Given an organic formula, how do you assess the number of double bonds in an organic molecule?

Jul 20, 2017

Explanation:

An unsaturated organic compound has formula ${C}_{n} {H}_{2 n + 2}$. Each degree of unsaturation REDUCES the hydrogen count by 2, and corresponds to a RING JUNCTION or an olefinic bond. When oxygen appears in the formula, we assess the degree of unsaturation directly (and of course a degree of unsaturation could correspond to $C = O$). Halogens count for $1 \cdot H$; we subtract $N H$ from the formula if nitrogen is present.

And thus $H C \equiv C H$ has 2 degrees of unsaturation, it is an alkyne.

And ${H}_{3} C - C {H}_{3}$ has 0 degrees of unsaturation.....an alkane.

And ${H}_{2} C = C {H}_{2}$ has 1 degree of unsaturation.....an olefin, or alkene.

And ${H}_{3} C - C {H}_{2} N {H}_{2} \equiv {C}_{2} {H}_{6}$...no degrees of unsaturation.

And ${C}_{3} {H}_{6}$ has 1 degrees of unsaturation.....an alkene, i.e. ${H}_{3} C - C H - C {H}_{2}$ OR $\text{cyclopropane}$.

And ${C}_{3} {H}_{8}$ has 0 degrees of unsaturation.....an alkane.