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Author: Subject: Stoichiometry for Al(Ga) amalgam reduction of nitroalkane
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[*] posted on 1-4-2022 at 20:57
Stoichiometry for Al(Ga) amalgam reduction of nitroalkane

So I've never done a reduction with an amalgam yet because mercury salts scare the bedoodles out of me, but luckily I received some gallium, so was going to give it a try.

I read everything I could find on how it works and how to do it, but no matter where I look I can't seem to find the actual reaction mechanism or even information about stoichiometry.

So from what I gather, you need 3 moles on an alcohol (I will use methanol) per mole of aluminum. And you need about 0.1 moles of gallium to make the Al(Ga) amalgam (which serves to make the aluminum's surface more reactive towards the methanol to form Al(OMe)3). Plus apparently you need some extra water or methanol as a proton source.

But at this point I get a bit lost, because I couldn't find out anywhere how many mores of nitroalkane will be reduced to the amine by one mole of Al(OMe)3). Also, how much extra water/methanol is actually needed to have sufficient protons. For instance does the H2 produced when the Al(OMe)3) forms participate in the reduction as a source of protons? What happens to the oxygens on the nitro that is being reduced?

A simple reaction equation (or if you even know where to find the actual mechanism) would help immensely, even if it's the mercury version (since I'd guess the mechanism is still more or less the same).

(Cool stuff gallium anyway, btw, I played with some last night and rolled a piece of aluminum foil around it until it was literally gone! And then the gallium (amalgam?) began to melt all over my gloves lol). Anyway, I'm looking forward to this.
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[*] posted on 7-4-2022 at 07:01

The amalgam reduction of nitroalkanes must be one of the more popular reactions and it suprises me you cant find any reaction info online.
I think its done in several variants depending on what substrate to be reduced.

Attachment: Aluminum Amalgam.pdf (320kB)
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[*] posted on 7-4-2022 at 21:49

That's aluminum/mercury amalgam, I'm assuming what this person wants information on is aluminum/gallium. I'm interested in that amalgam as well, but I think Al/Ga would be different than Al/Hg.
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[*] posted on 8-4-2022 at 02:06

Sure there is a stoichiometry, but the question is how relevant it is. Al(OMe)3 doesn't reduce, as the aluminium is already oxidized, otherwise there would be no need for the amalgam to begin with. What you need to do is having the compound to be reduced available while the amalgam is in a state where it can still reduce. I don't think there is a real difference between the mercury and gallium amalgams.

Have a look at sodium in ethanol reduction for example given oximes. The sodium is oxidized by the ethanol, but sodium ethoxide doesn't reduce anything. There is a state in between where the ethanol makes the reduction of the oxime possible.

All an amalgam is is alloy of two metals, exposing one of the metals to the compound to be reduced. In the example of sodium in ethanol there is no need for mercury or gallium as the surface is already exposed.

[Edited on 8-4-2022 by Tsjerk]
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