Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: cswarth on September 01, 2010, 03:48:45 AM
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I am having a tough time understanding how to draw resonance structures beyond the very basic examples.
For instance, when asked to draw all of the plausible resonance structures for this molecule,
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimgur.com%2FX0sy8.png&hash=e61e179fed8ad9187af5d96da783e06e0dd8323d)
the answer is supposed to be,
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimgur.com%2FfLa80.png&hash=22521f04304252ba44df419c0377cf067e4bc4c3)
What I don't understand is why there are not more structures that turn the double bonds in the cyclic portion of the molecule into lonepairs on carbon. Each double bond reduced to a single bond would result in a carbon with a +1 charge and another carbon having a -1 charge so the net charge difference would be zero.
Why doesn't that happen?
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The main reason is that there is no electronegativity difference between two carbon atoms. Therefore, it is extremely unlikely that there would be a resonance contributor that has charge separation between two carbon atoms.
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Whilst tamim83 has given a simple answer, it may also help to appreciate that you are, to an extent right, there are resonance forms with charge separation on the carbons as suggested. However, these are relatively high in energy so not as accessible.
What a question like this [usually] really means is identify the low energy forms.
I would suggest that the electronegativity of the amide carbon is different to that of the tertiary centre here, and different again for both the olefinic carbons, it is not an absolute value of x for all carbons, y for all nitrogens etc.
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That is why you considering the formal charges of the atoms is useful.
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Thanks very much, all, that helps!