If i have an aromatic ring with methoxy gp on it and 3 aromatic protons in ortho, meta and para position ..
Is there an explanation for why the ortho proton appears at lower chemical shift than meta and para knowing that methoxy is electron withdrawing gp
Dear Sherif Arafa,
protons ortho with respect to an OMe group at an aromatic ring are influenced by the +M effect (mesomeric, electron-donating) of this substituent. This effect is much stronger than any inductive electron-withdrawing (-I) effect, hence the shielding effect of OMe upon the ortho-proton.
Best wishes,
Ludger Ernst
The reply from Dr Ludger about the contribute of methoxy substituent is correct. But, you have three protons on aromatic ring. What about the other two substituents? These also could have a role in the observed chemical shifts pattern.
Among the causes of disability that result in reducing the aromatic molecule required in this preparation
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