MeAndMissEmma
Greenlighter
- Joined
- Jun 13, 2017
- Messages
- 21
Hi,
As you can may infer from my other posts, I have been thinking of increasingly impractical means of separating the alkaloids (mainly morphine) from poppy tea; my latest idea is to separate the alkaloids by electrolysing acidified poppy tea (in the anode half-cell), with pure hydrochloric acid in the cathode half-cell, the cells being joined with an ion-bridge.
The idea is that the positive alkaloid ions will travel toward the cathode and be reduced, forming alkaloid free-bases, which will immediately react to form alkaloid hydrochlorides.
Eventually, I theorise, one should end up with a reasonably pure solution of alkaloid hydrochlorides in the cathode half-cell, while non-ionic compounds and negative-ions will remain in the anode half-cell.
As far as I know, the composition of poppy tea (solution of pod putty extracted from dried opium poppy pods) is something like:
Latex (non-ionic)
Oils (non-ionic)
Waxes (non-ionic)
Alkaloids (morphine, codeine, thebaine, papaverine, &c) (positive ions when acidified)
Meconic acid (negative meconate ions)
Lactic acid (negative meconate ions)
Water
Chlorophyll?
Other stuff
So the "impurities" in poppy tea are mostly non-ionic compounds or organic acids, therefore if we indiscriminately extract positive ions, we will be extracting only the organic alkaloid bases (there may be traces of other bases, but not too much to worry about I shouldn't think)
This is why it is possible to use a cationic exchange resin to purify poppy tea, and this is what led me to think of this method.
I believe that the alkaloids are mostly present in their non-ionised, free-base form, although there may be some alkaloid meconates and lactates present.
So we need to fully ionise the alkaloids by adding an acid, making them form alkaloid hydrochlorides; with positive alkaloid ions.
I confess to knowing very little about electrochemistry, so my idea is based solely on the knowledge that morphine does form ions, and that a cationic exchange resin can be used to chemisorb it.
As I said above, but in more detail, my idea would involve an electrolytic cell wherein:
The anode half-cell contains a solution of acidified poppy tea, hydrochloric acid would be ideal in my opinion as anyhdrous HCl is a gas, so easily removed if the poppy tea needs to be re-used.
The cathode half-cell contains pure aqueous HCl.
The two half-cells are joined with an ion-bridge soaked in hydrochloric acid.
The electrodes, especially the anode, would be made of graphite or an unreactive metal such as platinum.
Morphine is used to represent all of the alkaloids in the reactions below:
Reactions at the anode:
2Cl- --> Cl2 + 2e-
4OH- --> O2 + 2H2O + 4e-
Some of the chlorine will be evolved while some will react with the water, forming HCl and HOCl with the evolution of oxygen and hydrogen:
Cl2 + 2H2O --> 2HCl + H2 + O2
Cl2 + 2H2O --> 2HOCl + H2
Reactions at the cathode:
2H+ + 2e- --> H2
C17H19NO3H+ + e- --> C17H19NO3
The free-base morphine will then react with the aqueous hydrochloric acid, forming morphine hydrochloride:
C17H19NO3 + HCl --> C17H19NO3-HCl
Since one alkaloid ion requires one electron to reduce it, each mol will need to be reduced with 1 mol of electrons (96,485 Coulombs)
1 mol of morphine = 285G
So to reduce 2G of morphine, we will need 677 Coulombs of electricity.
1A of current corresponds to 1 Coulomb per second, so:
Assuming a current of around 100mA, and ignoring energy losses due to heat and other things, we would need to electrolyse it for 2 hours, which sounds reasonable.
Another interesting possibility is that, while morphine, and codeine to an extent, can form negative phenoxides in basic solution, the other alkaloids cannot, so it might be possible to further purify the alkaloids, isolating morphine and codeine, by electrolysing them in a solution of a base such as ammonia, the negative morphine and codeine phenoxides being oxidised at the anode, while the other alkaloids remain as free-bases...
I have already tried and failed to extract the alkaloids chemically, so I have been planning to attempt chemisorption with a cationic exchange resin, but this idea came to me so I thought it would be worth consideration.
Also it would be easy to test it as the solution at the cathode would progressively become more bitter. Haha.
Does anyone think that this might work?
I know that there are perhaps much quicker or more efficient methods, but I would give this a try if you think it might work..
