Author Topic: Finally... the reasons for low Al/Hg yields...  (Read 2895 times)

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methchembee

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Finally... the reasons for low Al/Hg yields...
« on: August 22, 2001, 06:59:00 AM »
First look at the structures of amphetamine. I would imagine that the secondary carbon holding the amino group would be susceptible to a free radical attack due to resonance possibilities, hyperconjugation, tautomerism and the imposing presence of the lone pair of the neighboring nitrogen. All these electronic factors contribute to a relatively stable  amphetamine monoradical . This also means that a relatvely small amount of energy is needed to create the amphetamine radical because the delta G between the amphetamine molecule and the transition state amphetamine monoradical is small. I'll repeat for emphasis that these factors make our beloved product susceptible to free radical degradations.

And what happens during free-radical reactions for which the Al(Hg) is a prime example? I would liken free radical reaction to a nuclear fission. With some slow moving neutrons the uranium core is bombarded and an intense chain reaction ensues. With a carefully contolled chain reaction,useful energy is harnessed as what happens inside nuclear reactors.With enough pure fissionable material and without careful control of the chain reaction a Hiroshima-like disaster could happen.

A very careful control of parameters is therefore very essential in free radical reactions because an undesirable chain reaction (dimerizations and polymerazations) may ensue.And this is where the difficulty lies because there are soooo many variables that can fuck up a free radical reaction. Talk about the initial presence of free radicals in your reaction mixture, the degree of acidity and basicity during A/B extractions, the oxygen from the air, the peroxide impurities in your solvent,the lighting in your lab.,etc.etc. etc.

Now let's go back to the mechanism of the Al/Hg reduction. First, the P2P and the methylamine combine in a reversible reaction to yield the imine and water. The meth imine, due to similar electronic factors discussed above, can be easily persuaded by the Al/Hg to receive one of the electrons it is yearning to donate. The nitrogen end of the C=N bond of the imine receives the electron(N being more electronegative than C ). This could give rise to a methamphetamine free radical with the unpaired electron on the secondary carbon atom.This meth radical would then abstract a hydrogen radical from water or the solvent to become a meth freebase.

Everyone should bear in mind that what happened above IS THE IDEAL SCENARIO that we are praying would happen. But actual conditions would greatly vary . Consider the following non-ideal scenarios:

1. AMOUNT OF WATER DURING THE REACTION
    The imine formation is reversible with the formation of water as one of the products thus excessive water will shift the reaction to the reactant side as per Le Chateliers's Principle. But water is needed in the reaction as a convenient source of hydrogens to saturate the imine double bond. I also think that the presence of water serves to moderate the tendency of the free radicals to run amuck in a chain reaction which could produce dimers, polymers , peroxides and hydroperoxides. I would certainly be wary of any peroxide formed if the reaction is approaching anhydrous conditions. I can forgive the presence of dimers and polymers as byproducts but peroxides are a no-no because they can easily trigger a chain reaction which can destroy your product in a rampage when the right opportunity comes like being exposed to light and oxygen.
It's basically a tradeoff here I think. The more water in the reaction the lesser the conversion to products but the lesser risk of peroxides also. I am begging to be corrected if I am wrong.

On the reduced risk of forming peroxides in the presence of water think of how powerful 100% H2O2 would be and how tame they are in the form of 3% solutions. The former is feared and rightly so but the latter is revered as a friend for its antiseptic properties.

Now again: which would be more likely to form peroxides a wet ether or an anhydrous ether?

Is this the reason why Shulgin  flooded his post reaction mix? Is this the reason why Ritter is all praises with flooding? Is this the reason why Labtop preferred flooding as an initial workup procedure?



2. THICKNESS OF THE FOIL
   I would like to think that the thicker, the better so that the release of the electrons from the Al/Hg would be much slower than the rate of imine formation otherwise pinacols and peroxides(AGAIN) will be likely side products. If the reaction is too fast, the Al/Hg may be forced to inject its electrons in the wrong place creating rogue peroxides and free radicals. The problem is if it is TOO SLOW you are giving more time for the oxygen, light and other impurities inside your reaction mix to eat your freebase. So again there is still a delicate balance here.

3. SOLVENT
   I would imagine that the best solvent would be diethyl ether saturated with distilled water, azeotropic IPA next best, azetropic ethanol, and then wet methanol.
   Still, the primary conderation for my choice is the control of the formation of the rogue peroxides.  I am so afraid with them. They can attack your product in a very treacherous way. WIth the use of wet diethyl ether your precious freebase will have a buffer from free radical attacks. Because ether is also susceptible to free radical attacks it would preferentially receive the free radical abuses of the free radicals more easily than the freebase.
I wouuld imagine the order of the ease of free radical attacks as follws: anhydrous diether>imine>wet diether>Wet IPA>freebase.

4. A/B CLEANUP
   
    What would happen if a peroxide was left after the flooding? It may wake up from its slumber and start a rampaging chain reaction again if light of the exact wavelength is absorbed or too much heat or too much oxygen.
THe shorter the wavelenth of the incident light, the more risky. Exposing the freebase especially in unprotected(unsolvated)state to light of the blue-violet-ultraviolet region may rupture the weak O-O bond of the peroxide
that was left unremoved .

I would prefer to add toluene first before adding base during A/B so that the sensitive freebase will be protected from atmospheric oxygen. And I would not certainly advise a vigorous shaking because that would exponentially increase the surface contact of the freebase (or worse ...peroxide)with oxygen .

