Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xshadow on May 29, 2019, 01:50:10 PM
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Someone can explain this mechanism?
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Never seen the carbinyl oxygen of a RCOOH attacking the C=O of an anhydride
I usually use enols(as a nucleophilic) for this reaction
Thx
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If it helps you think, it could be the oxygen of the OH is attacking. They are tautomers of one another.
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http://www.orgsyn.org/demo.aspx?prep=CV4P0630
Very old procedure
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If it helps you think, it could be the oxygen of the OH is attacking. They are tautomers of one another.
Can it be a nucleophilic substitution Sn1(COOH weak nucleophilic)on C=O of the anhydride R1CO-O-R2CO?
With acyl as ""RCOO-""good leaving group?
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http://www.orgsyn.org/demo.aspx?prep=CV4P0630
Very old procedure
Indeed on 't see it in my textbook...
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Blanc Reaction-Blanc Rule
https://www.lookchem.com/Chempedia/Basic-Chemical/Chemical-Reaction/8223.html
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They may have just been sloppy with their arrows....
But you can conceive of electron flow where the acid carbonyl lone pair goes to the anhydride carbonyl, the bond to the anhydride bridging oxygen breaks and the associated electrons take up the acid hydrogen, the electrons of the bond to the acid hydrogen move to make the acid OH oxygen the new carbonyl. There is a cyclic flow of electrons and it all goes in one step. Possibly concerted.
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Never seen the carbinyl oxygen of a RCOOH attacking the C=O of an anhydride
And if the oxygen of the carbonyl group is protonated? The acetic acid in acetic anhydride is a superacid.
In the first step form a mixed anhydride with one of a carboxylic group of your dicarboxylic acid
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If it helps you think, it could be the oxygen of the OH is attacking. They are tautomers of one another.
Can it be a nucleophilic substitution Sn1(COOH weak nucleophilic)on C=O of the anhydride R1CO-O-R2CO?
With acyl as ""RCOO-""good leaving group?
For an SN1 you would have to form an acylium ion, which usually requires a strong Lewis acid (e.g. Friedel Crafts acylation) so I think SN1 is unlikely.
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Think to have understood
Thanks you all!