Mass Spectrometry (MS) Fragmentation Patterns (HL) (DP IB Chemistry): Revision Note
Mass spectrometry (MS) fragmentation patterns
To analyse a sample in a mass spectrometer:
The molecules are vaporised
Then bombarded with a beam of high-speed electrons
The high speed electrons knock an electron off some of the molecules, creating molecular ions (M+):
molecule molecular ion (M
) + e-
The detected ions produce a mass spectrum
This is a molecular fingerprint used to identify compounds via a spectral database
The peak with the highest m/z value is the molecular ion (M+) peak
This gives the molecular mass (Mr) of the compound
In addition to the molecular ion peak, many compounds show a smaller [M+1] peak due to the presence of the carbon-13 isotope (13C)
The M+1 peak
The [M+1] peak is a smaller peak next to the molecular ion peak
Mass spectrum of pent-1-ene

The height of the [M+1] peak depends on the number of carbon atoms in the molecule:
More carbon atoms means a higher [M+1] peak
For example, hexane (6 carbon atoms) has a larger [M+1] peak than ethane (2 carbon atoms)
Worked Example
Determine whether the following mass spectrum belongs to propanal or butanal.

Answer:
The molecular ion of propanal is CH3CH2CHO+
Mr = (3 x 12.01) + (6 x 1.01) + 16.00 = 58.09
The molecular ion of butanal is CH3CH2CH2CHO+
Mr = (4 x 12.01) + (8 x 1.01) + 16.00 = 72.12
So, the mass spectrum corresponds to propanal as the molecular ion peak is at m/z = 58
Fragmentation patterns
The molecular ion peak gives the molecular mass (Mᵣ) of the compound
However, different compounds can have the same molecular mass
To confirm the structure, fragmentation analysis is used

Fragments arise due to:
The formation of characteristic fragments
CH3+ at m/z = 15
C2H5+ at m/z = 29
The loss of small molecules
H2O at m/z = 18
CO at m/z = 28
CO2 at m/z = 44
The IB Chemistry data booklet (Section 22) contains a list of common neutral fragments lost:
Mass lost (Mr) | Possible neutral fragment lost |
---|---|
15 |
|
17 |
|
18 | H2O |
28 | CH2=CH2 |
29 |
|
31 |
|
45 |
|
Alkanes and alkyl cations
In alkanes, fragmentation typically occurs by breaking C–C bonds
The resulting fragments are alkyl cations with identifiable m/z values:
Mass lost (Mr) | Possible fragment lost |
---|---|
15 | CH3+ |
29 | C2H5+ |
43 | C3H7+ |
57 | C4H9+ |
71 | C5H11+ |
85 | C6H13+ |
Fragmentation in a mass spectrum

Halogenoalkanes
Halogenoalkanes often show multiple peaks around the molecular ion
This is caused by the presence of different isotopes of the halogen
Mass spectrum of a bromine-containing compound
Bromine exists as two major isotopes: 79Br and 81Br
These are present in roughly equal abundance
So compounds containing bromine show two molecular ion peaks of similar height:
M (with 79Br)
M + 2 (with 81Br)

Alcohols
Alcohols often undergo dehydration
This produces a peak 18 units below the molecular ion
The CH2OH+ fragment is also commonly observed at m/z = 31
Fragmentation in propan-1-ol
The following fragmentation patterns are commonly observed in propan-1-ol:
Mass lost (Mr) | Possible fragment lost | Reason |
---|---|---|
29 | C2H5+ | Loss of |
31 | CH2OH+ | Characteristic alcohol fragment |
42 | C3H6+ | Loss of H2O |
59 | C3H7O+ | Loss of H |

Worked Example
Alcohol fragmentation
Which alcohol is not likely to have a fragment ion at m/z at 43 in its mass spectrum?
A. (CH3)2CHCH2OH
B. CH3CH(OH)CH2CH2CH3
C. CH3CH2CH2CH2OH
D. CH3CH2CH(OH)CH3
Answer:
The correct answer is option D - CH3CH2CH(OH)CH3 (2-butanol)
A fragment at m/z = 43 typically indicates one of the following ions:
CH3CH2CH2+
(CH3)2CH+
D is unlikely to form either of these fragments
Therefore, it is not expected to produce a peak at m/z = 43
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