Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: tasnim rahman on April 26, 2012, 01:27:52 PM
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I found the mechanism for ester, but I think carboxylic acid has a different mechanism.Searching the web I found this
http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/c_acid/reaktionen_organoli_carbons.vlu/Page/vsc/en/ch/12/oc/c_acid/organoli/organoli.vscml.html (http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/c_acid/reaktionen_organoli_carbons.vlu/Page/vsc/en/ch/12/oc/c_acid/organoli/organoli.vscml.html).
But I am not sure whether its the right one. Could anyone verify this?
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Yes, the reduction of carboxylic acids to primary alcohol mechanism on that webpage is correct.
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It is sort of right, if you aren't too concerned with details.
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Thanks everyone.
Which part might you be referring to, orgopete?
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The leaving group is an H2AlO(-) and the reductant is AlH4(-) to make the intermediate. While I like that as my leaving group, I also like to write a mechanism that clearly shows how all of the hydrogens can be transferred. You would see that I have written my BH4(-) and AlH4(-) mechanisms to show the regeneration of tetravalent borate and aluminate species. However, as I said, the mechanism is basically correct and I like my mechanisms to explain everything or as much as possible.
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I understand, but this was the best I could find after a thorough search of the web.Thank you very much orgopete. Could you please post links of any websites that you know that show the detailed mechanism?Thanks again.
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Let me just tell you how I wrote them in my book.
R2C=O + AlH4(-) :rarrow: R2CHO(-) + AlH3 :rarrow: R2CHOAlH3(-)
R2C=O + R2CHOAlH3(-) :rarrow: R2CHO(-) + R2CHOAlH2 etc.
What I did was use one of the aluminate (tetravalent aluminum with neg charge) to reduce a carbonyl group. If you do this with a acid reduction, then you will generate the trivalent aluminum needed for the elimination.
It is a similar reaction with NaBH4. I wanted a complete mechanism that showed how all hydrides are transferred.
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I understood the carbonyl group reduction mechanism. What seemed new was the formation of hydrogen from H- and H+, and a trivalent aluminum species as a leaving group. Thanks again orgopete.
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The aluminum bonded to the oxygen in the carboxylic acid reduction mechanism is trivalent right? :-[
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The aluminum bonded to the oxygen in the carboxylic acid reduction mechanism is trivalent right? :-[
If you note in the mechanism that I wrote, the electrons of a tetravalent aluminum are held less tightly and thus that is the reducing agent. Upon loss of the electrons (with the proton), aluminum is now trivalent. In order for it to donate electrons it must again become tetravalent. It can do that from an alkoxide present in the reaction mixture. For example, I imagine a reaction of AlH4(-) with a RCOOH to give H2 plus AlH3 and RCOO(-). If the carboxylate adds to the AlH3, it gives a new tetravalent aluminum, RCOOAlH3(-). This should also be a reductant, either intramolecularly or intermolecularly. The reduction will generate a new alkoxide and trivalent aluminum. Just keep cycling these species until the hydrides are depleted or no reducible groups are present.
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The eliminated alkoxide is OAlH2(-), and the aluminum is trivalent, right?
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The eliminated alkoxide is OAlH2(-), and the aluminum is trivalent, right?
Um, (-)OAlH2 :resonance: O=AlH2(-)
pick em.
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Sorry I was trying to mean (-)OAlH2 .
Those are resonance structures, and aluminum in O=AlH2(-) is tetravalent, right?
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Yes, O=AlH2(-) is tetravalent, but because it is a resonance structure, it is also sort of not tetravalent as the other resonance structure.
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I guess, I get it now. Thanks. ;D