The carbonyl stretch of an amide is typically found around 1,675 wavenumbers; while, the carbonyl stretch of an ester is typically found around 1,725 wavenumbers. Using complete sentences, explain the observed difference in carbonyl stretches for these two functional groups.

Explanation:

In the attached file , the resonance structure of amide and ester has been given . One can find that due to oxygen being more electronegative than nitrogen , the contribution of second resonance structure is more . Hence the double bond character of carbonyl bond ( =C=O ) as double bond is less . It is more in the form of single bond . ( see the resonance structure )

On the other hand it is not so in ester because of absence of nitrogen atom. Here carbonyl group is more in the form of double bond . The stretching wave number for double bond is more because it has greater energy .

"the periodic table is arranged in order of atomic number."

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