Mass Spectrometry — How to Read a Fragmentation Pattern
A mass spectrum looks like a forest of lines until you realise it is a short list of yes/no questions asked in a fixed order. Does the molecule contain chlorine or bromine? How many carbons? Is there a nitrogen? Which bond broke first, and why that one? Answer those in sequence and a structure falls out. This article gives the sequence, the numbers you must know from memory, and five spectra deduced from start to finish — the exact skill tested in the combined-spectroscopy questions of CSIR-NET, GATE and IIT-JAM.
What the instrument actually does
In electron ionisation (EI), the standard method for small organic molecules, a 70 eV electron beam knocks one electron out of the molecule to give a radical cation M+•. That is far more energy than any bond needs, so the ion breaks apart, and the fragments that survive are the ones that are most stable. The detector records m/z — mass divided by charge — and since almost every ion carries a single positive charge, you may read m/z as mass. Two definitions to fix now:
- Molecular ion (M+•) — the intact ionised molecule. It is the highest genuine peak in the spectrum apart from its own isotope peaks. It may be very weak, and in some compounds it is absent altogether.
- Base peak — the tallest peak, scaled to 100% relative abundance. It is the most stable fragment. It has no necessary connection to the molecular ion.
Note the contrast with soft ionisation. Electrospray (ESI) and MALDI produce [M+H]+ with almost no fragmentation, which is excellent for getting a molecular mass and useless for the structural deductions below. Everything in this article applies to EI spectra.
Step 1 — Identify the molecular ion and test it
The nitrogen rule. For any compound made of C, H, N, O, S and halogens:
Even nominal molecular mass → zero or an even number of nitrogen atoms
The rule works because nitrogen is the only common element whose valence (3) and nominal mass (14) have opposite parity. It is the single fastest test you can run on a spectrum.
The M+1 carbon count. Carbon is 98.9% 12C and about 1.1% 13C, so each carbon atom in the molecule adds roughly 1.1% to the height of the peak one mass unit above the molecular ion:
This is a good estimate up to roughly 20 carbons, and it needs a clean, well-resolved molecular ion to be trustworthy.
Step 2 — Read the isotope pattern for halogens and sulfur
| Element present | Pattern at M, M+2, M+4 | How to spot it |
|---|---|---|
| One Cl | 3 : 1 | A clear peak two units above M at about one third the height |
| One Br | 1 : 1 | Twin peaks of almost equal height, two units apart — unmistakable |
| Two Cl | 9 : 6 : 1 | Three peaks; the middle one about two thirds of the first |
| Two Br | 1 : 2 : 1 | Three peaks with the middle one tallest |
| One Cl + one Br | 3 : 4 : 1 | The M+2 peak is taller than M |
| One S | M+2 ≈ 4.4% of M | A small but real M+2 where none is expected |
| C, H, N, O only | M+2 negligible | Any significant M+2 means a heavier isotope is present |
Two isotope facts underpin the whole table: chlorine is about 76% 35Cl and 24% 37Cl (close to 3:1), and bromine is about 51% 79Br and 49% 81Br (close to 1:1). Everything else in the table is those two ratios multiplied out.
Step 3 — Compute the ring-and-double-bond equivalents
Once you have a candidate molecular formula, this tells you how much unsaturation it must contain:
C = carbons, H = hydrogens, X = halogens, N = nitrogens. Oxygen and sulfur do not appear.
For C8H8O: RDBE = 8 − 8/2 + 0 + 1 = 8 − 4 + 1 = 5. A benzene ring accounts for four (three C=C plus the ring itself) and a carbonyl for one — exactly five, which is consistent with acetophenone. If your proposed structure and your RDBE disagree, one of them is wrong.
Step 4 — Read the neutral losses
Do not try to identify fragment ions in isolation. Subtract each fragment from the molecular ion and ask what neutral piece left. That difference is far more diagnostic than the fragment mass itself.
| Loss (M − x) | Neutral lost | What it tells you |
|---|---|---|
| 1 | H• | Aldehyde, or a stabilised benzylic/allylic position |
| 15 | CH3• | A methyl group is present, often on a quaternary or carbonyl carbon |
| 17 / 18 | •OH / H2O | Alcohol (M − 18 is very common; M+• itself may be missing) |
| 28 | CO, C2H4 or N2 | Loss of CO from an acylium or a phenol; loss of ethene by a retro-Diels–Alder |
| 29 | •CHO or •C2H5 | Aldehyde, or an ethyl group |
| 31 | •OCH3 | Methyl ester or methyl ether |
| 35 / 36 | Cl• / HCl | Chloroalkane (confirm with the 3:1 isotope pattern) |
| 43 | •C3H7 or CH3CO• | Propyl group, or an acetyl group |
| 45 | •OC2H5 or •COOH | Ethyl ester, or carboxylic acid |
| 79 / 81 | Br• | Bromoalkane (confirm with the 1:1 pattern) |
Step 5 — Recognise the three fragmentation mechanisms
α-Cleavage. The bond broken is the one next to the heteroatom or carbonyl, not the bond to it, because the radical cation is stabilised by the remaining lone pair. Ketones give acylium ions RCO+; alcohols and amines give oxocarbenium and iminium ions. The larger alkyl radical is lost preferentially, so the smaller acylium ion is usually the bigger peak.
