Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: mrsandman on February 25, 2009, 07:57:24 PM
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I have a question about deprotecting tboc group with tfa
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg3.imageshack.us%2Fimg3%2F9397%2Ftfaremovaloftboc.jpg&hash=b9f968b7548d5066ee54a76059088a97983dddac)
I'm thinking that the Nitrogen will attack the carbonyl on TFA and form a tetrahedral intermediate. Then the OH on TFA would be in proximity to the the carbonyl to the amide and that would attack and you would have like a cyclobutane like structure and the lone pairs would come down and in the end you woud have the tboc group attached to the tfa + the deprotected molecule.
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nvm got it
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I don't know what you've found. Please share with us. I'm sure TFA protonates carbonyl group in CO(OBu-t) and the positive charge moves to t_Bu group that among alkyls stabilizes the positive charge the best. t_Bu+ cleaved from the molecule and gets quenched by any nucleophile available around, even if it'd CF3COO-. Obviously, CO2 released at the next stage.
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Did you get isobutene as one of your products?
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You do, by loosing proton from Me3C+, if no good nucleophiles around.
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(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fc%2Fc5%2FTBOC_deprotection_scheme.png&hash=eddb2bbbca6b76e5b4e6b90b13ac1622e9fc569a)
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Is the nitrogen of an amide the most basic part? How would that change the mechanism you drew?