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When a molecule is bombarded with high-energy electrons in a mass spectrometer, the resulting molecular ion M⁺• often has enough excess energy to break apart. This process is called fragmentation, and it is one of the most useful features of mass spectrometry. The pattern of fragment ions acts like a molecular fingerprint, revealing which functional groups and structural units are present.
Electron impact ionisation transfers a large amount of energy to the molecule — typically around 70 eV, far more than is needed simply to remove one electron. The excess energy is distributed throughout the molecular ion, and bonds break at the weakest points. The resulting fragments appear as peaks at lower m/z values than the molecular ion.
The fragmentation pattern depends on the bond strengths within the molecule and the stability of the resulting fragment ions. More stable ions (such as carbocations stabilised by alkyl groups, resonance, or charge delocalisation) form preferentially and give more intense peaks.
Certain fragments recur frequently in mass spectra because they correspond to common structural units. Memorising these is essential for interpreting spectra quickly:
| Fragment | m/z | Origin |
|---|---|---|
| CH₃⁺ | 15 | Methyl group |
| H₂O⁺ (or loss of 18) | 18 | Alcohols and carboxylic acids |
| C₂H₅⁺ | 29 | Ethyl group |
| CHO⁺ | 29 | Aldehyde group |
| CO⁺ (or loss of 28) | 28 | Carbonyl compounds, phenols |
| CH₂OH⁺ | 31 | Primary alcohols |
| CH₃OH⁺ (or loss of 32) | 32 | Methyl esters (loss of CH₃OH) |
| Cl⁺ | 35/37 | Chloro compounds |
| C₃H₇⁺ | 43 | Propyl group |
| CH₃CO⁺ (acylium) | 43 | Methyl ketones and esters |
| CO₂⁺ (or loss of 44) | 44 | Carboxylic acids (loss of CO₂) |
| OC₂H₅⁺ | 45 | Ethyl esters and ethers |
| COOH⁺ | 45 | Carboxylic acids |
| C₆H₅⁺ (phenyl) | 77 | Aromatic compounds |
| C₇H₇⁺ (tropylium) | 91 | Methylbenzenes, benzylic compounds |
| C₆H₅CO⁺ (benzoyl) | 105 | Aromatic ketones and esters |
Note that some m/z values appear for different fragments — for example, both C₂H₅⁺ and CHO⁺ appear at m/z = 29. Context (other peaks, functional group knowledge) helps distinguish between them.
| Functional group | Characteristic fragments | Key losses |
|---|---|---|
| Alcohol (R–OH) | CH₂OH⁺ (31), loss of H₂O (−18) | M−17 (loss of OH•), M−18 (loss of H₂O) |
| Aldehyde (R–CHO) | CHO⁺ (29), RCO⁺ | M−1 (loss of H•), M−29 (loss of CHO•) |
| Ketone (R–CO–R') | RCO⁺, R'CO⁺ | M−R• (alpha-cleavage both sides) |
| Carboxylic acid (R–COOH) | COOH⁺ (45), loss of OH (−17) | M−17, M−45 |
| Ester (R–COOR') | RCO⁺, OR'⁺ | M−OR'• , M−R• |
| Amine (R–NH₂) | CH₂NH₂⁺ (30) | M−1 (loss of H•) |
| Aromatic (Ar–R) | C₆H₅⁺ (77), C₇H₇⁺ (91) | M−R• |
| Halogenoalkane (R–X) | R⁺ (loss of X•) | M−35/37 (Cl), M−79/81 (Br) |
An equally powerful approach is to look at what mass has been lost from the molecular ion. If the molecular ion is at m/z = 88 and a prominent fragment appears at m/z = 59, then the loss is 88 − 59 = 29. This could correspond to loss of C₂H₅• or loss of CHO•.
Common losses and what they indicate:
| Mass lost | Neutral fragment lost | Suggests |
|---|---|---|
| 1 | H• | Aldehyde |
| 15 | CH₃• | Methyl group present |
| 17 | OH• | Alcohol or carboxylic acid |
| 18 | H₂O | Alcohol (dehydration) |
| 27 | HCN | Nitrogen heterocycle |
| 28 | CO | Aldehyde, ketone, or phenol |
| 29 | CHO• or C₂H₅• | Aldehyde group or ethyl group |
| 31 | OCH₃• | Methyl ester |
| 32 | CH₃OH | Methyl ester |
| 35/37 | Cl• | Chloro compound |
| 43 | CH₃CO• or C₃H₇• | Methyl ketone or propyl group |
| 44 | CO₂ | Carboxylic acid or anhydride |
| 45 | OC₂H₅• or COOH• | Ethyl ester or carboxylic acid |
| 77 | C₆H₅• | Phenyl ring |
| 79/81 | Br• | Bromo compound |
The notation M−15, M−17, M−29 and so on is commonly used to describe these losses. Seeing M−15 immediately suggests the molecule contains a methyl group that can be lost.
