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

1 Answer
Dec 10, 2017

Answer:

Well, we assess the #"degree of unsaturation.."#

Explanation:

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

AND so #"FULLY saturated"# alkanes have a general formula of #C_nH_(2n+2)#. Try this out for #"methane,"# #"ethane,"# .........#"pentane, etc."# The presence of oxygen does not reflect the degree of unsaturation...EXCEPT in the case of a carbonyl function; #"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, #"ethane"# has the saturated formula of #H_3C-CH_3#, but #"ethylene"#, #H_2C=CH_2#, and #"acetaldehyde"#, #H_3C-C(=O)H# has #1^@# of unsaturation. #"Acetylene"#, #HC-=CH# has #2^@# of unsaturation. Halogen atoms count for one hydrogen; for nitrogen atoms, substract #NH# from the formula before assessing unsaturation; i.e. for #"ethylamine,"# #H_2NCH_2CH_3# #rarr C_2H_6#, i.e. #"no degrees of unsaturation"#. For pyridine, #C_5H_5N#, we assess #C_5H_4#...i.e. #"four degrees of unsaturation"#, THREE double bonds, and the ring...

And so here we got the third alkane, i.e. #H_3C-CH_2CH_3#, which has only the one isomeric form, or #"propylene"#...with the one degree of saturation, i.e. #H_2C=CHCH_3#, and again this #C_3H_6# formula will support this isomer, and also #"cyclopropane"#. The ring junction, like an olefin bond, REDUCES the hydrogen count by 2 with respect to the saturated formula...