There is this weird information I found in my organic chemistry book ("Essential Organic Chemistry" by Ranjeet Shahi)

The question asks us to give reason as to why the following occurs.

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Also, don't bother giving reasons for the first (above) reaction. I would like to know the reason for the below one. Why would ortho be major in presence of acetic anhydride ?


1 Answer 1


It is complicated, as this paper from Olah et al. here shows, but this passage offers a possible explanation:

With the reactive nitronium salts, the isomer distribution of the nitration of anisole shows the highest ortho/para ratio (2.7-2.4), reflecting the "early" (i.e., starting aromatic-like) nature of the transition state of highest energy. Indeed, the calculated electron density in the ground state of anisole is the highest around the ortho positions, even higher than in the case of toluene. As the reactivity of the nitrating system decreases (e.g., with $\ce{CH3ONO2}$/$\ce{BF3}$ or acetylnitrate), the amount of para-substitution increases. Because the para-methoxy substituent can stabilize the arenium ion intermediate better than a para-methyl group, these reactions seem to show increasing intermediate ion-like nature of the transition state.

$\ce{HNO3}$/$\ce{Ac2O}$ has a similar ratio to the formal salts of $\ce{NO^+2}$ so may be classified as a reactive nitronium source.

A later passage in the same paper suggests that the nitrating species in the $\ce{HNO3}$/$\ce{H2SO4}$ may be the less reactive $\ce{NO+}$

If this is the case, it reflects the much weaker electrophilic nature of $\ce{NO+}$ (as compared to $\ce{NO^+2}$) and its reaction through an intermediate areniumion-like transition state of highest energy.

It would also be worth reading some of the papers referenced by Olah for further discussion.


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