elfspice - 6-5-2004 at 19:21
I have been thinking about this recently, and posted on the hive, but nobody commented directly, and quite frankly, i'm tired of the tweakers and
their bad moods, so here's my idea in a more chemistry oriented forum.
Phase Transfer Catalysis, doing reactions in two phase systems with a nonpolar and polar phase, utilising solubilities of compounds to use the phase
boundary as a reaction area.
now, here's the set of reactions that i'm interested in:
HI + R-OH --> R-I + H2O (plus possible ring closure in the case of an amine)
R-I + OH- --> R-OH + HI (the reverse of the above reaction)
Na+ + ring closed amine (aziridine) + (H+)2 --> Na+ + amine
in this case, i am considering the reduction of the ephedrine molecule.
The reaction conditions i was thinking of was a mixture of NaI and Na, probably so as to result in an ionic mixture of 2:1 Na+:I-. This would be a
weight ratio of I2 and NaOH of 80:254 elemental iodine to sodium hydroxide.
In order to ensure that the HI is in the correct concentration for its reactivity, there needs to be the appropriate amount of water. HI is
concentrated at the water to HI ratio of 1:4 (54:128 by molar weights), which is exactly 56.25%, not 55, not 57. Now, in order to ensure that the I2
is in the right ratio to water, the sodium ions need to have the right amount of hydration too. The ratio given for maximum conc NaOH is 50:50 by
weight, so that means two water molecules for each NaOH.
The ratios for the whole mixture are then, as I:NaOH with the water for each part in brackets: 254(8xH2O):80(4xH2O) or, 1:2:12 moles I2:NaOH:H2O, as
weight, 245:40:216
The idea i had was that if this mixture of reagents was sitting at the bottom in the nonpolar, then add freebase ephedrine, and the idea is the
ephedrine stays in the nonpolar, so it can react at the phase interface with the free ions.
the HI reacts with the ephedrine iodinating it, then the ring closes because of the alkaline conditions, and then the aziridine formed reacts with a
free sodium ion which pulls the ring apart and draws the hydrogen from the aliphatic hydrocarbons in the solvent. (since sodium ions moderately
soluble in nonpolars, there would be somewhat of a phase transfer, at least some distance past the boundary)
This is the thoughts i've had so far, it's not a complete thought nor has it been tested as yet, but it should be. it's basically like
the ker plunk method described at the hive. I would like to emphasise that it is my hunch that the saturation of the salts to their perfect hydration
level is essential to it working, as this can allow the ions to be somewhat active by themselves (as happens with iodine ions for example) more
thought is needed as yet. maybe zinc instead of sodium, dunno...
any thoughts?
Blind Angel - 6-5-2004 at 19:25
I'm not a specialist in the case, but i think that if your clear the E word in your post and replace it by "a certain ß-ketone" or
something similar, it will make the mod happier
Geomancer - 7-5-2004 at 08:17
There are several problems here. First, you really need an acid environment. The alcohol group is a lousy leaving group, it needs to be modified to
something better (say, protonated in acid solution). This reaction looks SN1-ish, so the activity of the iodide ion is irrelevant ayway. Also, your
assertion that signifigant phase transfer will occur is highly questionable. Try to dissolve sodium chloride in your non-polar solvent. Finally,
I'm not sure of the wisdom of using elemental iodine and NaOH in lieu of NaI. After all, using strong oxidizers to do a reduction seems rather
odd. At best, you waste half your iodine. At worst, the iodine/iodate will start to oxidize stuff, and all hell will break loose.
how confusing!
Polverone - 7-5-2004 at 08:32
How does the acidic reducing agent HI form in this alkaline, oxidizing environment?