Can it form m/z = 43 ( C₃H₇⁺ )? | Outcome |
|---|---|---|---|---|
| A | 3-methylbutan-1-ol (CH₃)₂CH–CH₂–CH₂OH | ❌ No ethyl ( –CH₂CH₃ ) group present. | ✅ Yes: isopropyl cation, [(CH₃)₂CH]⁺ . | Incorrect |
| B | 2-methylbutan-1-ol CH₃CH₂–CH(CH₃)–CH₂OH | ✅ Yes: ethyl cation, [CH₃CH₂]⁺ . | ❌ No propyl or isopropyl group can form via a single C–C cleavage. | Incorrect |
| C | pentan-2-ol CH₃CH₂CH₂–CH(OH)CH₃ | ✅ Yes: cleavage of C3–C4 gives [CH₃CH₂]⁺ . | ✅ Yes: cleavage of C2–C3 gives [CH₃CH₂CH₂]⁺ . | Correct ✅ |
| D | 3-methylbutan-2-ol (CH₃)₂CH–CH(OH)CH₃ | ❌ No ethyl group; only methyls and an isopropyl group. | ✅ Yes: isopropyl cation, [(CH₃)₂CH]⁺ . | Incorrect |
❌ Common Misconceptions & Traps
- Confusing Isopropyl and Propyl: Options A and D easily form m/z = 43 (isopropyl), tempting students to pick them. However, neither contains an ethyl group ( –CH₂CH₃ ), so neither can directly produce m/z = 29 by simple fragmentation.
- Assuming m/z = 29 is [CHO]⁺ : While [CHO]⁺ has a mass of 29 (12 + 1 + 16), it is typical of aldehydes, not simple cleavage of secondary/primary saturated alcohols without oxidation/rearrangement.
- Forgetting to check the "AND": The question demands all three fragments: 15 and 29 and 43. An isomer that produces only two of them is incorrect.
Topics
Module 4: Core organic chemistry · 4.2 Alcohols, haloalkanes and analysis
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.