I have plenty of left over dihydrocodeine painkillers left over from a prescription and as the pharmacy can not take them back to be reused i thought
i would do something with them.
The pills (100x20mg) were extracted with water, the solvent then evaporated to give 3.5g of solid (containing ~2g of dihydrocodeine). I figured the
other contaminent was an inert filler and so would be fine left in and not purified furthur (might be wrong?)
The entire 3.5g of crude dihydrocodeine was dissolved in 40ml of conc aqueous hydrobromic acid and heated. This is when the problems started.
-First, once the contents started boiling it turned dark black, and bubbled up through the 100ml rbf, partly into the condenser. The flask was removed
from heat and this was returned to normal.
- I then continued heating for a furthur 30 mins, constantly shaking by hand and removing from heat to make sure overboiling was not happening. It was
clear a black (tarry) ppt was formed.
-After 30mins the flask was diluted to 100ml with water and filtered to get rid of the crap. What was left in the filter was a considerable amount of
black powder, far greater in volume than the origional crude dihydrocodeine. Can anyone explain this?
-The liquid was then neutralised to PH 8 but no preicipitate of dihydromorphine formed. What have i done wrong?
The flask is now being left ovenight in the fridge to hopefully bring on crystallisation.
Anyone know how to get this to work? I do not have any more dihdyrocodeine so this would be purely academic. Nicodem - 18-12-2011 at 10:30
Have you even bothered looking at the structure of dihydrocodeine before heating it up with concentrated hydrobromic acid? What are the references to
back up such an experiment?kavu - 18-12-2011 at 10:43
It might be that the HBr cleaves the bridge ether too. Protonation of the aryl etheral oxygen would lead to formation of phenol, a good leaving group.
Further reactions of such products might lead to formation of all sorts of nasty products when heated. Some alkylation of the amine ring due to formed
methyl bromide might also occur. The reaction has been succesfully carried out with Lewis acids, such as BBr3, with good yields (http://www.erowid.org/archive/rhodium/chemistry/codeine2morp...). Activation of the -OMe site by coordination with Lewis bases, such as pyridine,
might also work.
With prolonged heating the bromination and elimination of alcohols are also possible... There is a myriad of things that could go on when such
intricate compounds are exposed to as harsh conditions as refluxing with HBr.
[Edited on 18-12-2011 by kavu]Picric-A - 18-12-2011 at 11:25
Have you even bothered looking at the structure of dihydrocodeine before heating it up with concentrated hydrobromic acid? What are the references to
back up such an experiment?
Of course i have. The cyclic ether is supposedly stabilised to attack from HBr due to the single carbon bonds either side of it. This is not true in
codeine's case however as the double bond renders the cyclic ether suseptible to attack.
Alkylation of the amine is unlikely in such polar/acidic conditions so i doubt that could be occuring.
Nucleophilic substitution of the alcohol with HBr followed by elimination is one possibility i though of but didnt think it would occur so quickly?
My ref is poste below. The internet is covered with other ones stating >80% yields often however a few of them are drug related sites so are not so
trust worthy.
All unused medicines should be taken back to a pharmacy where they can be disposed of. All you gotta do is hand them over.Nicodem - 21-12-2011 at 10:48
You are comparing quite different conditions. What you did is not even similar to what is described in the article. You heated your substrate in 48%
HBr(aq) at its reflux (124 °C). How were you to expect that the other ether functionalities in the molecule would remain intact? There is plenty of
literature confirming that ethers and alcohols succumb to bromination at such conditions (and very much has been discussed about it on this forum).
In the article they heated the compound in 15% anhydrous solution of HBr in acetic acid at 100 °C. I do not know the pKa of HBr in acetic acid, but I
would expect HBr is mostly undissociated in it. The alcohol gets immediately acetylated under such conditions which partially protects it from
bromination (esters are considerably less basic than alcohols, so they don't get protonated as easily). The ethers of secondary alcohols, like the
dihydrobenzofuran moiety in dihydrocodeine, cleave via SN1 which is more difficult in absence of water. Only the SN2 demethylation of the anisol
moiety is favoured in such media. For what I know, in aqueous media, the ease of cleavage is actually reversed (sec-alkyl aryl ether cleave faster
than anisols). Nevertheless, I'm quite confident that even anhydrous HBr in acetic acid might brominate the alcohol and cleave all the ether groups if
the temperature was increased to 124 °C. Not as rapidly as in 48% HBr(aq) which is a monster in regard to bromination of alcohols and ether cleavage.
JibbyDee - 7-1-2012 at 16:25
Nicoderm: The pKa of acetic acid is around 4.5 while the pKa of HBr is around -8. There is no way acetic acid keeps HBr protonated. Anyhow, I bet this
reaction would work using less concentrated aqueous HBr and lower temperatures. Why not use a milder acid such as HCl? Dr.Bob - 10-1-2012 at 13:54
Using HBr to remove an O-methyl group is usually not a very clean reaction in my experience, even is best cases, it has side products, and the optimum
conditions are very different from substrate to substrate, so this is not surprising.
HCl does not typically work for these demethylations, as Cl- is not as good a nucleophile as Br-. And water is usually bad for more complex
reactions. Using BBr3 in dry DCM seems for work best if you have more heteroatoms, but even that does progressively worse with more non-carbon
atoms, and 3 is starting to get high. A simple anisole might do well in these reactions, but once you add more oxygens or nitrogen, they all create
side reactions which compete with the desired reaction. GreenD - 16-1-2012 at 10:08
Being a moron I would assume that the tertiary amine group is going to undergo some sort of cleavage . Once that happens you're going to have a quat.
I didn't think the ether ring would break, but Nicodem seems pretty sure.
At any rate it looks more like you put a bunch of shit in a pot with the "more is better" attitude.
