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Topic: Aromatic stabilization  (Read 7434 times)

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Offline a confused chiral girl

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Aromatic stabilization
« on: May 29, 2007, 02:30:42 AM »
Hi,

I am wondering if the resonance sturctures that I drew for the given structure is correct, or should there be more than a total of 3?  Also, I know that the given structure is aromatic compound, but I don't understand the reasons behind why it will show aromatic stabilization.

Thanks!

Offline Custos

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Re: Aromatic stabilization
« Reply #1 on: May 29, 2007, 03:06:38 AM »
There's one more. If you share one of the oxygen's lone pairs of electrons into the ring you get the pyrylium cation. You can see this offers aromatic stabilisation, right?

Offline a confused chiral girl

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Re: Aromatic stabilization
« Reply #2 on: May 29, 2007, 03:30:51 AM »
that's what I thought as well, but then I reviewed what resonance structures are supposed to be, and there should be same number of double bonds and lone pairs in all structures, so how come there can be a positive charge on the O atom, as it will add 1 more double bond to a total of 3..?

Offline a confused chiral girl

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Re: Aromatic stabilization
« Reply #3 on: May 29, 2007, 09:36:56 PM »
So you mean, there would be a total of 4 resonance structures? or, 5? because O can either donate the pair of electrons to the right and to the left as well...?

Offline english

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Re: Aromatic stabilization
« Reply #4 on: May 30, 2007, 02:24:22 AM »
There's actually far more than that, but many of them are insignificant for this problem.

I've noticed that when you draw your problems it looks as if you're using MS paint.   :P


Try a real drawing program. It's free (and safe).

Offline a confused chiral girl

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Re: Aromatic stabilization
« Reply #5 on: May 30, 2007, 02:35:26 AM »
There's actually far more than that, but many of them are insignificant for this problem.

I've noticed that when you draw your problems it looks as if you're using MS paint.   :P


Try a real drawing program. It's free (and safe).

The question says, draw all resonance structures of the given structure, so how come some of them..you're saying that there're insignificant? You mean there are more than 5?
 
;D you're right! I've been using Paint to draw the structures all this time~ actually quite a hassle as my hand isn't so steady..and the drawings are .. ::)
thanks for the new program~ can't wait to try it  ;)

Offline english

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Re: Aromatic stabilization
« Reply #6 on: May 30, 2007, 04:06:13 AM »
The question says, draw all resonance structures of the given structure, so how come some of them..you're saying that there're insignificant? You mean there are more than 5?

Yours are correct, along with Custos.

The most important contributors are Custos', because not only does this give an aromatic compound, but most resonance structures that are made from concerted movement of electrons give more stable species than separate delocalization and relocalization of electrons.





Alternatively, you could delocalize the pi electrons as lone pairs, then relocalize them again as a pi bond.  These structures give carbanions as well as separated charges, however.  Which method do you thinks looks better?





As for the "other" resonance structures I mentioned, they're insignificant either because they're too unstable to even be considered resonance structures, or they're high energy species (like radicals).  This is an example of a diradical.





But I'm just rambling.  These aren't significant.  Maybe I went a little ahead of myself by saying they were true "resonance" structures.  If a structure is too high in energy, it does not hold any structural merit to the compound. 

Don't be fooled by the general "must be stable to hold merit" rule.  There are lots of examples in which unstable species are used to explain certain mechanisms, and some of them are truly bizarre things to behold.

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