01-16-02 07:59
No 257227
In the Al/Hg reductive amination reaction in an abs alcohol solvent, the H2 comes from... the alcohol. SWIM sure doesn't know where else it would come from... So what is the stoichiometry with this part of the reaction? SWIM sure is interested in finding out what 'end product' from the alcohol is formed.
Also, SWIM see's that absolute MeOH is the #1 choice for solvent in the Al/Hg nitromethane amination. Why is this prefered over denatured etOH? Is this purely a matter of a more prefered reflux temperature?
xoxo
Julia
(Old P2P Cook)
01-16-02 18:21
No 257383
I expect that the reaction would be
Al + 3 ROH __> Al(OR)3 + 3 H
where Al(OR)3 is an aluminum alkoxide.
(Hive Prodigy)
01-16-02 19:00
No 257392
I thought the reaction was:
Ketone + MethylAmine --> MethylImine + H2O
Al + 3H2O --> Al(OH)3 + 1.5H2
Imine + H2 --> Amine
Isn't this why a small amount of water must be added? Because the water produced in imine formation is enough to help the reaction, but not enough to completely fuel it?
PrimoPyro
Vivent Longtemps la Ruche!
(Chef d'Equippe)
01-16-02 20:38
No 257425
A small amount of water MUST NOT be added is more like it.
The alcohol already has trace amounts of H2O, thats plenty for this reaction regarding MDMA production.
GODISNOWHERE
(Hive Prodigy)
01-16-02 20:44
No 257426
The traces of H2O in absolute alcohol are enough, baal?
PrimoPyro
Vivent Longtemps la Ruche!
(Junior Service Representative)
01-16-02 22:36
No 257449
How about anhydrous MeOH, when using nitro as methylamine source?
PB
"A taste of honey is worse than none at all"
~ Sublime
(Stoni's sexual toy)
01-17-02 03:05
No 257529
IMHO performing the reaction without any added water doesn't work nearly as well, the water is necessary. One of the driving forces in this reaction is the formation of the insoluble Al oxides/hydroxides, and yields achieved by me and described in the literature prove that a little water will not affect the reaction at all. On the other hand I never used MeOH as the solvent, and MeOH is just as acidic as H2O, so Al(OMe)3 might be just as favorable as the hydroxides, but I'm still sceptical. I have stored amalgamated sheet Al for a day under EtOH and there was almost no reaction. When water was added the whole mother gradually heated up to the boiling point, boiling off the EtOH. This makes me believe that the oxide/hydroxide formation is a major driving force in this reaction.
Pyro: wrong wrong wrong!
The reaction taking place is an electron transfer from the Hg to the imine, and the hydrogen (actually the protons) necessary to form the end product are then supplied by the solvent.
H2 gas produced in this reaction is completely worthless. It will not contribute anything to this reaction, and when done properly there is no almost no H2 produced until the very end, when nearly all imine is reduced and the only possible reaction left for the Al/Hg is reaction with the solvent. The reason for this is the overvoltage(?), the difficulty of H2 production on Hg surfaces, the same effect that allows sodium amalgam to be formed in NaCl electrolysis, even though this reaction is electrochemically very unlikely to happen when you look at the standard potentials of the reactions taking place.
Another important observation: when doing the Al/Hg with thick (=low surface area), well-amalgamated sheet Al there is almost nothing happening until you start adding the ketone. The imine is formed, and as soon as there is something to be reduced around the reduction takes place. You stop adding ketone and the reaction dies down. Once all ketone has reacted the reaction cools down and pretty much comes to an end, leaving the leftover unreacted Al/Hg behind. This means you will need less Al in such a reaction compared to the one using foil, and you will end up with much less sludge to deal with.
(Hive Bee)
01-17-02 07:34
No 257566
Osmium-
What about the nitromethane variation? Are you saying that protons, rather than H2 gas is responsible for this reduction? If water was present, would this hinder the reduction of the nitro group?
Also, is the conversion from nitromethane to methylamine quantitative? SWIM has always wondered if more nitromethane added to the reaction would not result higher yields.
xoxo
julia
(Stoni's sexual toy)
01-17-02 09:48
No 257600
> What about the nitromethane variation? Are you saying
> that protons, rather than H2 gas is responsible for this
> reduction?
