October 28, 2020, 05:27:46 AM
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Topic: Why does the delocalized lone pair on the bottom-left structure move right?  (Read 157 times)

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Offline Roddy3

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As the title states, the bottom-left resonance structure has the lone pair move to the right carbon in the structure to its immediate right--why is this the case instead of the lone pair moving back towards the OH to form a structure similar to the top-left? Does it have something to do with the C=C bond being favored over the C=O bond? Hopefully I'm not missing something obvious, but that's the best explanation I can think of right now.

I included a picture of the solution that I marked up to help visualize what I'm talking about, just in case I'm using improper terminology. Or if I'm not making any sense.

Offline spirochete

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Re: Why does the delocalized lone pair on the bottom-left structure move right?
« Reply #1 on: September 14, 2020, 06:58:50 PM »
The lone pair actually goes both directions. Technically, that list of resonance structures is missing one structure: the one where the all three pi bonds in the benzene ring have simply shuffled over one carbon. This resonance structure is less "interesting" because it doesn't move a charge. Sometimes, this resonance structure is omitted because the person who wrote the answer key considers it to be relatively obvious and unimportant, because this structure doesn't move the location of a formal charge. But technically, it should be included for a full picture of all the resonance structures.

If anything, the lone pair "prefers" to be on oxygen due to the relative electronegativity of oxygen. But the goal for this type of problem is to draw all of the minor resonance contributors.

Offline Roddy3

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Re: Why does the delocalized lone pair on the bottom-left structure move right?
« Reply #2 on: September 14, 2020, 08:09:49 PM »
Thank you, I truly appreciate the clarification.

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