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I am currently working on a mechanism for forming paracetamol (or N-(4-hydroxyphenyl)-ethanamide). The typical reaction to nitrate the phenol (starting product) is $\ce{NaNo3}$ with a catalyst of sulfuric acid. The sulfuric acid is required as a catalyst to form the nitronium cation - an electrophile - to attack the aromatic phenol. Unlike the nitration of toluene (for TNT) or benzene, it is possible to use $\ce{NaNO3}$ because the oxygen in the hydroxy group is activating.

My proposed mechanism for the formation of the nitronium is this: $$\ce{NaNO3 + H2SO4 + 2H2O <=> Na+ +NO3- + SO4- + 2H3O+}$$ I think this is a likely equilibrium between the starting ions. Sulfuric acid is a strong acid, and therefore to use a mechanism with it undissociated I think unlikely - the reaction would be slower, due to lower concentrations of the undissociated sulfuric acid. I know that $\ce{NaNO3}$ is fairly soluble, so would dissociate into ions.

I suggest attack of the $\ce{NO3-}$ by $\ce{H3O+}$, protonating one of the oxygens: $$\ce{NO3- + H3O+ -> HNO3 + H2O}$$ $\ce{HNO3}$ has a Ka of ${24 * 10}$ - 24 moles of $\ce{HNO3-}$ present for every $\ce{NO3}$ we can attribute this to the stability of the $\ce{NO3-}$ ion, but it also is fairly low compared to the Ka of sulfuric acid - ${1*10^3}$. By adding sulfuric acid, a strong acid, it increases the concentration of $\ce{H3O+}$, therefore shifting the equilibrium to the right. This favors the reaction.

In the $\ce{HNO3}$ molecule the oxygen-hydrogen bonds are all different lengths - $\ce{N-O}$ is 119.9pm, or 121.1pm, and the $\ce{N-O-H}$ bond is 140.6pm. The $\ce{N-O-H}$ bond is clearly weaker than the other two bonds. This is attributed to the resonance effect - a valence electron from each oxygen interacts with a lone pair on nitrogen favorably, shortening and strengthening the bonds. Protonating the oxygen in $\ce{NO3-}$ thus weakens the bond. Protonating the oxygen a second time: $$\ce{HNO3 + H3O+ <=> NO2+ + H2O + H2O}$$ Would probably break that bond.

The hydronium protonates the oxygen already protonated - resulting in an unstable intermediate. The positive charge pulls electron density from the $\ce{N-O}$ sigma bond onto the oxygen (inductive effect). This breaks the bond, and stabilizes the positive charge on the oxygen, and it becomes $\ce{H2O}$

However, the hydronium may protonate one of the other two oxygens. Why is the one already protonated more basic? The other two oxygens have a higher electron density, so would be more nucleophilic - better for the electrophilic $\ce{H3O+}$ to attack. A fraction would surely react in the way I have described, but I would think a much higher proportion would react on other oxygens.

Is this mechanism likely/correct? Could you explain how the second protonation occurs?

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The electrons on the two other oxygens are delocalized making them much less basic.enter image description here

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    $\begingroup$ Even if the charge is delocalized, those oxygens should be more basic. For comparison, acetate anion is around a million times more basic than water. $\endgroup$ – jerepierre Aug 30 '14 at 4:39
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As you described, the reaction depends on formation of the nitronium intermediate. I think you are right that the (partially) negatively charged oxygens are more basic. However, nitronium formation is much much faster by using water as a leaving group rather than hydroxide, which is what would be required otherwise.

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  • $\begingroup$ So you're saying that while the proportion forming on the already protonated oxygen is lower, but it still proceeds to a significant enough extent that nitronium ion is formed? $\endgroup$ – Swedish Architect Aug 30 '14 at 14:01
  • $\begingroup$ Yes, that's right. $\endgroup$ – jerepierre Aug 30 '14 at 14:10

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