No, the reducing agent, a.k.a. Al/Hg is responsible for it. You can bubble H2 through nitromethane for weeks without producing any methylamine, and you can also acidify it (that's what protons are, acids) without producing any methylamine.
> If water was present, would this hinder the reduction of
> the nitro group?
Not at all.
> Also, is the conversion from nitromethane to methylamine
> quantitative?
Yes. Or at least close to quantitative.
> SWIM has always wondered if more nitromethane added to
> the reaction would not result higher yields.
If you add more nitromethane you also have to add more Al. Maybe that will raise your yields.
(Junior Service Representative)
01-17-02 12:01
No 257627
Would pre-amalgamation be detrimental to yields when using nitro for methylamine source? Haven't seen anyone mention whether or not this has been tried...
PB
"A taste of honey is worse than none at all"
~ Sublime
(Hive Bee)
01-17-02 12:17
No 257638
Thanks Osmium!
So then if methylamine was desired, one could run this type of reaction by dripping nitromethane into a three neck flask with a long condenser running 20C water through it. The methylamine could pass through this condenser and be piped into cold alcohol. Do you think this would work?
xoxo
Julia
(Hive Bee)
01-17-02 12:36
No 257645
Why are you worried obout the nitro yeilds. Generaly the loss with yeilds in that procedure are from the ketone not being toatly pure.
Vl_
(Junior Service Representative)
01-17-02 12:43
No 257648
That's not really what I meant; would pre-amalgamation prevent nitro from being transformed into methylamine?
Thanks,
PB
"A taste of honey is worse than none at all"
~ Sublime
(Stoni's sexual toy)
01-17-02 16:02
No 257736
Of course not! But you shouldn't store sheet Al/Hg, and you CANNOT store foil Al/Hg. It takes only a few minutes to amalgamate, so why do you ask this?
(Junior Service Representative)
01-17-02 17:22
No 257763
Right, one must rinse the Al with H20, then MeOH, and store under anhydrous MeOH? The reason I ask is because SWIP will be attempting to amalgamate using Hg in iodine tincture, and to advoid any side rxns, he figured washing out the Hg/I (HgI?) first would be prefered...
PB
If this doesn't work, he will go through the HgCl2 synth...
"A taste of honey is worse than none at all"
~ Sublime
(Old P2P Cook)
01-17-02 19:33
No 257800
The reaction taking place is an electron transfer from the Hg to the imine, and the hydrogen (actually the protons) necessary to form the end product are then supplied by the solvent.
H2 gas produced in this reaction is completely worthless. It will not contribute anything to this reaction, and when done properly there is no almost no H2 produced until the very end, when nearly all imine is reduced
...
Another important observation: when doing the Al/Hg with thick (=low surface area), well-amalgamated sheet Al there is almost nothing happening until you start adding the ketone. The imine is formed, and as soon as there is something to be reduced around the reduction takes place. You stop adding ketone and the reaction dies down. Once all ketone has reacted the reaction cools down and pretty much comes to an end, leaving the leftover unreacted Al/Hg behind.
All of which indicates to me that the only reason for water would be, as you said, to drive the reaction by forming insoluble aluminum (hyd)oxides, other than that methanol would be just as good of a hydrogen donor. Also, all the necessary water can come from the imine formation and/or the nitromethane reduction.
(Stoni's sexual toy)
01-18-02 01:51
No 257865
To reduce 1mol of nitromethane you need two moles of Al. I don't think that the two moles of water produced in this reduction are enough to form the hydroxide. Also, the imine formation produces 1mol of water, but requires 2/3 mol of Al.
I had great yields with 94% EtOH and 40%MeNH2. The water didn't interfere at all.
(Stoni's sexual toy)
01-18-02 01:53
No 257866
> one must rinse the Al with H20, then MeOH, and store
> under anhydrous MeOH?
No. Just dump it all into the reaction flask before you start the reaction. No need for washing, pre-etching and all that bullshit.
(Junior Service Representative)
01-18-02 09:46
No 257970
Thanks for the advise, Osmium. Sorry for these questions in your thread, Julia...
PB
"A taste of honey is worse than none at all"
~ Sublime