The Image for reference

For example, in the given image, I have starred the step where the relatively unstable resonating structure proceeds forward in the reaction.


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    $\begingroup$ These aren't two states, only two depictions of one. $\endgroup$ – Mithoron Mar 21 '16 at 19:05
  • $\begingroup$ Oh yeah. Let me clarify the question. $\endgroup$ – Reeshabh Ranjan Mar 21 '16 at 19:06
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    $\begingroup$ I would avoid the term 'resonance', since we're not dealing with two states that 'resonate' one between another. In reality, the electron densities are delocalised, making it that you can draw depictions of how this delocalized state would look like. $\endgroup$ – L3ul Mar 21 '16 at 20:01
  • $\begingroup$ Indeed, in each case there is only one real structure, and the resonance variants are but artifacts of our representation. That being said, though, your observation might actually make some sense. A stable structure has no incentive to react, so to say. $\endgroup$ – Ivan Neretin Mar 21 '16 at 20:59

Imagine a case where the resonance form with the negative charge localized on the oxygen went forward and attacked the second carbonyl: That charge needs to go somewhere so the double bond is broken and the electrons go up to the second oxygen (the one that started as the second carbonyl). Those electrons would look to come back and reform the double bond, and in doing so would force one of the groups to leave. And the bond that leaves is the one you just made. This reaction happens with both atoms acting as a nucleophile except only one (the carbanion) goes forward and doesn't immediately leave again.

  • $\begingroup$ So you mean that the negative charge should be on Carbon, so that the electrons can be 'easily passed on' to next step in the reaction, otherwise Oxygen is more electronegative and won't lose 'pass on' electrons that easily? This seems acceptable, just confirm if you meant this. $\endgroup$ – Reeshabh Ranjan Mar 22 '16 at 6:57

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