A Elements Hydrogen, Fluorine, Phosphorus, and Iodine

4 A elements hydrogen, fluorine, phosphorus, and iodine After the number of each A + 2 and A + 1 element has been fixed (or esti-  

Illustrating this with the previous spectra, the base peak at m/z 36 of Unknown 2.1 must contain one chlorine, based on the abundance of its A + 2. Subtracting the mass of chlorine, 35, from the mass of this peak, 36, leaves a difference of one, which of course must arise from the presence of one hydrogen. The m/z 94 peak for Unknown 2.2 contains one bromine and one carbon  

94 — 79 — 12 = 3, so the elemental composition of the base peak is CHjBr. For Unknown 2.3, the m/z 64 peak was assigned the elements SOj 32 + (2 x 16) = 64, so there are no A elements. For Unknown 2.4, the m/z 43 and 58 peaks contained Cj and C4, respectively thus 43 — 36 and 58 — 48 indicate the number of hydrogen atoms in each. Note that only hydrogen atoms can be used to account for this difference until the value is as large as 19, the mass of fluorine. Additionally, the number of hydrogens can often be inferred from the proton NMR spectrum, and the minimum number of carbon atoms from the C-13 NMR spectrum. 

Try this partial spectrum representing a single fragment ion  

No A + 2 or A + 1 elements are possible. All the A elements but phosphorus are monovalent so this grouping must be due to a single monovalent (A) element (which ), or a pair, or a combination of these with phosphorus (thus H3P4 is a conceivable assignment). 

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