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I am told by Dr. Klein that in the nitration of benzene, hydrogen sulfate anion acts as the base and deprotonates the sigma complex.

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My question is why would the hydrogen sulfate anion be doing the deprotonation? We've just established that sulfuric acid is a stronger acid than nitric acid; hence sulfuric acid is able to protonate nitric acid to some extent. Now are we expected to believe that the conjugate base of a stronger acid is stronger than the conjugate base of a weaker acid? Why not have nitrate anion do the deprotonation? It is a stronger base than hydrogen sulfate anion.

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    $\begingroup$ There's usually a lot of water present with these acids, why couldn't water remove the proton? $\endgroup$
    – ron
    Commented Aug 19, 2014 at 18:25
  • $\begingroup$ Would water removing the proton be the kinetically favored product? And thermodynamically favored product now that I think of it ... water is the strongest base of all three options. $\endgroup$
    – Dissenter
    Commented Aug 19, 2014 at 18:26
  • $\begingroup$ Sorry, I don't understand "be the kinetically favored product." $\endgroup$
    – ron
    Commented Aug 19, 2014 at 18:30
  • $\begingroup$ As in there's more water than either hydrogen sulfate ion and nitrate ion, so water abstracting a proton is more likely than either of the other two doing the same. $\endgroup$
    – Dissenter
    Commented Aug 19, 2014 at 18:31
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    $\begingroup$ Yes, removal of the proton by water would be kinetically and thermodynamically favored. $\endgroup$
    – ron
    Commented Aug 19, 2014 at 18:34

1 Answer 1

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It's not advisable to consider these types of reaction schemes in a slavishly literal fashion. In reality, there are multiple proton transfer equilibria occurring simultaneously, and the reactions will generally be quite rapid. While it's logical and technically correct to depict the most basic of the molecules (among those present in sufficiently large concentration) performing the proton abstraction, it's also of literal practical interest or importance in many cases. I suspect the reason that $\ce{HSO4-}$ was chosen is to emphasize that the sulfuric acid is actually catalytic, so it should be regenerated in the course of the depicted reaction mechanism. Of course, water is produced as a byproduct, so the sulfuric acid will not exist primarily in its neutral molecular form.

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  • $\begingroup$ "It's not advisable to consider these types of reaction schemes in a slavishly literal fashion." It (HSO4- doing the deprotonation) was stated in the author's commentary, not implied through a mechanism. $\endgroup$
    – Dissenter
    Commented Aug 19, 2014 at 18:26
  • $\begingroup$ Also what do you mean water is produced as a byproduct? $\endgroup$
    – Dissenter
    Commented Aug 19, 2014 at 18:32
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    $\begingroup$ @Dissenter, well, the preliminary reaction is protonation of nitric acid to give $\ce{H2NO3+}$, which dehydrates to give $\ce{NO2+ + H2O}$. $\endgroup$
    – Greg E.
    Commented Aug 19, 2014 at 18:35
  • $\begingroup$ Oh okay. Is it true that pure sulfuric acid is rarely used as a reagent? I remember Martin saying something to that effect. Something about sulfuric acid dehydrating organic material. $\endgroup$
    – Dissenter
    Commented Aug 19, 2014 at 18:36
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    $\begingroup$ @Dissenter, I've used sulfuric acid at very high (around $95\%$, if I remember correctly) concentration, as well as fuming sulfuric acid, but in general it's true that it's an extremely aggressive dehydrating agent and relatively hazardous in those forms. Furthermore, w/r/t this reaction, my understanding is that anhydrous nitric acid is very difficult to prepare and doesn't have a great shelf-life due to its tendency to decompose. I've never used anything more concentrated than the standard azeotropic $68\%$. $\endgroup$
    – Greg E.
    Commented Aug 19, 2014 at 18:48

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