The Vespiary
The Hive => Novel Discourse => Topic started by: josef_k on September 01, 2002, 12:51:00 PM
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I just read about some chemistry reactions and came across the beckmann rearangement which is the conversion of oximes to amides:
(https://www.thevespiary.org/rhodium/Rhodium/hive/hiveboard/picproxie_docs/000351262-file_dwoi.gif)
It seems to me that if you took alpha-methylhydrocinnamaldehyd and made a oxime out of it and ran it through this reaction you would get N-formylamphetamine, which you could then boil in HCl to get the good stuff. Am I wrong? I my self think that perhaps it wouldn't work with an aldehyde so you would need the corrensponding methyl ketone, in which case you would get N-acylamphetamine which is a bit harder to hydrolize.
Another question, how do you know which side the N moves to? If it moved to the "wrong" side you would get something totaly useless :) .
Has anyone got any experimential procedures for a beckmann rearangement?
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aldoximes are rearranged to nitriles, but ketoximes will go. the final product depends upon the oxime(if it's (E) or (Z)).
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Ok, so it has to be a ketone. But could you please further elaborate on how you know which side the N goes to, E and Z doesn't mean anything to me, hehe.
Take 4-phenyl-3-methyl-2-butanone as an example perhaps. Would you really get N-acylamphetamine or would the nitrogen go to the other side?
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ok think this way:
an oxime has a double bond, now look to the hydroxyl attached to the nitrogen and the main group attached to the carbon they could have two forms cis(Z) and trans(E).
About the ketone you said, you will have to get the (E) isomer of the oxime.(don't ask me how) :)
i don't know futher details ::)