SWIM has read various articles on the topic of mostly codeine->oxycodone, but SWIM's final goal is not OXY, but rather hydrocodone (aka vicodin, norco). Dope_amine has written the fairly easy to understand workup of codeine->oxycodone, and arrives there with a middle step that produces "14-Hydroxycodeinone". Now is this Hydroxycodeineone what I am after? or is it another substance?
Thanks in advance
14-Hydroxycodeinone
is definitly not Hydrocodone (3-Methoxy-7,8 dihydromorphinon- 6)
The pos. 14 in codeine/morphine molecule with an Hydrogen on it is oxidated to give a Hydroxy there.
Hydrocodone has only an H there.
So you need to make Dihydrocodeine(3-Methoxy-7,8 dihydromorphine) from Codeine (3-Mehoxy- Morphine)
via hydrogenation.
Then oxidate this substance to have an double bonded
oxygen at pos. 6
If you would hydrogenate 14-Hydroxycodeinone you had
Oxycodone (3-Methoxy-14-Hydroxy-7,8 dihydromorphinon-6)
A Rapid, Nearly Quantitative Conversion of Codeine to HydrocodoneBlack, T. H.; Forsee, J. C.; Probst, D. A.Synth. Commun. 30(17), 3195-3201 (2000) (https://www.thevespiary.org/rhodium/Rhodium/chemistry/hydrocodone.html)
(https://www.thevespiary.org/rhodium/Rhodium/chemistry/hydrocodone.html)
(https://www.thevespiary.org/rhodium/Rhodium/hive/hiveboard/picproxie_docs/000521499-file_6tq6.gif)Besides that recent article, there is also a series of infamous german patents from 1936 detailing a one-pot isomerization of Codeine to Hydrocodone using either palladium or platinum, available at
https://www.thevespiary.org/rhodium/Rhodium/chemistry/dihydromorphinones.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/dihydromorphinones.html)
