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Hepworth, Waring and Waring (2002) mentions that:

In general, electron-withdrawing substituents make the arene more susceptible to reduction, while the opposite applies for electron-donating substituents.

They then go on to say:

Methoxybenzene (anisole) is reduced more rapidly than benzene under Birch conditions. This is an exception to the rule, the reasons for which are not clear.

Why is this the case? Does anyone know of any possible explanations?

enter image description here

Source: Wikipedia Commons

Also, I was thinking, since the methoxy group donates electron density mesomerically to the ortho and para positions, wouldn't it be unfavourable for the carbanion to form that the ortho positions? This seems to happen in the mechanism shown above.

Lastly, I understand that the dispute between whether it is "Mechanism O" or "Mechanism M" was settled a while ago but I've read that the evidence pointing to "Mechanism O" seems to be computational studies and experiments. However, there has yet to be any qualitative explanations, based on chemistry principles and concepts, as to why "Mechanism O" is the one which occurs to a greater extent. I was wondering if anyone knows of any such explanations?

Reference

Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

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    $\begingroup$ You have very good questions but in my opinion it would be better to split them up. Two of your three questions are research level (which may not actually have an answer!) and fully deserve to stand on their own. $\endgroup$ – orthocresol Feb 18 '18 at 19:37
  • $\begingroup$ @orthocresol I understand the first and third questions are somewhat research-level. I don't really expect complete answers for these questions but I do hope to get some insightful input from some users. As for my second question, I definitely do expect an answer as it is more of a beginner-learning-a-new-topic kind of question. $\endgroup$ – Tan Yong Boon Feb 19 '18 at 1:08
  • $\begingroup$ People can feel free to just answer one of the three questions when responding and I will accept the most comprehensive and insightful of them. $\endgroup$ – Tan Yong Boon Feb 19 '18 at 1:10
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Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.

The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. Therefore, the rate-limiting step is the protonation of the radical anion 2 of anisole 1. Based on the observation that the dianion of anthracene is more basic than its radical anion, 2 the authors designed an experiment utilizing 2% d1-t-BuOH as the alcohol source. The argument is that the radical anion, which is less basic than the anion, will be more discriminating,3 favoring protonation rather than deuteration (isotope effect). The more basic anion, 4 or 7, which is less discriminating, will give rise to a higher D/H ratio. Employing deuterium NMR, the meta position in 8 was found to have 7-times the amount of deuterium than the ortho-position. Thus, the ortho-position (2a --> 3) is protonated first; mechanism O (ortho) predominates over mechanism M (meta). --continued--

![enter image description here

Calculations showed that the methoxy ortho-substituted cyclopentadienyl radical is more stable than the meta- and para-isomers (red box). But your request was for a qualitative rationale for preferring radical 3 over 6. Radical 3 has an advantage over radical 6 because radical 3 may be written with the radical on the ipso-carbon, the same radical that would be expected to be generated by auto-oxidation of 5-methoxycyclohexa-1,3-diene. Cross-conjugated radical 6 does not have this option.

1) H. E. Zimmerman and P. A. Wang, J. Am. Chem. Soc., 1993, 115, 2205.; H. E. Zimmerman, Acc. Chem. Res., 2012, 45, 164.

2) V. D. Parker, M. Tilset and O. Hammerich, J. Am. Chem. Soc., 1987, 109, 7905.

3) Consider by analogy that chlorine atoms are less selective than bromine atoms in the radical halogenation of alkanes.

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    $\begingroup$ Thanks for your very comprehensive answer! It is truly very insightful and does respond directly to my queries. If I may clarify, the reason for why having the radical being able to be generated at the ipso carbon is favourable is that the methoxy group can better stabilise it through donation of its lone pair? $\endgroup$ – Tan Yong Boon Sep 21 at 2:52
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    $\begingroup$ The radical is a resonance structure! A localized radical on the ipso carbon is just one of the canonical structures. I was trying to make an analogy to the ease with which diisopropyl ether or even diethyl ether forms hydroperoxides via a radical mechanism. Thank you for the complement. $\endgroup$ – user55119 Sep 21 at 12:08

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