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:

• Making single bonds with four different atoms, e.g. CH4, CCl4, CH3Cl, CH3OH, etc. In this saturated hydrocarbon system, compounds like methane, ethane, propane, butane, etc. are represented by the general formula CnH(2n+2).
• Making a double bond to one atom and a single bond to each of two other atoms, e.g. CCl2=CCl2, CH2=CH2, etc. This is an example of unsaturated system and compounds like ethene, are represented by the general formula, CnH2n.
• Making a triple bond to one atom and a single bond to another. e.g. acetylene.
• A not much useful case in structural elucidation is a case where carbon  form a double bond to each of two different atoms, e.g. CO2.

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

• For the molecular formula containing nitrogen atom:  CxHyNzOn
• rings + double bonds = x - (1/2)y + (1/2)z + 1

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.

Back to Spectroscopy Page

Back to Hompage