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I am told that a carboxylic ester would have a higher carbonyl stretching wave number than a carboxylic amide due to there being more double bond character in the carbonyl group of the ester.

This makes sense prima facie; the resonance structures suggest that there is somewhat more double bond character in the carbonyl of the ester. And nitrogen and oxygen are nearly identical in mass.

Is this actually true and applicable?

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    $\begingroup$ And what's your question? Nitrogen is less electronegative than oxygen, so it gives its electrons in conjugation more easily, if that's your question. $\endgroup$ – RBW Sep 5 '14 at 15:40
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Yes, it is true and applicable. In fact, the resonance structure you've drawn with the double bond between the carbonyl carbon and the amide nitrogen is so significant that one can actually observe restricted rotation about this bond on the nmr timescale. That is to say, if a methyl group was attached to the amide nitrogen, you can see distinct resonances for the syn and anti isomers. The rotational barrier about the C-N bond is around 20 kcal/mol, depending upon the substituents, etc. In the case of the ester, for the reason you provided, the barrier to rotation about the carbonyl carbon - ester oxygen bond is much lower, typically in the range of 10-12 kcal/mol.

So in the case of the amide, the second resonance structure is more important than it is for the ester. Consequently, the amide carbonyl bond is weaker than the carbonyl bond is for the ester.

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