[SheffieldWednesday4ever]

Hey everyone,

In my notes, it tells me that if there is an O atom beside an N atom, and there's a double bond, it's going to favor the more electronegative atom, the O. But later, it shows a N,N-dimethylforamide. O forms resonance with the N, but N is less electronegative..?           
                                     
       http://upload.wikimedia.org/wikipedia/commons/2/2e/DMF_resonances.png  

Can anyone tell me why this is the case?

Thanks for reading
                                 

[nj_bartel]

I'm not sure exactly what you're asking.  That's a valid and very common resonance structure though (amides).

[SheffieldWednesday4ever]

Sorry for not being totally clear. In my notes, it shows an O-C=N atom. Note:= is actually triple bond. VS O=C=N (both double bonds). The 1st O has 3 lone pairs of electrons around it. Because O is more electronegative, the molecule is more stable than the second because it more welling to accommodate more electrons than N is. I just wonder why there would still be resonance? it shows the first molecule being more stable than the second. i hope that is a bit clearer.

[cyborgarm]

Resonance structures don't technically exist in actual molecules because the electrons are flicking about so quickly.
They are just possible states for the bond configurations in a "freeze-frame" of time.
Think of them as contributing portions to the overall character of the actual molecule, rather than being independent molecules.
The more stable structure(s) will contribute more to the actual character of the molecule than the less stable ones.
Thus, even though it is not favored, the less stable one is represented in the molecule as well, simply because it is a plausible configuration of the bonds.