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2

I believe that if we ignore the inductive effects of the substituents for the time being, then the electron-donating group(which donates electrons via resonance effect) would manage to activate a few positions in one or more resonating structures, and the electron-withdrawing group can only take up this delocalized pair of $\pi$ electrons when it is present ...


7

The product at the para-position is more stable as it allows the compound to exist in the quinoid form, rather than only in the benzoid form: This is important as in your case, if the substituent $\ce{X}$ is $\ce{H}$, then depending on the solvent, as much as 89% of it may exist in the quinoid form ($\bf{2}$). References: Satoshi Kishimoto, Shinya Kitahara,...


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