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

1 Answer
Jul 20, 2017

Answer:

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.