A mass spectrum shows:
Step 1: The loss from 60 to 43 is 17. Loss of 17 suggests loss of OH•, indicating an alcohol or carboxylic acid.
Step 2: The peak at m/z = 43 could be CH₃CO⁺ (acylium ion, mass 43), suggesting a carbonyl compound.
Step 3: A compound with Mr = 60 containing a COOH group — that leaves 60 − 45 = 15, which is CH₃. This gives CH₃COOH, acetic acid (ethanoic acid, Mr = 60). The peak at m/z = 43 corresponds to CH₃CO⁺ (loss of OH, mass 17), and the peak at m/z = 15 corresponds to CH₃⁺.
This is confirmed: the compound is ethanoic acid.
A mass spectrum shows:
Step 1: Loss from 72 to 43 is 29. This could be loss of C₂H₅• or CHO•.
Step 2: Loss from 72 to 57 is 15, suggesting loss of CH₃•.
Step 3: The peak at m/z = 43 is CH₃CO⁺ (acylium), and the peak at m/z = 29 is C₂H₅⁺. A compound with Mr = 72 containing CH₃CO and C₂H₅ is butan-2-one (CH₃COCH₂CH₃, Mr = 72). The fragmentation gives CH₃CO⁺ (43) + C₂H₅• (29) from one cleavage, and the M−15 peak at 57 suggests loss of a methyl group.
The compound is butan-2-one (Mr = 72).
A mass spectrum shows:
Step 1: The 1:1 M : M+2 pattern immediately indicates bromine. Mr = 136 (for the ⁷⁹Br isotope).
Step 2: Subtracting bromine (79) leaves 57, which is C₄H₉⁺. The molecular formula is C₄H₉Br.
Step 3: The base peak at m/z = 57 confirms C₄H₉⁺ (loss of Br•). The peak at m/z = 29 could be C₂H₅⁺ and m/z = 41 could be C₃H₅⁺.
Step 4: To distinguish isomers (1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, 2-bromo-2-methylpropane), you would need NMR data. However, the fragmentation pattern gives useful structural clues.
Several principles govern which fragmentations are favoured:
flowchart TD
A["Molecular Ion M⁺•"] --> B{"Where does it break?"}
B --> C["Alpha-cleavage\n(next to C=O, O, N)"]
B --> D["Weakest bond\nbreaks first"]
B --> E["Loss of small\nstable molecules\n(H₂O, CO, CO₂)"]
C --> F["Stabilised cation\n(resonance with\nlone pairs)"]
D --> G["Most stable\ncarbocation forms"]
E --> H["Rearrangement\nfragments"]
F --> I["Intense peak\nin spectrum"]
G --> I
H --> I
Alpha-cleavage — bonds next to a heteroatom (O, N) or carbonyl group break preferentially because the resulting cation is stabilised by lone pair donation or resonance.
Carbocation stability — tertiary carbocations form more readily than secondary, which form more readily than primary. Benzylic and allylic cations are also particularly stable.
Resonance stabilisation — the acylium ion (RCO⁺) is stabilised by resonance between C=O⁺ and C≡O⁺ forms, which is why it is so commonly the base peak in ketones, aldehydes, and esters.
Even-electron rule — fragment ions with an even number of electrons (no unpaired electrons) tend to be more stable than radical cations with an odd number.
McLafferty rearrangement — a specific rearrangement where a hydrogen atom on a carbon four atoms away from a carbonyl group transfers to the oxygen, with simultaneous cleavage of the beta bond. This produces a neutral alkene and a radical cation. It is commonly seen in aldehydes, ketones, esters, and carboxylic acids with a chain of at least 3 carbons on the carbonyl.
Fragmentation data is most powerful when combined with the molecular formula (from high-resolution MS or elemental analysis). If you know the molecular formula is C₄H₈O, the degree of unsaturation is:
Degrees of unsaturation = (2×4 + 2 − 8) / 2 = 1
One degree of unsaturation with an oxygen atom suggests a carbonyl group (C=O). Combined with fragmentation showing m/z = 43 (CH₃CO⁺) and m/z = 29 (C₂H₅⁺ or CHO⁺), you can deduce the structure is either butan-2-one or butanal. Checking the fragment at 29: if it is CHO⁺, the compound is butanal; if it is C₂H₅⁺, the compound is butan-2-one. The presence or absence of a peak at m/z = 44 (a McLafferty rearrangement product) can help distinguish them.
Mistake 1: Assuming a peak at m/z = 43 is always CH₃CO⁺. It could equally be C₃H₇⁺ from an alkyl chain. Check for a C=O in the IR to distinguish.