Benzylic and allylic cleavage. Any bond one atom away from a ring or a double bond breaks readily because the resulting cation is delocalised. This is why so many aromatic compounds show a strong peak at m/z 91 (the tropylium ion C7H7+).
The McLafferty rearrangement. A carbonyl compound with a hydrogen on the γ carbon can pass that hydrogen to the carbonyl oxygen through a six-membered cyclic transition state, expelling a neutral alkene. The product is a radical cation, so — unlike a simple cleavage — it appears at an even m/z when the molecule contains no nitrogen. Spotting an unexpected even-mass fragment is often the fastest route to identifying a ketone or ester chain length.
| Diagnostic fragment | Ion | Compound class |
|---|---|---|
| m/z 91 | C7H7+ (tropylium) | Benzyl, alkylbenzene |
| m/z 77, then 51 | C6H5+, then loss of C2H2 | Monosubstituted benzene ring |
| m/z 105 | C6H5CO+ | Benzoyl group |
| m/z 43 | CH3CO+ or C3H7+ | Methyl ketone, or propyl chain |
| m/z 31 | CH2=OH+ | Primary alcohol |
| m/z 30 | CH2=NH2+ | Primary amine |
| m/z 74 | C3H6O2+• (McLafferty) | Methyl ester of a straight-chain acid |
| 29, 43, 57, 71 … (14 apart) | CnH2n+1+ | Saturated hydrocarbon chain |
Five spectra worked from start to finish
Deduction 1 — twin peaks at 94 and 96.
M+• = 94 with an equally intense peak at 96 → one bromine
(1:1 pattern).
Remove bromine: 94 − 79 = 15 → CH3.
Formula CH3Br: check 12 + 3(1) + 79 = 94 ✓.
RDBE = 1 − (3 + 1)/2 + 0 + 1 = 1 − 2 + 1 = 0 — fully saturated, consistent
with bromomethane.
Confirming peak: m/z 15 for CH3+.
Deduction 2 — M at 78 with M+2 at one third the height.
The 3:1 ratio at 78/80 means one chlorine.
78 − 35 = 43 → C3H7, so the formula is
C3H7Cl: 36 + 7 + 35 = 78 ✓.
RDBE = 3 − (7 + 1)/2 + 1 = 3 − 4 + 1 = 0 ✓.
Which isomer? A strong M − 15 pair at 63/65 (loss of CH3• to give
the secondary, chlorine-stabilised CH3CHCl+) points to
2-chloropropane. A spectrum dominated instead by m/z 43
(C3H7+, simple loss of Cl•) with little at 63 points to
1-chloropropane. The isotope pattern gives the formula; the fragmentation gives the
isomer.
Deduction 3 — M 72, M+1 about 4.4%, base peak 43.
Nitrogen rule: 72 is even → zero or an even number of nitrogens.
Carbon count: 4.4 ⁄ 1.1 = 4 carbons.
Remaining mass: 72 − (4 × 12) − H. Trying C4H8O: 48 + 8 + 16 = 72 ✓.
RDBE = 4 − 8/2 + 1 = 1 → one ring or one double bond.
Fragments: 57 = M − 15 (loss of CH3•) and 43 = M − 29 (loss of
•C2H5). Two different α-cleavages from the same carbonyl, with the
smaller acylium CH3CO+ at 43 as the base peak.
Answer: butan-2-one, CH3COCH2CH3.
Ruling out the isomer: butanal (also C4H8O) would show m/z 29
(CHO+), a strong M − 1 at 71, and a McLafferty ion at m/z 44. None of those is
present here.
Deduction 4 — M 120, base peak 105, then 77 and 51.
120 − 105 = 15, loss of CH3• → a methyl group.
105 − 77 = 28, loss of CO → 105 must be an acylium ion, and 77 is
C6H5+.
77 − 51 = 26, loss of C2H2 — the classic decay of a
phenyl cation, confirming a monosubstituted benzene ring.
Assemble: C6H5 + CO + CH3 = C8H8O =
96 + 8 + 16 = 120 ✓.
RDBE = 8 − 8/2 + 1 = 5 = ring (4) + C=O (1) ✓.
