How do you handle formulas with all five atoms? Consider the formula C8H10BrClN2O3. First we see that there are 2 halogen atoms and 2 nitrogens. The number of hydrogens is even. By ignoring the oxygen atoms we have C8H10BrClN2. Substituting 2 hydrogens for the halogens the formula becomes C8H12N2. Finally, The addition of 2 carbons and 2 hydrogens in place of the 2 nitrogens gives C10H14. The most saturated C10 hydrocarbon is C10H22. Thus the degree of unsaturation of C8H10BrClN2O3 is (22-14)/2 = 4. The following compounds all fit the data. Draw some more structures with this formula.
Can the compound C4H11Cl2NO exist? Based upon the presence of 2 chlorines and 1 nitrogen, the odd number of hydrogen atoms is also odd. So far; so good. The formula reduces as follows: C4H11Cl2NO ---> C4H11Cl2N ---> C4H13N ---> C5H14. This structure is more saturated than C5H12. The formula is incorrect.
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