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Two different species obeying the octet rule

  1. Aug 26, 2004 #1
    I began taking Organic Chemistry one week ago, so my question should be very easy for most of you.
    I am having a lot of problems determining which specie, of a group of given species, is more stable. I can determine it for resonance conformers; for example I know that one conformer with a separation of charges is less stable that one with no separation of charges. I also know that the lone pairs must be in the most electronegative atom...etc. However I am having problems when I am asked to determine which of two or three species (not resonance conformers) is more stable.
    I read my book and I didn't find any kind of rule for comparing the stability of two different species obeying the octet rule and with no separation of charges. I would reeeaaaallly appreciate if someone could explain me what I must consider.
    One example of the problems that I cannot do is the following:

    CH3C(O-)HCH=CH2 --> (the O has formal charge -1, I didn't know how
    to write it here)



    The most stable is the second one, but why?
  2. jcsd
  3. Aug 26, 2004 #2
    CH3-CO(-)=CHCH3 is more stable because of the resonance structure CH3-CO-C(-)HCH3 in which the lone pair of the oxygen forms a pi bond with the carbonyl carbon while the adjacent pi bond is thrown onto the carbon.

    With the other structure there is no resonance structure because their can't be any pi bond formed from the oxygen's lone pair. If this were to happen the carbonyl carbon would have 5 bonds, which is illegal and can be punishable by death.
  4. Aug 26, 2004 #3


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    Let's try to answer this question


    I think you should look in the term "most stable carbocation" in your books. Tertiary carbocations do not possess hydrogen at all, whereas secondary carbocations have one, and primary ones have two on them.

    With these basics, let me answer your question. Please view the attachments first. In the first attachment, you will find the first resonance hybrid. The compound on the left is not an enol; so is highly reactive, especially through its vinylic terminal. The one on the right has a secondary carbocation, which is weaker than tertiary carbocation. So if you can write an isomer consisting a tertiary carbocation, then you may conclude that this one is more stable. You may even suggest that the first compound will try to convert itself to the second with intramolecular shifts.

    When you look at the second attachment, there seems to be an enol tautomer of a ketone; the ketone is written in its "neutral" form. If you want to write its anianic form, you should take an alfa hydrogen from the ketonic carbon.

    To sum up, it can be explained by the relevant stabilization effects of tertiary carbocations versus secondary ones.

  5. Aug 26, 2004 #4


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    Other related things you may find in your textbook:

    "enolates" (or just "enols")
    "allylic anion"
    "allylic cation"
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