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Topic: reactions with NaOCH3, CH3OH  (Read 40054 times)

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Offline illegalcheese

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reactions with NaOCH3, CH3OH
« on: May 03, 2010, 12:55:33 AM »
I have a problem where I have to show why a certain reaction will not occur, and the reagents include NaOCH3 in CH3OH. The problem is, I don't know what those reagents do in a general situation. I've seen them used similarly to H20 in that they provide protons, but I don't see the mechanism behind it and I can't find it in my book. Could anyone get back to me and quickly explain the purpose of these two?

For context, we just learned about Robinson annulation, and the problem has to do with replacing the oxygen in a ketone with a double carbon bond through and intramolecular reaction. The reaction is given, as are the reagents. I'm guessing, GUESSING, that the reaction is trying to go through a Robinson annulation using NaOCH3 as a base. Is that the only purpose of it? Thanks y'all.

Offline MissPhosgene

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Re: reactions with NaOCH3, CH3OH
« Reply #1 on: May 03, 2010, 01:10:15 AM »
NaOCH3 is a base called sodium methoxide. When it is in solution with CH3OH (methanol) , you have a basic solution.

Methanol is a protic solvent.
Stereograms of the 32 crystallographic point groups: little bike wheels of cold, hard, pure rationality.

Offline illegalcheese

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Re: reactions with NaOCH3, CH3OH
« Reply #2 on: May 03, 2010, 01:36:26 AM »
Is that the only reaction it facilitates, a redox reaction, or is there any specific reaction caused by methanol or sodium methoxide.

Also, thanks. I figured out the problem, and apparently it IS just acting as a base, like you said. I'm just curious though because its completely thrown me off on other problems I've seen and screwed me over pretty bad on a test recently. And knowledge is my catharsis.

Offline OC pro

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Re: reactions with NaOCH3, CH3OH
« Reply #3 on: May 03, 2010, 11:59:01 AM »
Since sodium methoxide is a base it faciliates an ACID/BASE reaction. So acidic protons will give a reaction with NaOMe. This is the start point for your Robinson annulation.
Have you understand the mechanism? For me it seems that you donĀ“t..

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