Logic crash on electron withdrawing and donating

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Which type of molecule contains the SHORTEST carbonyl (C=O) bond?

A. An aldehyde B. An amide C. Carbon dioxide C is the best answer. An aldehyde C=O bond is composed of an sp2-hybridized carbon and an oxygen. An amide C=O bond and an ester C=O bond are also composed of an sp2-hybridized carbon and an oxygen, but the other oxygen in an ester and the nitrogen in an amide are capable of donating electron density to the C=O bond through resonance, which lengthens the C=O bond. This eliminates choices B and D. The C=O bond in carbon dioxide is composed of an sp-hybridized carbon and an oxygen. Because the sp-hybridized orbital is shorter than the sp2-hybridized orbital, the C=O bond of carbon dioxide must be shorter than the C=O bond of an aldehyde, so choice C is a better answer than choice A. The best answer is C. D. An ester


Donating e- from alkoxy and amide to electron withdrawing carbonyl carbon group lengthens the bond with oxygen?? This doesn't sound right to me.

My logic goes as follows.
Lengthening the bond --> weaker bond --> better leaving group for the electron withdrawing side --> meaning stronger electron withdrawing group.

But giving e- to electron withdrawing group does not strengthen the electron withdrawing.

The answer is CO2 cause it has the mighty sp which wins over sp2. But I still don't see why amides and esters have longer carbonyl than aldehyde.
If they are weaker then they should be reduced to alcohol by NaBH4 and other weaker reducing agents as well not just with LiAlH4 and Hydrazine.

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If you draw the resonance structures for amides and esters, you see that resonance gives the carbonyl bond some single-bond characteristic. Aldehydes have pure double-bond carbonyls. The single-bond character in amides/esters makes for a longer bond.
 
If you draw the resonance structures for amides and esters, you see that resonance gives the carbonyl bond some single-bond characteristic. Aldehydes have pure double-bond carbonyls. The single-bond character in amides/esters makes for a longer bond.


wow.... how did I not think this way!!!!! However, is my logic correct? Donating e- to EWG weakens the EWG, so it becomes a worse leaving group.
 
wow.... how did I not think this way!!!!! However, is my logic correct? Donating e- to EWG weakens the EWG, so it becomes a worse leaving group.

I'm not sure, I usually think of something as either an EWG affecting another part of the molecule, or as a leaving group, but not both. I think the trick is to not to read too much into the problem. The longer bond will be weaker, but I don't think bond length relates to whether the LG is good.

I suppose adding an e- to an EWG would no longer make it an EWG, but if it's through resonance then it's supposed to do that.
 
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