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Author: Subject: Dissolving Metal Reductions of Aromatic Nitro Groups
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[*] posted on 4-10-2002 at 17:06
Dissolving Metal Reductions of Aromatic Nitro Groups

Mechanics of them of course. Here is another part of that once-to-be-posted writeup.

The Mechanics Behind Reduction of Aromatic Nitro Functions to Anilines via Dissolving Metal Reductions

Sometimes It's useful to be able to reduce a nitro group to an amino group, and there are myriads of ways to do this. One of the easiest and cheapest ways to do this is via dissolving metal reductions. Not only is this method usually high yielding, but it is incredibly easy to perform, and the material cost is near zero. Nice.

It is literally as easy as adding your substrate to a solution of mineral acid and an electropositive metal, such as zinc. We will use zinc here as an example. I am not 100% certain, but I believe it is important that the metal be multivalent, that is, have a valency of more than one, like all the alkalai metals. The most common metals employed for this are zinc, nickel, and iron. A similar but different mechanism uses aluminum. I personally believe that aluminum can be used as is in a dissolving metal reduction just like these other metals are.

Here are the mechanics:

Don’t get lost on me, now. Don’t feel overwhelmed, loom at each step individually, I’ll explain it all. Our first molecule is nitrobenzene, our substrate. The first step is oxidation of the zinc atom by the positively polarized nitrogen center, to give a monovalent Zn+ ion intermediate. The neutral nitrogen radical allows the double bonded oxygen to withdraw an electron completely from the nitrogen, reforming the positive nitrogen center, and a second oxide ion, O-, to give the intermediate compound in our second bracket.

Next, aqueous HCl which is dissociated to H+ and Cl-, enter the equation. The proton combines with one of the oxide ions, forming a nitroxyl group, and the chloride is attracted to the zinc ion. Next, we see that the original negative oxide ion lends it’s electron to the nitrogen, allowing for the removal of the hydroxide ion. The zinc atom lends another electron to the nitrogen, neutralizing it from having yet another electron stolen from it by the leaving hydroxyl. This forms a nitroso group, and a Zn+2 ion. The hydroxide ion is neutralized by a second proton. This leaves Zn+2 and a second chloride ion, which gives us neutral ZnCl2 + H2O + our nitrosobenzene intermediate.

Notice how in our original molecule, the original oxide ion did not get protonated, but rather instead formed the double bonded oxygen of the nitroso group, and the newly formed oxide got protonated. I am unsure if this was stated deliberately, or if it is a generalization, and either reaction can occur. In any case, the result is the same, and it is worth mentioning.

Next, our nitrogen is positively polarized by the oxygen of the nitroso group, and a second zinc atom lends two electrons to the nitrogen cation, forming a negative oxide ion and a negative nitrogen ion as well, and a Zn+2 ion. Then two HCl molecules step in, the protons reacting with the negative ions in the group formerly known as nitoso (:P) and the chlorides pairing up with Zn+2 to form a second molecule of ZnCl2. Our product is the hydroxylamine pictured in the bottom left corner of the picture.

The hydroxyl acts as a leaving group, yielding –OH and a positive nitrogen intermediate ion. This ion is very electronegative and reacts with a third electropositive zinc atom to give Zn+2 and the negative nitrogen ion. The hydroxide is then neutralized by a proton from another HCl, as is the negative nitrogen ion, forming the amine, water, and more ZnCl2.

The reason this works is because of the very low electronegativity of the zinc or other metal used in the reducing system. The acid is to supply protons for neutralization. Each arrow in the picture represents the movement of one proton, or one single electron if the arrow does not originate from an H+ ion.

Useful, eh? Definitely.

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