zmth
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the OH group vs the ketone replacement in for example codeine etc.
It seems that when they talk about putting something at the 14 carbon on say basically codeine they always start out with codeine that has been
converted at the 6 carbon from the initial codeine which has -OH there to one that has been converted eg as in hydrocodone or oxycodone or anyway has
the ketone that
is O , oxygen , at the carbon 6 position. Is that always necessary? and why ? Is it because with the initial OH that what one wants to put at carbon
14 would go to the 6 position?
Also not only that it also seems they always start with OH as in
oxycodone at the 14 position rather than just the simple hydrogen as initially there in say codeine(and morphine) and hydrocodone(as opposed to
oxycodone) ? and is this also necessary? In short seems they always start with oxycodone -
with the O replacing the OH to be a ketone (and i guess with one less H on that ring) , the saturation of the initial double bond in that pertinent
ring of the 6 carbon(not at the 6 carbon but nearby) and the OH replacement of initially just H at the 14 position and similar type thing in the
morphine analog. Are all 3 of these necessary say if one wants to put a cinnamyl group at the 14 carbon position ?
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Sandmeyer
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It is much easier to follow if you draw pictures and comment them... 
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ScienceSquirrel
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Mood: Dogs are pets but cats are little furry humans with four feet and self determination! 
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I cannot work out what you are asking at all.
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mr.crow
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Maybe he's high?
Double, double toil and trouble; Fire burn, and caldron bubble
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Bolt
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I really haven't looked into the chemistry or derivatives of opium poppy alkaloids, but from the looks of it you need to #) oxidize the 6-hydroxy to a
ketone and then #) isomerize, for instance, codiene in order to get the 8,14 dedihydro isomer and then perform the markovnikov addition of water
across the double bond to achieve the 14-hydroxy which can then be acylated with, e.g., cinnamyl chloride.
You need to oxidize the 6-hydroxy to the ketone before you try to acylate the 14-hydroxy because you will also acylate the 6-hydroxy. Also, if you
were to attempt a thermal or irradiational isomerization, there may be a competing isomeriztion pathway in which the pi bond electrons form the enol
which then tautomerizes to the more stable ketone. This pathway would be undesirable because of the loss of functionalization, so the hydroxy should
probably be oxidized to ketone first.
You have a lot to learn.
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