...............A very useful idea that is used to rationalize the formulae of organic compounds is the #"degree of unsaturation"#. An alkane is fully #"saturated"#, and it contains the MAXIMUM ALLOWABLE number of #C-H# bonds.
Alkanes are #"FULLY saturated"# and have a general formula of #C_nH_(2n+2)#. Try this out for #"methane,"# #"ethane,"# .........#"pentane, etc."#
Each double bond, each olefinic bond or carbonyl group, OR a 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. 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"#.
Given the former, a formula of #C_4H_10O#, the which is SATURATED, specifies FOUR alcoholic isomers, one of which can generate stereoisomers. And there are also THREE ethereal isomers...and some punter has kindly given the structures on the web..
