Why does Birch reduction on benzene derivatives with an electron donating group happen at ortho and meta positions? When the substitute is electron withdrawing, ipso and para reduction take place because the electron withdrawing group stabilizes the anion in these positions. What's the explanation for electron donating group directivity?
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The explanation lies in the distribution of electron density in the intermediate radical anions that appear during the Birch reduction. I use the following reactions as guiding examples:
Electron-withdrawing groups stabilize electron density at the ipso and para positions through conjugation and so the negative charge will mainly be found in these positions; subsequent protonation occurs para.
Electron-donating groups destabilize a negative charge (when compared to simple benzene) through conjugation and this effect would be strongest at the ipso and para positions.
Thus the negative charge will mainly be found at the ortho and meta position.
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$\begingroup$ But why so? How does it destabilize ipso and para more than meta and ortho? $\endgroup$– EJCSep 6, 2014 at 13:17
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$\begingroup$ @Marko If you would have the same para*/*ipso pattern for a electron-donating group (EDG) as you have for an electron-withdrawing group (EWG) you would have more electron density next to the EDG or at the para position. But the EDG has a free electron pair, i.e. a high electron density, which in this case could be effectively engage in conjugation with the electron density at the ipso and para positions. And, loosly speaking, when two high electron densities interact they "repel" each other thus raising the energy. $\endgroup$– PhilippSep 6, 2014 at 13:45
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$\begingroup$ @Marko With the ortho / meta pattern, the orbital interaction and overlap is different and the two high electron densities can better "go out of each others way". In the case at hand it is rather difficult to show this in terms of resonance structures. Orbital pictures would be more suitable. $\endgroup$– PhilippSep 6, 2014 at 13:48