Question

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

Answer 1

We assess the [degree of unsaturation.](https://socratic.org/questions/how-are-cycloalkanes-different-from-aromatic-compounds)

Explanation:

An unsaturated organic compound has formula `C_nH_(2n+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*H`; we subtract `NH` from the formula if nitrogen is present.

And thus `HC-=CH` has 2 degrees of unsaturation, it is an alkyne.

And `H_3C-CH_3` has 0 degrees of unsaturation.....an alkane.

And `H_2C=CH_2` has 1 degree of unsaturation.....an olefin, or alkene.

And `H_3C-CH_2NH_2-=C_2H_6`...no degrees of unsaturation.

And `C_3H_6` has 1 degrees of unsaturation.....an alkene, i.e. `H_3C-CH-CH_2` OR `"cyclopropane"`.

And `C_3H_8` has 0 degrees of unsaturation.....an alkane.