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.
Answer
The very idea of oxidation states is not all that useful within the realm of organic chemistry, so you may just as well leave it at the door.
That being said, it still can be used, and your question concerning the oxidation state of carbon in propene has a meaningful answer, which you gave yourself: the oxidation state of the methyl carbon is -3, that of the middle one is -1, and that of the methylene one is -2. (There is no such thing as 2-propene, but that's another story.) Asking for anything above that is pretty much like pointing your finger at three random guys in the street and asking what is their name. What are they supposed to answer, if they are not relatives and don't have a name common to all three?
Well, numbers are not quite like names, so you may calculate an average... but I guess you already see why the organic chemists tend not to think in terms of oxidation states.
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