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Why do we not need $\ce{H2SO4}$ as an acid catalyst for the di-nitration of veratrole?

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Why does the 2nd nitro group go to positions 4 or 5, following the directing effects of the methoxy-group? Doesn't the first nitro group, that has been substituted onto the compound, have any effect on its position?

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  • $\begingroup$ You'll get 3,5 substituted dinitro because of 1st nitro group effect. $\endgroup$ – Mithoron Jun 28 '15 at 22:01
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When there are activating and deactivating groups on an aromatic ring, the direction of the electrophilic aromatic substitution is generally controlled by the activating groups. Think that activating groups improve the substitution in certain positions by increasing its electron density. On the other hand, deactivating groups favor one position over another, not by activating it, but deactivating it less than the other positions. So, the directing effect of a deactivating group is masked by the directing effect of the activating ones, which is stronger.

You most likely don't need the sulphuric acid due to the presence of two activating groups. In contrast, as an example, last nitration step of trinitrotoluene is so disfavored that it has to be carried out in boiling sulphuric acid.

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