Chlorine atoms, mass spectrum

The peak at m/z 77 m the mass spectrum of chlorobenzene m Figure 13 41 is attributed to this fragmentation Because there is no peak of significant intensity two atomic mass units higher we know that the cation responsible for the peak at m/z 77 cannot contain chlorine... 

In a process similar to that described in the previous item, the stored data can be used to identify not just a series of compounds but specific ones. For example, any compound containing a chlorine atom is obvious from its mass spectrum, since natural chlorine occurs as two isotopes, Cl and Cl, in a ratio of. 3 1. Thus its mass spectrum will have two molecular ions separated by two mass units (35 -i- 2 = 37) in an abundance ratio of 3 1. It becomes a trivial exercise for the computer to print out only those scans in which two ions are found separated by two mass units in the abundance ratio of 3 1 (Figure 36.10). This selection of only certain ion masses is called selected ion recording (SIR) or, sometimes, selected ion monitoring (SIM, an unfortunate... 

Naturally occurring isotopes of any element are present in unequal amounts. For example, chlorine exists in two isotopic forms, one with 17 protons and 18 neutrons ( Cl) and the other with 17 protons and 20 neutrons ( Cl). The isotopes are not radioactive, and they occur, respectively, in a ratio of nearly 3 1. In a mass spectrum, any compound containing one chlorine atom will have two different molecular masses (m/z values). For example, methyl chloride (CH3CI) has masses of 15 (for the CH3) plus 35 (total = 50) for one isotope of chlorine and 15 plus 37 (total = 52) for the other isotope. Since the isotopes occur in the ratio of 3 1, molecular ions of methyl chloride will show two molecular-mass peaks at m/z values of 50 and 52, with the heights of the peaks in the ratio of 3 1 (Figure 46.4). 

A diagrammatic illustration of the effect of an isotope pattern on a mass spectrum. The two naturally occurring isotopes of chlorine combine with a methyl group to give methyl chloride. Statistically, because their abundance ratio is 3 1, three Cl isotope atoms combine for each Cl atom. Thus, the ratio of the molecular ion peaks at m/z 50, 52 found for methyl chloride in its mass spectrum will also be in the ratio of 3 1. If nothing had been known about the structure of this compound, the appearance in its mass spectrum of two peaks at m/z 50, 52 (two mass units apart) in a ratio of 3 1 would immediately identify the compound as containing chlorine. 

The mass spectrum of the unknown compound showed a molecular ion at m/z 246 with an isotope pattern indicating that one chlorine atom and possibly a sulfur atom are present. The fragment ion at m/z 218 also showed the presence of chlorine and sulfur. The accurate mass measurement showed the molecular formula to be C]3FI7OSCl R + DB = 10. 

The molecular ion is apparent in the mass spectrum of DDT (Figure 25.2) at m/z 352 with the classic isotope pattern for five chlorine atoms (see Appendix 11). The major fragment ion is the loss of CCI3 at m/z 235. 

The mass spectrum of CTC, shown in Figure 14 (42), is characterized by a reasonably intense molecular ion at m/e 478 with the concomitant isotope peak at P+2 representing one chlorine atom in the ring system. Although it has been suggested that this chlorine atom be employed as a tracer via the isotope ratio for detection of species containing... 

Even if the analyte is chemically perfectly pure it represents a mixture of different isotopic compositions, provided it is not composed of monoisotopic elements only. Therefore, a mass spectrum is normally composed of superimpositions of the mass spectra of all isotopic species involved. [11] The isotopic distribution or isotopic pattern of molecules containing one chlorine or bromine atom is listed in Table 3.1. But what about molecules containing two or more di-isotopic or even polyisotopic elements While it may seem, at the first glance, to complicate the interpretation of mass spectra, isotopic patterns are in fact an ideal source of analytical information. 

If sulfur or silicon, is present, the M + 2 will be more intense. In the case of a single sulfur atom, 34S contributes approximately 4.40% to the M + 2 peak for a single silicon in the molecule, 30Si contributes about 3.35% to the M + 2 peak (see Section 2.10.15). The effect of several bromine and chlorine atoms is described in Section 2.10.16. Note the appearance of additional isotope peaks in the case of multiple bromine and chlorine atoms. Obviously the mass spectrum should be routinely scanned for the relative intensities of the M + 2, M + 4, and higher isotope peaks, and the relative intensities should be carefully measured. Note that F and I are monoisotopic. 

This example illustrates how m/e values of ions that differ only in isotopic composition can be used to determine elemental compositions. The important isotopes for this purpose in addition to those of chlorine are the stable isotopes of natural abundance, 13C (1.1%), 15N (0.37%), 170 (0.04%), lsO (0.20%). As a further example, suppose that we have isolated a hydrocarbon and have determined from its mass spectrum that M + = 86 mass units. In the absence of any combination reactions there will be an (M + 1)+ ion corresponding to the same molecular ion but with one 13C in place of 12C. The intensity ratio (M + 1 )+/M+ will depend on the number of carbon atoms present, because the more carbons there are the greater the probability will be that one of them is 13C. The greater the probability, the larger the (M + 1 )+/M+ ratio. For n carbons, we expect... 

Members of the last group, and especially chlorine and bromine, are most easily recognised from the characteristic patterns of the peaks, spaced at intervals of two mass units, which they produce in the spectrum. Typical patterns for combinations of bromine and chlorine atoms are shown in Figs 3.77 and 3.78. It may be difficult to estimate the number of oxygen atoms due to the low natural abundance (0.20%) of lsO. 

Due to the distinctive mass spectral patterns caused by the presence of chlorine and bromine in a molecule, interpretation of a mass spectrum can be much easier if the results of the relative isotopic concentrations are known. The following table provides peak intensities (relative to the molecular ion (M+) at an intensity normalized to 100%) for various combinations of chlorine and bromine atoms, assuming the absence of all other elements except carbon and hydrogen.1 The mass abundance calculations were based on the most recent atomic mass data.1... 

For an organic compound the first step is usually to find the molecular formula, probably from the mass spectrum, and to calculate the number of double bond equivalents (DBEs). An acyclic saturated hydrocarbon has the formula where M = 2N+2. Each double bond or ring in the molecule reduces the value of M by two. So if M = 2N the molecule has one DBE we cannot tell from the formula whether it is in the form of a ring or unsaturation. A benzene ring corresponds to 4 DBEs three double bonds and a ring. The presence of oxygen or other divalent elements does not affect the value of M. Each monovalent atom such as chlorine can be treated as a proton for the purpose of calculation, while one proton has to be subtracted for each trivalent atom such as nitrogen. 

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