91Ó°ÊÓ

The rule of thirteen is employed to determine the molecular formula of a compound based on its molecular ion peak. According to the rule, the base molecular ion m/z value divided by the molar mass of CH, i.e. 13, gives the base formula.

The relation between the base formula and n and r is;

$\frac{\text{M}}{\text{13}}\text{ = n + }\frac{\text{r}}{\text{13}}$

where n and r are the subscripts of carbon and hydrogen. Accordingly, the molecular formula is expressed as ${\text{C}}_{\text{n}}{\text{H}}_{\text{n+r}}$ .

The degree of unsaturation of a compound arises from the number of pi-bonds or ring systems present in the compound.For a compound whose molecular formula is derived from the rule of thirteen, the DBE (double bond equivalent) is related to the subscripts of carbon and hydrogen, as shown below:

$\text{¶Ù.µþ.·¡=​â¶Ä‹â¶Ä‰}\frac{\left(\text{²Ô -°ù​â¶Ä‹â¶Ä‰+ 2}\right)}{\text{2}}$

The molecular ion peak is found at m/z: 204. The molecular ion peak is the molecular mass of the compound.

Thus, M = 204

Use the rule of thirteen to determine the base formula, as shown below:

$\begin{array}{c}\frac{\text{M}}{\text{13}}\text{ = n + }\frac{\text{r}}{\text{13}}\\ \frac{204}{\text{13}}\text{ = }15\text{ + }\frac{9}{\text{13}}\end{array}$

Thus, putting n = 15 and r = 9, the molecular formula is obtained as:

$\begin{array}{l}{\text{C}}_{\text{15}}{\text{H}}_{\left(\text{9+15}\right)}\text{ }={\text{ C}}_{\text{15}}{\text{H}}_{\text{24}}\\ \end{array}$

Substituting n= 15 and r = 9 in the below formula,

$\text{¶Ù.µþ.·¡=​â¶Ä‹â¶Ä‰}\frac{\left(\text{²Ô -°ù​â¶Ä‹â¶Ä‰+ 2}\right)}{\text{2}}$

The D.B.E is obtained as:

$\begin{array}{c}\text{¶Ù.µþ.·¡=​â¶Ä‹â¶Ä‰}\frac{\left(15\text{ -9​â¶Ä‹â¶Ä‰+ 2}\right)}{2}\\ =\text{ }4\end{array}$

Thus, the molecular formula is ${\text{C}}_{\text{15}}{\text{H}}_{\text{24}}$.