[soggybread]

So I've been doing some NMR analysis in my lab class, and there's a concept I'm not sure I've got my head wrapped around.

Take p-anisic acid (p-methoxybenzoic acid).

H-NMR data:

    Letter     Shift(ppm)
      A             13.0
      B             7.914
      C             7.027
      D             3.836

Okay, so I'm not quite sure why the (B) Hydrogens are further downfield and more deshielded than the (C) Hydrogens. I thought from my understanding that a C-O bond would be more electronegative than a C=O bond, and thus, be further upfield.

Could someone help explain why this is the case? Or do I have my all electronegativities mixed up?

Thanks!

[Borek]

You are trying to hotlink pictures and they are not visible. You can see them because they are in your cache.

[soggybread]

Oops! Sorry about that, it's been fixed now.

[stewie griffin]

First off, bonds are electronegative... atoms are. So I'm not sure what you mean when you say C=O bond is electronegative. Perhaps you mean electron-withdrawing??
Regardless, shielded = more electron density = upfield in the NMR. Deshielded means less electron density and thus more downfield in the NMR spectrum. So you need to explain why there's more electron density at the C protons than at the B protons (since C is obviously more upfield than B).
Anytime you have an NMR question regarding aromatic rings or conjugated double bonds the answer will almost always have to do with resonance. Draw some resonance structures of your compound and hopefully you'll see why C is more shielded than B.

[soggybread]

Ah! It makes sense now after I drew out the resonance! and yes I meant more electron-widthrawing, bad choice of words.

Thanks!