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.
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3 Answers
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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.
answered Mar 14, 2017 at 19:40
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$\begingroup$ I understood my mistake about 2-propene, sorry for that. As concerns your answer, I conclude that oxidation states of organic molecules can help us in conversion of molecules such as alcohol to aldehydes etc but in general, this concept is not very useful. $\endgroup$ Mar 15, 2017 at 6:37
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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.
answered Mar 14, 2017 at 17:03
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As others have stated, assigning absolute charges to carbon atoms in organic chemistry is not particularly useful or productive. What is important is the relative oxidation level of two related structures and whether or not an oxidizing or reducing agent may be required to interconvert the two structures.
The simplest method of comparison is to consider the polarity of bonds undergoing reactions. Since hydrogen is more electropositive than carbon, the two C-H bonds in propane (1a) are as shown above. To form propene (2b) from propane, two protons and a pair of electrons (equivalent to a proton and hydride = hydrogen molecule) must be removed from propane. In so doing one is left with 2a which is nothing more than a canonical resonance structure of propene. The carbon framework is oxidized while the protons are reduced to hydrogen. This process is a "thought experiment" independent of the reagents employed. In the same way, the oxidation of propene (2c) to propyne (3b) may be analyzed in a similar fashion. [In structures 2a and 3a the positions of the positive and negative charges have been placed arbitrarily where the more stable carbocation or carbanion are expected. The methodology works if the charges are switched.] For a more detailed discussion, click here.
answered Jun 13, 2018 at 19:52