General Note on Interpretation of Mass Spectral Data
Below is a rather simplified way of mass spectral data interpretation approach:
1. Identify the molecular ion peak
The molecular ion peak may not appear in the spectrum, BUT if it does, it is the highest mass peak in the spectrum. Watch out the isotope peaks (see below).
The molecular weight of a compound (a rounded number) is even number if it contains only C, H, O, S, Si.
The molecular weight of a compound (a rounded number) is odd number if it contains an odd number of N (1,3,5,..).
2. Making sense of the isotope peaks
Carbon 13 is an isotope of carbon 12.
Carbon 12 has an abundances of - 12C=100%, 13C=1.1%. i.e. for every 100 (12)C atoms, there are 1.1 (13)C atoms.
If a compound contains 20 carbons, then each atom has a 1.1% abundance of (13)C.
If the molecular ion peak of this compound is 100%, then its isotope peak (1 mass unit higher) would be 20 x1.1%=22%.
If the molecular ion peak is not 100%, then we can calculate the relative abundance of the isotope peak from the ion peaks. For example, if the molecular ion peak were 65% and the isotope peak 15.5%: (15.5/65)x100 = 23.8% as the relative abundance of the isotope peak to the ion peak. Let's now divide the relative abundance by the isotope abundance and we will get - 23.8/1.1 = 22 carbons.
3. The "double bond equivalent" rule - total number of rings and/or double bonds in the molecule
This is one of by far the most interesting and useful rule to follow in structural interpretation exercises. Fortunately, mastering this rule does not require you to know much about mass spectrometry. Carbon is tetravalent and hence forms four bonds with other atoms in order to fill its valence shell. This can be fulfilled in one of the following ways:
In the above examples the degree of unsaturation and/or ring systems can easily be calculated since one double bond or a ring is equivalent to a loss of two protons from the saturated system. Similarly, a triple bond in a molecule would have a 2 double bond equivalent.
Thus 'double bond equivalent' for the molecular formula of CxHyOn is: [(2x + 2)-y]/2
4. Having all the above accounted for, try to postulate the molecular structure consistent with the abundance and m/z of fragments. Remember that we use data coming from other spectroscopic methods, such as NMR, too.
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