Isotopic Purity of Samples

Mass spectrometry provides a unique opportunity to estimate the presence of heavy isotopes in a sample molecule. A chemist faces this problem rather often, for example, studying reaction mechanisms. Usually 2H, 13C, 14C, 15N, and lsO isotopes are used to synthesize the labeled compounds. Other isotopes are also used. 

Electron ionization is a perfect method for the analysis of labeled molecules as in this case ion-molecular reactions are suppressed. It is better to use for the calculations the most intense spectral peaks with the highest m/z values. Molecular ion is the best choice. However, if notable [M + H]+ or [M — H]+ peaks are present in the spectrum of the unlabeled compound the correct calculation will be problematic. To eliminate [M + H]+ peaks it is helpful to record a spectrum with the minimum quantity of sample. To consider interference with [M — H]+ ions one should know from what position the hydrogen atom is lost and whether deuterium could be in this position. 

TABLE 5.6. Calculation of the Abundance of Isotopic Peaks in the Mass Spectrum of Deuterated Acetophenone 

If the answer to these questions is not obvious, one can use for the calculations another ion of known composition, for example, [M CH3]+, [M CO]+, or [M — Cl]+. It is relevant to avoid the presence of fragment ions 1 to 2 Da lighter or heavier than the selected ion. 

Let us calculate the intensities of the molecular ion cluster of an acetophenone sample, which contains 10% of unlabeled molecules, 60% of molecules with three deuterium atoms, 20% with two deuterium atoms, and 10% with one deuterium atom. Let us create a table (Table 5.6), taking into account that the molecular ion cluster of pure unlabeled acetophenone is as follows 120 (100%), 121 (8.8%), 122 (0.54%). The calculations with the help of Table 5.6 are quite straightforward, while the final line represents the labeled sample spectrum. 

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