When an alkane is bombarded by high energy electrons it will lose an electron to form a radical cation. This radical cation has the same mass as the parent compound and produces the molecular ion (M+) peak. The type of radical formed follows the stability of radicals:
The alkane molecular ion can further fragment to form a homologous series of neutral alkyl radicals usually beginning with the methyl radical. The methyl radical has a mass of 15 and the next largest peak in the mass spectrum usually corresponds to the loss of methyl radical (M-15). Ethyl radical can also be lost (M-29) and so forth. Figure 7 shows the mechanism of fragmentation for butane; the corresponding mass spectrum for butane is given in Figure 8. Notice the appearance of M-15 and M-29 peaks in the Figure. Table 1 shows the typical fragments lost by acyclic alkanes and their respective masses.
Figure 7. Mechanism of fragmentation for butane.
Figure 8. Mass spectrum for butane.
Table 1. Typical fragments lost from straight chain alkanes.
| Molecular Ion - | Fragment Lost |
| 1 | H· |
| 2 | 2 H· |
| 15 | CH3· |
| 29 | C2H5· |
| 43 | C3H7· |
| 57 | C4H9· |
| 71 | C5H11· |
Peaks in the mass spectra of straight chain alkanes will usually appear in groups of 14 mass unit intervals (corresponding to one CH2 group). The most intense fragmentation peak is usually the 3 carbon fragment, with the intensities of the peaks decreasing with increasing mass. Often, the M-15 peak (loss of methyl radical) will be absent. Fragments to look for in these spectra correspond to CnH2n+1+, CnH2n+, and CnH2n-1+. The mass spectrum for n-hexadecane (Figure 9) illustrates these points.
Figure 9. Mass spectrum for n-hexadecane.
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