IS THIS THE REASON WHY THE VENERABLE LABTOP ,with all his years in this field, ABHORS doing an A/B  as part of his purification arsenal?

Is the almost total absence of free-radicals and peroxides in the boro and the HI/P the reason their popularity among those in the know? IN other words, is this why Worlock's and Labtop's products  the very best?
  
I assume the NABH4 and HI/P reductions proceed via SN2-like mechanisms unlike the Leuckart, Al/(Hg), Birch, electrolytic reductions, and even catalytic hydrogenations proceed through free -radical mehanisms of some kind. Now there is nothing wrong with the Al/(Hg), and other free -radical mechanisms . They are good pathways to our beloved products only that they should be done with utmost care....

Osmium

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #1 on: August 22, 2001, 08:41:00 AM »
Good theory! But I´m sorry I have to inform you that there are no free radicals in that reaction.If there were then we would see a huge difference between benzo-made and other ketones, which we don´t. And no H2O2 being produced either under such strong reducing conditions.
And a few more reasons why all this won´t happen.

Rhodium

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #2 on: August 22, 2001, 09:18:00 AM »
Because the water present hinders the formation of the imine to be reduced. Amine + ketone => imine + water. But water is necessary to turn all the Al into Al(OH)3 and H2.

What if the reaction was run in only methanol, would Aluminium methoxide be formed instead? What if the stoichiometric amount of water was added dropwise to the reaction mixture instead of all at once in the beginning? Would it increase the yield?

The non-amine leftovers from the reaction, could they be distilled to give an amount of MDP2Pol, which could then be oxidized back to MDP2P?

methchembee

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #3 on: August 22, 2001, 10:23:00 PM »
Os, I agree with you that it is very difficult for H2O2 to form during the reduction proper but how about during workup especially if solution is overbasified, too much oxygen in contact with freebase because of very vigorous shaking, and the nonpolar solvent is contaminated with trace amounts of peroxides?

I disagree with you on the point that there are no free radicals being formed in the Al(Hg) reduction. Why is the Al(Hg) classified under the free-radical mechanism? And what results when a molecule accepts one electron from another entity? That molecule receiving the single electron is itself transformed into a free -radical, isn't it? Now, how long that newly formed free-radical survives as a free-radical depends on its stability. But I guess that the free radicals formed in the Al(Hg) reduction would not last longer than 3 seconds. It would have to abstract something with a single electron to stabilize itself . Nevertheless, a free radical was formed albeit for a brief moment in time.

sunlight

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #4 on: August 23, 2001, 04:22:00 AM »
Well, I would say that around a 80 % is not a bad yield, and it's usual for MD ketone + methylamine.

b159510

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #5 on: August 23, 2001, 03:18:00 PM »
Why is the Al(Hg) classified under the free-radical mechanism?...I believe this is a radical reaction of the dissolving metal type. A clue would be that heat must be controlled and it runs at atmospheric pressure.Now, how long that newly formed free-radical survives as a free-radical depends on its stability...How easy a radical is formed depends on the stability of the radical. Unless you're talking about some internal reaction that leads to decomposition of the radical, stability plays only a minor role in the lifespan of a radical..It would have to abstract something with a single electron to stabilize itself ..By definition, you are saying that two radicals are reacting together. This would be a termination step in the reaction. I think the Al/Hg reduction should terminate with the hydrocarbon radical abstracting a proton from the solvent, not by reacting with another radical...Os, I agree with you that it is very difficult for H2O2 to form during the reduction proper but how about during workup especially if solution is overbasified, too much oxygen in contact with freebase because of very vigorous shaking, and the nonpolar solvent is contaminated with trace amounts of peroxides?.. This should be a non-issue and likely has nothing to do with low yield.

Back to the Primitive

PolytheneSam

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #6 on: August 23, 2001, 05:26:00 PM »
If there was no water in the reaction mixture maybe you would have this reaction.
Al(Hg) + 5HCl --> AlCl3 + 2.5H2 + HgCl2
Also, AlCl3 is probably formed first in the presence of water which is then converted to Al(OH)3.
Look up AlCl3 in the Merck Index for insight into this reaction.

http://www.geocities.com/dritte123/PSPF.html

b159510

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #7 on: August 24, 2001, 10:07:00 AM »
I think the Al/Hg reduction should terminate with the hydrocarbon radical abstracting a proton from the solvent, not by reacting with another radical..OOPS. I think the hydrogen comes from the solvent in the form of a proton, but the termination step would be from the metal donating a second electron to the carbon radical.

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Semtexium

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #8 on: September 19, 2001, 08:10:00 PM »
Was it the IPA that contained the water...?  Are you sure, cause I seem to remember that he also used 40% MeAm in WATER... :P

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Lino

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Re: Finally... the reasons for low Al/Hg yields...
« Reply #9 on: September 20, 2001, 07:54:00 PM »
Low Yeilds!!!!! Bad technique. Wise, careful bees get >weight of amine out than ketone in. 
Os' and others use 40% MeAm in this reaction… works a treat w/cut up pie tins. 30% weight of Al to ketone, expect some unreacted Al after 12- 24 hrs. Vigorous stirring is good (mechanical) mag stirring doesn't. Needs additional heat. Solvent… MeOH. Drip ketone down cond. diluted 50-50 w/MeOH, over 2-3 hrs.

WISE BEEs… would toluene or hexane make a suitable solvent for this rxn? Would be v.easy w/up if it did.

Linoleum: the 13th element!  Now available at all good DIY stores… & Walmart.