2-1 100 CH D 75 a 50 D 25 mlz Rel. Int. 17 16 100.0 15 85.0 14 15.9 7.8 2.6 12 5

Question

2-1 100 CH D 75 a 50 D 25 mlz Rel. Int. 17 16 100.0 15 85.0 14 15.9 7.8 2.6 12 5 10 20 The spectrum of methane (above) is unremarkable except for the small peak at m/z 17. This peak is not an artifact, nor is it due to an impurity. Where does it come from? A clue is found in the spectrum in Problem 2-1 below. What structural feature do these two compounds have in common? What features do their mass spectra have in common? (Both figures reprinted from R.M.Smith, Understanding Mass Spectra - A Basic Approach. Copyright John Wiley and Sons, New York.) Problem 2-1 mlz Rel. Int. 46 0.4 0 75 4 100.0 9.8 1.9 9.6 16 25 12 16 12 87 28 20 30 40 50 m/z

Explanation / Answer

The first spectrum is of methane where we observed 1.2% of M+1 peak at m/z 17

The second spectrum is of CO2 which also has one carbon with M+1 peak at 1.2

They both have in common only one carbon atom which has a natural bundance of heavier isotope C13 with 1.08% .

Along with that methane has Hydrogen with a heavier isotope of 2H with 0.016% abundance and the second one has two oxygen atoms with heavy isotope O17 with 0.04% abundance .

Thus for methane the relative intensity of M+1 peak is calculated as

relaative intensity of [M+1] = [C x1.1]+ [Hx0.015] + [Ox0.04]

For methane intensity of [M+1] = (1x1.1) + (4x0.015)

= 1.16 clsoer to 2

similarly for CO2

[M+1] = (1x1.1) + (2x0.04)

= 1.18 closer to 1.2

thus the presence of natural isotopes of heaavier mass gives rise to M+1 and M+2 peaks and the ratio of M to M+1 always remains constant and can be calulated using the simplified formula roughly.

Thus the presence of M+1 and higher peaks in very small proportions is due to the presence of heavier isotopes of C,H and O naturally.

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