Oxidation State Of Propene I was just trying to understand the oxidation states of organic compounds. A more electronegative atom than carbon increases it's oxidation state by 1 and less electronegaitve atom decreases it's oxidation state by 1. Consider the molecule  propene $\ce{CH3-CH=CH2}$  the oxidation state of left carbon is -3, that of the middle one is -1 and that of the right one is -2? So, what is the oxidation state of carbon in   propene? Same complexity arises in 2-propyne or any such molecule.
Reconsider if a more electronegative element bound to carbon and bond to carbon will -- formally -- take the electrons, or donate the electrons in shared the bond with carbon, and how this is reflected in the oxidation state.  In each instance, watch out the signs, please.

After getting familiar with the basics, look up references focussing on organic chemistry and redox chemistry.  Albeit organic chemists tend to prefer moving electrons by hooks and arrows, there are references like this (hint: it includes an answer you may compare with the one you should find yourself, too) or even video classes like this.

Eventually, "left" and "right" will advantageously become obsolete when you gain enough credit to upload structural schemes on SE, too.