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Mass spectrometry separates isotopes

(2023-07-02 15:18:02) 下一個

You will know in theory that most elements naturally exist as mixtures of isotopes. If you didn’t believe it, now you will. Chlorine is normally a 3:1 mixture of 35Cland 37Cl (hence the obviously false relative atomic mass of ‘35.5’ for chlorine) while bromine is an almost 1:1 mixture of 79Br and 81Br (hence the ‘average’ mass of 80 for bromine!). Mass spectrometry separates these isotopes so that you get true not average molecular weights. The molecular ion in the E.I. mass spectrum of the bromo-amide below has two peaks at 213 and 215 of roughly equal intensity. This might just represent the loss of molecular hydrogen from a molecular ion 215, but, when we notice that the first fragment
(and base peak) has the same pattern at 171/173, the presence of bromine is a more likely explanation. All the smaller fragments at 155, 92, etc. lack the 1:1 pattern and also therefore lack bromine.

The mass spectrum of chlorobenzene (PhCl, C6H5Cl) is very simple. There are two peaks at 112 (100%) and 114 (33%), a peak at 77 (40%), and very little else. The peaks at 112/114 with their 3:1 ratio are the molecular ions, while the fragment at 77 is the phenyl cation (Ph+ or C6H5+).

The mass spectrum of DDT is very revealing. This very effective insecticide became notorious as it accumulated in the fat of birds of prey (and humans) and was phased out of use. It can be detected easily by mass spectrometry because the five chlorine atoms produce a complex molecular ion at 252/254/256/258/260 with ratios of 243:405:270:90:15:1 (the last is too small to see). The peak at 252 contains nothing but 35Cl, the peak at 254 has four atoms of 35Cl and one atom of 37Cl, while the invisible peak at 260 has five 37Cl atoms. The ratios need some working out, but the first fragment at 235/237/239 in a ratio 9:6:1 is easier. It shows just two chlorine atoms as the CCl3 group has been lost as a radical.

Many elements have minor isotopes at below the 1% level and we can ignore these. One important one we cannot ignore is the 1.1% of 13C present in ordinary carbon. The main isotope is 12C and you may recall that 14C is radioactive and used in carbon dating, but its natural abundance is minute.
The stable isotope 13C is not radioactive, but it is NMR active as we shall soon see. If you look back at the mass spectra illustrated so far in this chapter, you will see a small peak one mass unit higher than each peak in most of the spectra. This is no instrumental aberration: these are genuine peaks containing 13C instead of 12C. The exact height of these peaks is useful as an indication of the number of
carbon atoms in the molecule. If there are n carbon atoms in a molecular ion, then the ratio of M+ to [M + 1]+ is 100: (1.1) Xn.

The electron impact mass spectrum of BHT gives a good example. The molecular ion at 220 has an abundance of 34% and [M + 1]+ at 221 has 5–6% abundance but is difficult to measure as it is so weak. BHT is C15H26O so this should give an [M + 1]+ peak due to 13C of 15X 1.1% of M+, that is, 16.5% of M+ or 34 X 16.5 = 5.6% actual abundance. An easier peak to interpret is the base peak at 205 formed by the loss of one of the six identical methyl groups from the t-butyl side chains. The base peak (100%) 205 is [M—Me]+ and the 13C peak 206 is 15%, which fits well with 14 ´ 1.1% = 15.4%.

Other examples you have seen include the DDT spectrum, where the peaks between the main peaks are 13C peaks: thus 236, 238, and 240 are each 14% of the peak one mass unit less, as this fragment has 13 carbon atoms. If the number of carbons gets very large, so does the 13C peak; eventually it is more likely that the molecule contains one 13C than that it doesn’t. We can ignore the possibility of two 13C atoms as 1.1% of 1.1% is very small (probability of 0.000121).

 summarizes the abundance of the isotopes in these
three elements.

Notice that the ratio for chlorine is not exactly 3:1
nor that for bromine exactly 1:1; nevertheless you should use the
simpler ratios when examining a mass spectrum. Always look at the
heaviest peak first: see whether there is chlorine or bromine in it,

Notice that the ratio for chlorine is not exactly 3:1 nor that for bromine exactly 1:1; nevertheless you should use the simpler ratios when examining a mass spectrum. Always look at the heaviest peak first: see whether there is chlorine or bromine in it, and whether the ratio of M+ to [M + 1]+ is about right. If, for example, you have what seems to be M+ at 120 and the peak at 121 is 20% of the supposed M+ at 120, then this cannot be a 13C peak as it would mean that the molecule would have to contain 18 carbon atoms and you cannot fit 18 carbon atoms into a molecular ion of 120. Maybe 121 is the molecular ion.

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