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Topic: 3-nitroacetophenone reduction by HCl/Sn  (Read 41646 times)

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Offline IoanaD

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3-nitroacetophenone reduction by HCl/Sn
« on: October 11, 2009, 02:49:44 AM »
The product isolation of this product is quite confusing to me:

- once the reaction is complete the flask must be allowed to cool to room temperature and then lowered into an ice-water bath
A yellow paste of tin salts start to form (What tin salts form from this and how?)
- NaOH is added dropwise until the solution reaches a pH of 10 (I know this is apparently going to form SnO, but i have no idea what the mechanism is)
- the flask is then heated for another 15 minutes in boiling water bath (I'm not sure why this step is performed)

Some insight pls?  ???

Offline lavoisier

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Re: 3-nitroacetophenone reduction by HCl/Sn
« Reply #1 on: October 11, 2009, 05:05:51 AM »
Hi IoanaD,
in this reaction, Sn(0) is oxidized to Sn2+, and your organic ArNO2 is reduced to ArNH2. So you end up with a positively charged metallic species, and an organic species bearing an electron-rich nitrogen atom. Lewis acid an Lewis base.

I have the impression that the exact stoichiometry of the tin 'salts' that form is quite difficult to establish; I'd say that you get various complexes where the metal coordinates your amine, chloride anions, water, etc. Although at a very acidic pH your amine might be too strongly protonated to form complexes with tin at all (ArNH3+). But then solubility might come into play. Never mind.

When you add NaOH, the pH increases, and as water is present there will be an increase in the concentration of OH-. These hydroxide anions will gradually displace your amine from the complexes with Sn2+, and once you have something like Sn(OH)2 you just need to have dehydration to go to SnO.
But please note that the exact mechanisms probably are much more complex.

Concerning the boiling step, I suppose it's just a matter of kinetics. A reaction might be thermodynamically favoured but too slow; increasing the temperature usually helps, as you know from Arrhenius' equation.

Offline IoanaD

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Re: 3-nitroacetophenone reduction by HCl/Sn
« Reply #2 on: October 11, 2009, 02:28:03 PM »
Thank you! excellent explanation...i think i understand that.

One other problem:

For the same reaction:
After the reaction has taken place we had to do the following:

1. Allow the reaction mixture to cool slightly. Place 1 mL of NaHCO3 solution and 1 mL of ethyl acetate to a test tube. Using a pipette, add a few drops of your tin reaction mixture to this.
2. Mix the 2 phases by filling a pipette with the biphasic mixture and emptying it back into the test tube. Repeat this 4 or 5 times.

The question is, before taking the TLC the tin/HCl reaction, these 2 steps are required - the reaction mixture can't be taken directly.

Why is that? I suspect that the sodium bicarbonate is neutralizing any unreacted HCl from the mixture and perhaps deprotonate the positively charge 2-aminoacetophenone species. I just need a confirmation that that's the case.

What i have most trouble understanding is overall mechanism of this whole reaction (3-nitroacetophenone + Sn/HCl)..The textbook only presents the reactants and the products without showing any process. 

Offline lavoisier

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Re: 3-nitroacetophenone reduction by HCl/Sn
« Reply #3 on: October 12, 2009, 01:20:18 PM »
You're right about the HCl neutralization. If you didn't do that, you would have your amine as an ammonium cation ArNH3+, and this would just stick to silica and not be eluted by common solvents used for TLC.
Some amines are actually so basic that they still stick to silica even when they are spotted as freebases, and that's why mixtures of DCM and methanolic ammonia are sometimes used for the elution. But that's one step further. Your amine is not very basic.

As for the second question, well, most books report that the mechanisms of these 'dissolving metal' reductions are not completely understood. It seems established that the NO2 group is reduced stepwise, i.e. it goes to NO, then NHOH, then NH2 (I actually saw that myself when reducing a nitro cpd with Fe/NH4Cl - by LCMS I could detect the intermediate NHOH cpd forming and then disappearing). The coexistence of these species in the reaction mixture also accounts for some byproducts such as Ar-N=N-Ar (probably from condensation between ArNH2 + ArNO).

I think what you need to understand is the basic concept of the redox process. Your metal loses two electrons, which are transferred to the NO2 group, lowering the oxidation state of nitrogen.
Maybe you can try and write the balanced equation for the reaction.

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