Mass Spectrometry (MS) Fragmentation Patterns (HL) (DP IB Chemistry)
Revision Note
Mass Spectrometry (MS) Fragmentation Patterns
When a compound is analysed in a mass spectrometer, vaporised molecules are bombarded with a beam of high-speed electrons
These knock off an electron from some of the molecules, creating molecular ions:
The relative abundances of the detected ions form a mass spectrum: a kind of molecular fingerprint that can be identified by computer using a spectral database
The peak with the highest m/z value is the molecular ion (M+) peak which gives information about the molecular mass of the compound
This value of m/z is equal to the relative molecular mass of the compound
The M+1 peak
The [M+1] peak is a smaller peak which is due to the natural abundance of the isotope carbon-13
The height of the [M+1] peak for a particular ion depends on how many carbon atoms are present in that molecule; the more carbon atoms, the larger the [M+1] peak is
For example, the height of the [M+1] peak for a hexane (containing six carbon atoms) ion will be greater than the height of the [M+1] peak of an ethane (containing two carbon atoms) ion
Worked Example
Determine whether the following mass spectrum belongs to propanal or butanal
Answer:
The mass spectrum corresponds to propanal as the molecular ion peak is at m/z = 58
Propanal arises from the CH3CH2CHO+ ion which has a molecular mass of 58
Butanal arises from the CH3CH2CH2CHO+ ion which has a molecular mass of 72
The molecular ion peak can be used to identify the molecular mass of a compound
However, different compounds may have the same molecular mass
To further determine the structure of the unknown compound, fragmentation analysis is used
Fragments may appear due to the formation of characteristic fragments or the loss of small molecules
For example, a peak at 29 is due to the characteristic fragment C2H5+
Loss of small molecules gives rise to differences between peaks of, for example, 18 (H2O), 28 (CO), and 44 (CO2)
An alcohol can typically dehydrate in a MS, so one peak to look for is M-18
Alkanes
Simple alkanes are fragmented in mass spectroscopy by breaking the C-C bonds
m/z values of some of the common alkane fragments are given in the table below
m/z values of fragments table
Fragment | m/z |
---|---|
CH3+ | 15 |
C2H5+ | 29 |
C3H7+ | 43 |
C4H9+ | 57 |
C5H11+ | 71 |
C6H13+ | 85 |
Fragmentation in a mass spectrum
Mass spectrum showing fragmentation of alkanes
Alcohols
Alcohols often tend to lose a water molecule giving rise to a peak at 18 below the molecular ion
Another common peak is found at m/e value 31 which corresponds to the CH2OH+ fragment
Loss of H• to form a C3H7O+ fragment with m/e = 59
Loss of a water molecule to form a C3H6+ fragment with m/e = 42
Loss of a •C2H5 to form a CH2OH+ fragment with m/e = 31
And the loss of •CH2OH to form a C2H5+ fragment with m/e = 29
For example, the mass spectrum of propan-1-ol shows that the compound has fragmented in four different ways:
Fragmentation in a mass spectrum
The mass spectrum of propan-1-ol shows that the compound has fragmented in four different ways
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
Because a line at m/z = 43 corresponds to an ion with a mass of 43 for example:
[CH3CH2CH2]+
[(CH3)2CH]+
2-butanol is not likely to have a fragment at m/z = 43 as it does not have either of these fragments in its structure.
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