...............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 degree of unsaturation, an olefinic bond OR a ring junction, corresponds to a FORMULA that is TWO 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# 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"#. For #"acetylene"#, #HC-=CH#, and #"acrolein"#, #H_2C=CH-C(=O)H# we gots #2^@# degrees of unsaturation: the saturated formula would be #C_2H_6#, and #C_3H_8O# respectively.
So for hexane we gots #C_6H_14# (it does not matter which saturated isomer); but for a cycloalkane with formula #C_6H_12#, we gots ONE ring: cyclohexane, cyclopentane, cyclobutane, or cyclopropane, and the extra carbons are substituents on the ring. If you don't know that it is a ring, then you do know that IT IS A RING or AN OLEFIN.
And thus a formula of #C_6H_12# corresponds to a ring OR an olefin bond, i.e. #1^@# of unsaturation. If we put oxygen in the formula, #C_6H_12O#, we could have a carbonyl or formyl functionality, i.e. #C=O#, that accounts for the given #1^@# of unsaturation.
Are you following me? If I have been unclear, or have not addressed your query, try again.