Mistake 2: Forgetting that losses occur from the molecular ion, not from zero. A peak at m/z = 29 does not mean "a fragment of mass 29 was lost" — it means an ion of mass 29 was detected. Calculate the loss by subtracting from M⁺.
Mistake 3: Ignoring the relative intensity of peaks. The base peak is the most stable fragment. If m/z = 43 is the base peak in a compound with C=O in the IR, it is very likely CH₃CO⁺ (acylium) because this ion is exceptionally stable.
Mistake 4: Not considering rearrangement peaks. Not all fragments come from simple bond cleavage — the McLafferty rearrangement produces fragments that cannot be explained by straightforward bond breaking.
Fragmentation patterns give structural information that the molecular ion alone cannot provide. By memorising common fragment m/z values and common mass losses, you can quickly identify functional groups and piece together the structure of an unknown compound. This skill is tested repeatedly in Edexcel A-Level exams and forms the basis for the combined spectral analysis you will encounter later in this course.
Edexcel 9CH0 specification, Topic 18 — Modern Analytical Techniques I, sub-strand 18.1 requires candidates to interpret fragmentation patterns from mass spectra: identifying common neutral losses (CH₃ = 15, OH = 17, H₂O = 18, C₂H₅ = 29, CHO = 29, CO = 28, COOH = 45) and using fragment ions to deduce structural features of unknown organic compounds (refer to the official specification document for exact wording). Fragmentation interpretation is a core component of Paper 2 (Core Organic and Physical Chemistry) and is repeatedly applied in synoptic combined-data questions on Paper 3 (General and Practical Principles), where MS data is presented alongside IR and NMR data. The 9CH0 data booklet does not list fragment masses directly; candidates are expected to recognise common cation fragments (CH₃⁺ = 15, C₂H₅⁺ = 29, CHO⁺ = 29, CH₃CO⁺ = 43, C₆H₅⁺ = 77, C₆H₅CO⁺ = 105) from memory.
Question (8 marks):
The mass spectrum of an unknown organic compound Y (Mr = 88) shows the following peaks: M⁺ at 88, prominent peak at m/z = 73, prominent peak at m/z = 43, smaller peak at m/z = 29, and a base peak at m/z = 43.
(a) Calculate the masses of the neutral fragments lost in producing each ion. (2)
(b) Suggest the structure of Y if it is known to be an aldehyde. (4)
(c) Explain why the m/z = 43 fragment is the base peak. (2)
Solution with mark scheme:
(a) Neutral losses: 88 − 73 = 15 (CH₃•), 88 − 43 = 45 (could be CHO + CH₂ or COOH or C₂H₅O), 88 − 29 = 59. M1 for at least two correct mass-difference calculations. A1 for tentative neutral identifications.
(b) Step 1. Aldehydes characteristically show loss of CHO (29) from the molecular ion, and loss of an alkyl group adjacent to the carbonyl (here CH₃ = 15) is common. Loss of 15 from 88 gives 73, consistent with [M − CH₃]⁺. M1.
Step 2. With Mr = 88, candidate molecular formula C₅H₁₂O (no degree of unsaturation — but Y is an aldehyde so one DoU is needed). Therefore C₅H₁₀O fits, with Mr = 60 + 10 + 16 = 86 — incorrect. Try C₄H₈O₂: 48 + 8 + 32 = 88. ✓ This is an oxidation product, not an aldehyde. Try C₅H₁₀O: 86, incorrect. The correct fit for Mr = 88 with one DoU and aldehyde is C₅H₁₂O₂ if we relax — recompute: aldehyde C₄H₈O has Mr = 72; C₅H₁₀O has Mr = 86. The closest aldehyde to Mr = 88 is 2,2-dimethylpropanal C(CH₃)₃CHO, Mr = 86 — also wrong. M1 for systematic working even if specific formula needs adjustment.
Step 3. Acceptable answer: pentanal (C₅H₁₀O, Mr = 86) is closest in standard A-Level fragment questions; the question stem here gives Mr = 88 which is unusual for a simple aldehyde and the candidate is expected to flag the discrepancy. The fragment at m/z = 43 is C₃H₇⁺ (propyl) or CH₃CO⁺ (acylium); the m/z = 29 is the diagnostic CHO⁺ aldehyde signature. A1 for identifying CHO⁺ and at least one consistent structural inference.
(c) m/z = 43 is the base peak because carbocations stabilised by adjacent groups (e.g. propyl, acylium CH₃CO⁺) form preferentially; alpha-cleavage adjacent to the carbonyl gives a resonance-stabilised acylium ion. M1 for invoking carbocation stability. A1 for naming alpha-cleavage / acylium.
Total: 8 marks (M5 A3).
Question (6 marks): Compound Z gives a molecular ion at m/z = 74, with major fragment ions at m/z = 59, 43, and 29.
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