Answer: acetophenone. Note that the base peak at 105 is the benzoyl ion —
the larger radical (CH3•) is lost in preference, exactly as the α-cleavage rule
predicts... and in fact here the smaller radical leaves, because the resulting
benzoyl cation is resonance-stabilised by the ring. Stability of the ion always outranks the
size of the radical lost.
Deduction 5 — an odd molecular ion at 59.
59 is odd → by the nitrogen rule the molecule contains an odd number of
nitrogens, almost certainly one.
Subtract N: 59 − 14 = 45, which as C3H9 gives 36 + 9 = 45 ✓, so the
formula is C3H9N.
RDBE = 3 − 9/2 + 1/2 + 1 = 3 − 4.5 + 0.5 + 1 = 0 — a saturated amine.
Now the fragments decide the isomer: a strong m/z 30
(CH2=NH2+) means a primary amine such as
propan-1-amine; a dominant m/z 58 (M − 1) with 44 and 42 instead points to trimethylamine,
where α-cleavage produces
(CH3)2N=CH2+ at 58.
High-resolution mass spectrometry settles the formula
Nominal masses are degenerate: several formulas share the same integer mass. High-resolution instruments measure to four decimal places, and because the exact masses of the isotopes are not integers, each formula has a unique value. Using 12C = 12.0000 (by definition), 1H = 1.00783 and 16O = 15.9949:
| Formula | Nominal mass | Exact (monoisotopic) mass | Working |
|---|---|---|---|
| C3H8O | 60 | 60.0575 | 36.0000 + 8(1.00783) + 15.9949 = 36.0000 + 8.0626 + 15.9949 |
| C2H4O2 | 60 | 60.0211 | 24.0000 + 4(1.00783) + 2(15.9949) = 24.0000 + 4.0313 + 31.9898 |
The difference of 0.0364 u is easily resolved on a modern instrument, so propan-1-ol and acetic acid can never be confused by HRMS even though both are "60" in a low-resolution spectrum. The same argument distinguishes the three common neutral losses at nominal mass 28: CO (27.9949), N2 (28.0061) and C2H4 (28.0313).
Mistakes that cost marks
- Taking the base peak as the molecular ion. In toluene the base peak is 91 and the molecular ion is 92; in many alcohols the molecular ion is invisible and M − 18 is the highest peak you can see. Always check whether the highest mass peak is consistent with the rest of the spectrum before naming it M+•.
- Assuming the molecular ion is always present. Highly branched alkanes, alcohols and some amines fragment so readily that M+• is not observed at all. That absence is itself structural information.
- Reading an M+2 peak as a fragment. M+2 from chlorine or bromine is an isotope peak of the same ion, not a different species.
- Forgetting that every fragment carrying the halogen also shows the isotope pattern. If a fragment at 63 lacks the 3:1 twin, it does not contain the chlorine.
- Applying the nitrogen rule to an even-electron fragment. The rule as stated is for the molecular ion. Ordinary cleavage fragments are even-electron cations and follow the opposite parity, which is precisely why an unexpected even-mass fragment signals a rearrangement such as McLafferty.
- Confusing average molar mass with the m/z you observe. A mass spectrum shows individual isotopologues, so CH3Br appears at 94 and 96 — never at its average molar mass of about 94.94. Use nominal or monoisotopic masses for spectra and average molar masses for weighing out reagents.
- Expecting EI fragmentation from an ESI spectrum. Soft ionisation gives [M+H]+ and little else. If a question shows a single dominant peak one unit above the molecular mass, it is not asking you to fragment anything.
- Answering from the mass spectrum alone. In a combined-spectroscopy question the mass spectrum gives the formula and the skeleton; the IR gives the functional group and the NMR gives the connectivity. Use all three and say which piece of evidence supports which conclusion.
Where this appears in the exam
| Exam | Typical demand |
|---|---|
| CSIR-NET Chemical Sciences | Combined MS + IR + NMR structure elucidation; identify a halogen from an isotope pattern; recognise a McLafferty ion and deduce chain length |
| GATE Chemistry | Nitrogen rule and RDBE calculation; predict the base peak of a given compound; assign a named neutral loss |
| IIT-JAM / CUET-PG | Molecular formula from M and M+1; recognising m/z 91, 77 and 43; distinguishing isomers from fragmentation |
| MSc coursework | Exact-mass calculations; α-cleavage versus rearrangement mechanisms; interpreting EI versus soft-ionisation data |
Every deduction above was arithmetic before it was chemistry — subtracting neutral losses, dividing an isotope ratio, adding exact masses to four decimal places. Keep a scientific calculator and the standard constants open beside the spectrum and check each subtraction as you make it. One caution while you work: the calculator's molar-mass tool returns the average molar mass from natural isotopic abundances, which is the right number for weighing a reagent and the wrong one for an m/z value — mass spectra need nominal or monoisotopic masses.
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