# Given an organic formula, how do we know how many double bonds, triple bonds, and ring junctions, are present in the molecule?

Dec 10, 2017

Well, we assess the $\text{degree of unsaturation..}$

#### Explanation:

...............A very useful metric that is used to rationalize the formulae of organic compounds is their $\text{degree of unsaturation}$. An alkane is said to be fully $\text{saturated}$, and it contains the MAXIMUM ALLOWABLE number of $C - H$ bonds.

AND so $\text{FULLY saturated}$ alkanes have a general formula of ${C}_{n} {H}_{2 n + 2}$. Try this out for $\text{methane,}$ $\text{ethane,}$ .........$\text{pentane, etc.}$ The presence of oxygen does not reflect the degree of unsaturation...EXCEPT in the case of a carbonyl function; $\text{ethyl alcohol}$ is FULLY saturated by this criterion.

Each double bond, each olefinic bond or carbonyl group, OR ring junction, corresponds to 1""^@ of unsaturation; i.e. 2 hydrogens LESS than the saturated formula. So according to the scheme, $\text{ethane}$ has the saturated formula of ${H}_{3} C - C {H}_{3}$, but $\text{ethylene}$, ${H}_{2} C = C {H}_{2}$, and $\text{acetaldehyde}$, ${H}_{3} C - C \left(= O\right) H$ has ${1}^{\circ}$ of unsaturation. $\text{Acetylene}$, $H C \equiv C H$ has ${2}^{\circ}$ of unsaturation. Halogen atoms count for one hydrogen; for nitrogen atoms, substract $N H$ from the formula before assessing unsaturation; i.e. for $\text{ethylamine,}$ ${H}_{2} N C {H}_{2} C {H}_{3}$ $\rightarrow {C}_{2} {H}_{6}$, i.e. $\text{no degrees of unsaturation}$. For pyridine, ${C}_{5} {H}_{5} N$, we assess ${C}_{5} {H}_{4}$...i.e. $\text{four degrees of unsaturation}$, THREE double bonds, and the ring...

And so here we got the third alkane, i.e. ${H}_{3} C - C {H}_{2} C {H}_{3}$, which has only the one isomeric form, or $\text{propylene}$...with the one degree of saturation, i.e. ${H}_{2} C = C H C {H}_{3}$, and again this ${C}_{3} {H}_{6}$ formula will support this isomer, and also $\text{cyclopropane}$. The ring junction, like an olefin bond, REDUCES the hydrogen count by 2 with respect to the saturated formula...