Question

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8-21-2019 Spectroscopy Fall 2019 (4) Unknown 4 155 23 Relative intensity in 1232 miz L . (4-a) This is a poly-halogenated compound. How many of which halogen seem to be present in the molecular lon? (4-b) Based on (4-a), and assuming that the rest of the molecule consists of C and H, determine the molecular formula. (4-c) Account for the fragment cluster including m/2 = 155. 14.d) (A little curve ball) Account for the fragment cluster including m/t 75

Answer 1

4-a) If we consider the main peaks of the three "heavier" signals, we have signals around m/z: 236, 157 and 75, which correspond to differences of around 80 mass units, which are coherent with two Br atoms in the molecule.

4-b) The loss of the two Br atoms yields a fragment of mass = 76. If it is only made up of carbon and hydrogen, the fragment has a composition: C₆H₄, which means the whole molecular formula is C₆H₄Br₂.

4-c) This cluster is the result of the loss of a bromine atom from the molecular ion.

4-d) This fragment cluster results from the loss of both atoms of Br, since it represents a loss of around 160 units of mass respective to the molecular ion.