Thanks for all your help.
As you can may infer from my other posts, I have been thinking of increasingly impractical means of separating the alkaloids (mainly morphine) from poppy tea; my latest idea is to separate the alkaloids by electrolysing acidified poppy tea (in the anode half-cell), with pure hydrochloric acid in the cathode half-cell, the cells being joined with an ion-bridge.
The idea is that the positive alkaloid ions will travel toward the cathode and be reduced, forming alkaloid free-bases, which will immediately react to form alkaloid hydrochlorides.
Eventually, I theorise, one should end up with a reasonably pure solution of alkaloid hydrochlorides in the cathode half-cell, while non-ionic compounds and negative-ions will remain in the anode half-cell.
As far as I know, the composition of poppy tea (solution of pod putty extracted from dried opium poppy pods) is something like:
Latex (non-ionic)
Oils (non-ionic)
Waxes (non-ionic)
Alkaloids (morphine, codeine, thebaine, papaverine, &c) (positive ions when acidified)
Meconic acid (negative meconate ions)
Lactic acid (negative meconate ions)
Water
Chlorophyll?
Other stuff
So the "impurities" in poppy tea are mostly non-ionic compounds or organic acids, therefore if we indiscriminately extract positive ions, we will be extracting only the organic alkaloid bases (there may be traces of other bases, but not too much to worry about I shouldn't think)
This is why it is possible to use a cationic exchange resin to purify poppy tea, and this is what led me to think of this method.
I believe that the alkaloids are mostly present in their non-ionised, free-base form, although there may be some alkaloid meconates and lactates present.
So we need to fully ionise the alkaloids by adding an acid, making them form alkaloid hydrochlorides; with positive alkaloid ions.
I confess to knowing very little about electrochemistry, so my idea is based solely on the knowledge that morphine does form ions, and that a cationic exchange resin can be used to chemisorb it.
As I said above, but in more detail, my idea would involve an electrolytic cell wherein:
The anode half-cell contains a solution of acidified poppy tea, hydrochloric acid would be ideal in my opinion as anyhdrous HCl is a gas, so easily removed if the poppy tea needs to be re-used.
The cathode half-cell contains pure aqueous HCl.
The two half-cells are joined with an ion-bridge soaked in hydrochloric acid.
The electrodes, especially the anode, would be made of graphite or an unreactive metal such as platinum.
Morphine is used to represent all of the alkaloids in the reactions below:
Reactions at the anode:
2Cl- --> Cl2 + 2e-
4OH- --> O2 + 2H2O + 4e-
Some of the chlorine will be evolved while some will react with the water, forming HCl and HOCl with the evolution of oxygen and hydrogen:
Cl2 + 2H2O --> 2HCl + H2 + O2
Cl2 + 2H2O --> 2HOCl + H2
Reactions at the cathode:
2H+ + 2e- --> H2
C17H19NO3H+ + e- --> C17H19NO3
The free-base morphine will then react with the aqueous hydrochloric acid, forming morphine hydrochloride:
C17H19NO3 + HCl --> C17H19NO3-HCl
Since one alkaloid ion requires one electron to reduce it, each mol will need to be reduced with 1 mol of electrons (96,485 Coulombs)
1 mol of morphine = 285G
So to reduce 2G of morphine, we will need 677 Coulombs of electricity.
1A of current corresponds to 1 Coulomb per second, so:
Assuming a current of around 100mA, and ignoring energy losses due to heat and other things, we would need to electrolyse it for 2 hours, which sounds reasonable.
Another interesting possibility is that, while morphine, and codeine to an extent, can form negative phenoxides in basic solution, the other alkaloids cannot, so it might be possible to further purify the alkaloids, isolating morphine and codeine, by electrolysing them in a solution of a base such as ammonia, the negative morphine and codeine phenoxides being oxidised at the anode, while the other alkaloids remain as free-bases...
I have already tried and failed to extract the alkaloids chemically, so I have been planning to attempt chemisorption with a cationic exchange resin, but this idea came to me so I thought it would be worth consideration.
Also it would be easy to test it as the solution at the cathode would progressively become more bitter. Haha.
Does anyone think that this might work?
I know that there are perhaps much quicker or more efficient methods, but I would give this a try if you think it might work..
Thanks for all your help.
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