Molecular Ion Peak & Fragmentation (Cambridge (CIE) AS Chemistry)
Revision Note
Mass Spectrometry: Deducing Molecular Formula
Each peak in the mass spectrum corresponds to a certain fragment with a particular m/e value
The peak with the highest m/e value is the molecular ion (M+) peak which gives information about the molecular mass of the compound
The molecular ion is the entire molecule that has lost one electron when bombarded with a beam of electrons
MOLECULE MOLECULE+• + e–
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 an 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 corresponds to propanal or butanal.
Answer:
The mass spectrum corresponds to propanal as the molecular ion peak is at m/e = 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
Identifying Molecules using Fragmentation
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 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 peaks at 18 (H2O), 28 (CO), and 44 (CO2)
Alkanes
Simple alkanes are fragmented in mass spectroscopy by breaking the C-C bonds
M/e values of some of the common alkane fragments are given in the table below
m/e values of fragments table
Fragment | m/e |
---|---|
CH3+ | 15 |
C2H5+ | 29 |
C3H7+ | 43 |
C4H9+ | 57 |
C5H11+ | 71 |
C6H13+ | 85 |
Mass spectrum showing fragmentation of alkanes
Straight chain alkanes show characteristic peaks
Halogenoalkanes
Halogenoalkanes often have multiple peaks around the molecular ion peak
This is caused by the fact that there are different isotopes of the halogens
Mass spectrum for a bromine containing species
Mass spectra show different halogen isotopes in the molecular ion
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
For example, the mass spectrum of propan-1-ol shows that the compound has fragmented in four different ways:
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
Mass spectrum of propan-1-ol
Mass spectrum showing the fragmentation patterns in propan-1-ol (alcohol)
Worked Example
Which of the following statements about the mass spectrum of CH3Br is correct?
There is one peak for the molecular ion with an m/e value of 44
There is one peak for the molecular ion with an m/e value of 95
The last two peaks have abundances in the ratio 3:1 and occur at m/e values of 94 and 96
The last two peaks are of equal size and occur at m/e values of 94 and 96
Answer:
The correct answer is D as bromomethane (CH3Br) will fragment into 3 peaks
CH381Br → [CH381Br]+ + e− at m/e 96
CH379Br → [CH379Br]+ + e− at m/e 94
CH3Br → [CH3]+ + •Br at m/e 15
The last two peaks (which correspond to the molecular ion peak) therefore are equal in size and occur at m/e values of 94 and 96
Worked Example
Which alcohol is not likely to have a fragment ion at m/e at 43 in its mass spectrum?
(CH3)2CHCH2OH
CH3CH(OH)CH2CH2CH3
CH3CH2CH2CH2OH
CH3CH2CH(OH)CH3
Answer:
The correct answer is D because a line at m/e = 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/e = 43 as it does not have either of these fragments in its structure
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