[maakii]

The C=O stretching of esters is higher than normal carbonyls, in the range of 1740 as opposed to the usual 1720. This has been attributed to the electron withdrawing inductive effect of the oxygen atom of the alcohol part of the ester, which leads to a stronger C=O bond.

However, shouldn't resonance donation of the lone pair on O be more prominant than the inductive effect? For amides, this is the case, and C=O absorbtion is at 1690, due to the delocalisation of the lone pair on N, which leads to a weaker C=O bond. I also remember that in many cases, O is fairly good at accomodating a positive charge, so can anyone explain why C=O bonds in esters absorb at 1740?

Thanks loads!

[organoman]

Just a guess,
Just see the issue as competition of electronegativity in case of ester. A double bonded oxygen is definately going to loose electron when the oxygen of alcohol ( say OII) via pi bond.
But if N is present instead of (OII) [in case of amide] then carbonyl oxygen is still highly electronegative to pull away the electrons.
As I said just a guess....
Tell me if you are convinced or not.

[maakii]

Thanks for the reply, i am clearer now,

in amides,

O            O-
||           |
C-N   <-> C=N+ , thus C=O bond is weaker, so I am wondering if in esters

O          O-
||         |
C-O <-> C=O+, doesn't this occur to make the bond weaker too? If not just by a little? But i guess that O doesn't really like to have a + charge as much as N, so your explanation kind of makes sense =p

[kristo]

May want to keep in mind you have a resonance structure there, could be stability issues that go along with that.

[organoman]

hi Maakii,
You got it right. By the way how do you put those structures on page. I don't know I how to this. Please help if you can.
Regards,
ORG