Rings plus double bonds

The number of rings plus double bonds in a molecule of formula CjcHyN O is 

When atoms other than CHNO are present, the heteroatoms are matched according to their lowest absolute valence. Si takes 4 bonds, like C P is equivalent to N, while the halides, with a lowest absolute valence of 1, act like H. Hence the general equation is  

Because of the valences of the elements involved, the total number of rings and double bonds in a molecule of the formula C,(HyN O will be equal to x — 53/ + jz + I (Pellegrin 1983). For ions, the calculated value may end in j (indicating an even-electron ion, Section 3.2), and this fraction should be subtracted to obtain the true value. How to use this should be more obvious from inspection of the examples in Box 2.1. The value 4 found for pyridine represents the ring and three double bonds of this molecule. The 5.5 calculated for the benzoyl ion represents the ring, the three double bonds of benzene, and the double bond of 

For an even-electron ion (see Section 3.2), the true value will be followed by Examples  

C5H5N rings plus double bonds = 5 — 2.5 -I- 0.5 -I- 1 =4 For example, pyridine (odd-electron) 

Extensive experience has shown that practice is necessary to develop one s ability to calculate elemental compositions from isotopic abundances. (Perhaps this is because the chemist must become accustomed to the fact that the mass spectrum shows his/her carefully purified compound to be a mixture of isotopically different molecules.) It is important that you understand this procedure, since it is the key primary step in interpreting an unknown spectrum. To avoid confusion, follow the stepwise procedure below while you are learning. Table 2.3 illustrates this procedure with data from the region of the spectrum of rert-butylthiophene, C8H,2S. Note that this is not done for the mjz 140 abundance, because the 10% accuracy in abundance is relative to the abundance of that peak. 

Using this procedure, can you assign compositions and values for rings plus double bonds to the fragmentary spectra shown in Unknowns 2.9 through 2.14 The peaks of Unknowns 2.9, 2.10, and 2.12-2.14 contain the molecular ion 

Restriction to formulas of the general type CJIhNnOo reduces the expression to the commonly cited form  

other monovalent elements than hydrogen (F, Cl, Br, I) are counted as hydrogens , other trivalent elements such as phosphorus are counted as nitrogen and tetravalent elements (Si, Ge) are handled the same way as carbons. 

The r + d algorithm produces integers for odd-electron ions and molecules, but non-integers for even-electron ions that have to be rounded to the next lower integer, thereby allowing to distinguish even- from odd-electron species. 

Care has to be taken when elements of changing valence are encountered, e.g., S in sulfoxides and sulfones or P in phosphates. In such a case, Eqs. 6.9 or 6.10 have to be used, whereas Eq. 6.11 yields erroneous results. 

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List Probable Molecular Formula with Calculated Rings Plus Double Bonds... 

The following are examples of calculating rings plus double bonds. 

The number of electrons in an ion may be checked using a formula for the calculation of unsaturation degree (rings-plus-double-bonds, Equation 5.7) ... 

Fig. 3.13. Kendrick mass defect versus nominal Kendrick mass for odd-mass ions ([M-H] ions). The compound classes (O, O2, O3S and O4S) and the different numbers of rings plus double bonds (Chap. 6.4.4) are separated vertically. Horizontally, the points are spaced by CH2 groups along a homologous series. [29] By courtesy of A. G. Marshall, NHFL, Tallahassee.

D. (a) Find the number of rings plus double bonds in a molecule with the composition C 4H,2 and draw one plausible structure, (b) For an ion or radical, the rings + double bonds formula gives noninteger answers because the formula is based on valences in neutral molecules with all electrons paired. How many rings plus double bonds are predicted for C4Hl0NO+ Draw one structure for C4H 0NO+. 

Find the number of rings plus double bonds in molecules with the following compositions and draw one plausible structure for each (a) CMH18N203 (b) C 2Hl5BrNPOS (cl a fragment in a mass spectrum with the composition CjH. ... 

Hint This compound has 6 rings plus double bonds"... 

The amount of energy acquired during ionization and the structure of the analyte determine the complexity of the mass spectra obtained. Other features, including isotopic composition, the nitrogen rule, rings plus double bond considerations (see later), and accurate mass values, provide additional information. 

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