Q8(b). Mass spectrum: the M-1 peak of pentanal
Identify the compound, which can give M⁺-1 peak among the following. Discuss its fragmentation pattern with m/e values:
(i) CH₃CH₂CH₂CH₂CHO (ii) CH₃CHOHCH₂CH₂CH₃ (iii) CH₃CH₂COCH₂CH₃ (iv) CH₃COOCH₂CH₂CH₃
Answer in brief
Pentanal (i) gives the M-1 peak at m/z 85 (acylium ion); other peaks: 86 (M), 57 (C₄H₉⁺), 44 (McLafferty), 43, 41, 29 (HCO⁺).
Key points
- the radical cation loses the aldehydic H by α-cleavage; the acylium ion R-C#O⁺ <→ R-C+=O is stabilised by the oxygen lone pair. This peak is characteristic of aldehydes.
- pentanal has γ-hydrogens; through a six-membered transition state a γ-H moves to the carbonyl oxygen and the α–β bond breaks, giving the enol radical cation CH₂=CHOH⁺˙ (m/z 44) and propene. A strong 44 peak is typical of…
- 29 and 57: cleavage of the C1–C2 bond, the charge staying either on HCO⁺ (29) or on the butyl cation (57).
- pentan-2-ol cleaves next to C–OH to give CH₃CH=OH⁺ (m/z 45) and loses water (M-18, 70); pentan-3-one gives the acylium C₂H₅CO⁺ (m/z 57); propyl acetate gives CH₃CO⁺ (m/z 43) and the McLafferty-type ion at 61. None of them has…
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. All four compounds are C₅ oxygen compounds. An intense M-1 peak needs an easily lost hydrogen whose loss gives a stable cation. Only the aldehyde has an H on the carbonyl carbon: α-cleavage of that C–H gives a resonance-stabilised acylium ion. Answer: (i) pentanal, C₅H₁₀O, M=86.
Scheme: Loss of the aldehydic hydrogen gives the butylacylium ion C₄H₉C#O⁺.
| m/z | ion | how it forms |
| 86 | C₅H₁₀O⁺˙ (M) | loss of one electron (an O lone-pair electron) |
| 85 | C₄H₉C#O⁺ (M-1) | α-cleavage: loss of the aldehydic H atom |
| 57 | C₄H₉⁺ (M-29) | α-cleavage on the other side: loss of CHO• |
| 44 | CH₂=CH-OH⁺˙ | McLafferty rearrangement: γ-H transfer to O, loss of propene (42) |
| 43, 41 | C₃H₇⁺, C₃H₅⁺ | alkyl fragments of the chain |
| 29 | HC#O⁺ | α-cleavage with charge on the formyl group |
- M-1 (85): the radical cation loses the aldehydic H by α-cleavage; the acylium ion R-C#O⁺ <→ R-C+=O is stabilised by the oxygen lone pair. This peak is characteristic of aldehydes.
- McLafferty (44): pentanal has γ-hydrogens; through a six-membered transition state a γ-H moves to the carbonyl oxygen and the α–β bond breaks, giving the enol radical cation CH₂=CHOH⁺˙ (m/z 44) and propene. A strong 44 peak is typical of straight-chain aldehydes with a γ-H.
- 29 and 57: cleavage of the C1–C2 bond, the charge staying either on HCO⁺ (29) or on the butyl cation (57).
- The others: pentan-2-ol cleaves next to C–OH to give CH₃CH=OH⁺ (m/z 45) and loses water (M-18, 70); pentan-3-one gives the acylium C₂H₅CO⁺ (m/z 57); propyl acetate gives CH₃CO⁺ (m/z 43) and the McLafferty-type ion at 61. None of them has an aldehydic H, so none gives a significant M-1 peak.
Conclusion. Pentanal (i) gives the M-1 peak at m/z 85 (acylium ion); other peaks: 86 (M), 57 (C₄H₉⁺), 44 (McLafferty), 43, 41, 29 (HCO⁺).
Application. The M-1 and m/z 29 peaks, together with the McLafferty peak at 44, distinguish an aldehyde from an isomeric ketone (which shows acylium ions such as 57 or 43 instead) in GC–MS analysis of flavours and essential oils.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram
All parts of Question 8
- 8(a) Woodward–Fieser rules: which structure absorbs at 229 nm15 marks
- 8(b) Mass spectrum: the M-1 peak of pentanal10 marks
- 8(c) Equivalent protons and number of 1H NMR signals10 marks
- 8(d) SeO₂, NBS and Pb(OAc)₄ reactions15 marks