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Author: Subject: Birch reduction - why does it reduce OH from pseudoephedrine, rather than reducing the benzene ring?
Stibnut
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[*] posted on 16-5-2017 at 21:58
Birch reduction - why does it reduce OH from pseudoephedrine, rather than reducing the benzene ring?


First of all, this is strictly about the chemistry. I have no interest whatsoever in making meth. Even the one time I got a prescription for Adderall, I hated it and did not get it refilled - I never got anything useful done that was worth the horrible crash the next several days. I wouldn't make it even if it didn't mean a high likelihood of getting to know Bubba in certain unwanted but intimate ways for 5 or 10 years.

But, as we all know, one of the most popular ways of making meth involves Birch reduction. From what little I know, it appears that meth manufacturers just carry out a normal Birch, but for some reason it reduces the OH on pseudoephedrine, without altering the benzene ring. This flies in the face of how normal Birch reductions work, on e.g. benzene or toluene, where it is the ring that is reduced.

If someone were to do a Birch on, say, phenol, or benzyl alcohol, or 1-phenylethanol, would anything similar happen? Does the amine have anything to do with why the reaction works in the way it does in the case of pseudoephedrine?
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Melgar
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[*] posted on 16-5-2017 at 22:48


Remember, the shake-and-bake crown uses lithium, not sodium for their birch reductions, and lithium forms a weaker base. A more typical Birch reduction uses sodium.
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PirateDocBrown
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[*] posted on 17-5-2017 at 00:06


If it involves reduction of a hydroxyl group and not that of an aromatic ring, it's not a Birch reduction.

That notwithstanding, the answer to your question is simple: Aromatic rings are much more stable than the carbon-hydroxyl group bond.

Knowing what you should of organic chemistry, why do you think this is?
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Stibnut
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[*] posted on 17-5-2017 at 00:45


Okay, Birch-like reduction. Birchish reduction, if you will. A method involving using a dissolving-alkali metal (usually lithium, maybe sodium is more common in the lab) in anhydrous ammonia to produce solvated electrons, resulting in reductions that do not usually occur even if a strong reducing agent like LAH is used.

I know aromatic rings are very stable due to resonance stabilization, with delocalized electrons floating around the pi bonds at will. They all have bond orders of 1.5. I'll admit I don't know much more than that - namely, why things with even delocalized pi bonds should be so invulnerable to chemical attack while something involving a true double bond is much more reactive.

Still, though, it's also true that the -OH cannot be broken off by LAH or any similarly strong reducing agent. Do Birch(ish) reductions tend to favor reduction of a hydroxyl group over that of the aromatic ring whenever one is present? And is the amine contributing at all? Could Birch-like reductions be used in place of LAH in situations when LAH would also work? I'm not really clear on what is going on under the surface here.
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PirateDocBrown
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[*] posted on 17-5-2017 at 01:01


Hint: How much more electronegative do you think oxygen is than carbon?

Do the pi electrons contribute to bonding in the ring?

Do the oxygen's lone pair electrons contribute to the carbon-oxygen bond?

If there was no amine present, (or if it were protected), sure, LAH would reduce the hydroxyl to alkane. What effect do you think it would have on the aromatic ring?
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[*] posted on 17-5-2017 at 02:05


I don't know if this is Detritus or not, but I'm trying not to get involved.



I'm no longer involved in this forum.
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[*] posted on 17-5-2017 at 08:30


I thought that reaction was called the Benkeser reduction, not the Birch?
My understanding was that the birch reduction was specific to the use of sodium and NH3 as reactants/solvent, whereas Benkeser was generalized to redox reactions employing an alkali metal and amine solvent as a source of solvated electrons.

Wikipedia says Lithium/Potassium reactions reactions come under the Birch category though. So I'm a bit confused. Is the difference between them just the solvent employed(Birch when NH3 is used, Benkeser with organic amine solvents)
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Melgar
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[*] posted on 17-5-2017 at 08:45


I think Benkeser reduction is a variant of Birch reduction, using amines rather than liquid ammonia. I'd have to look it up to be sure though.
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PirateDocBrown
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[*] posted on 17-5-2017 at 09:13


A Birch reduction is indeed Na/NH3, (not LAH) but is done to an aromatic ring. Few substituent groups are reactive in these conditions; the reaction is fairly regiospecific.

The outcome is dihydrogenation of the ring, with a substituent group ending on one of the unsaturated carbons.

The OP is talking about the direct reduction of a benzylic hydroxy group (with a nearby amino group, which may or may not get involved), so it's not a Birch reduction.

[Edited on 5/17/17 by PirateDocBrown]
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[*] posted on 17-5-2017 at 10:31


Stibnut, the reductive cleavage occurs because the ring undergoes a Birch reduction. The net result is the reduction of the benzylic hydroxy group. Try depicting a Birch reduction on benzyl alcohol (or ephedrines), step by step, and you will see why it ends up this way.

Don't listen to PirateDocBrown. He is making things up. Nearly nothing of what he said is true.

And by the way, keep beginners questions in the Beginnings section.




…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)

Read the The ScienceMadness Guidelines!
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PirateDocBrown
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[*] posted on 17-5-2017 at 11:50


Sigh. I am making nothing up. This is exactly the Birch reduction. (see attached graphic). There's even a Wikipedia article on it.

https://en.wikipedia.org/wiki/Birch_reduction

Now to be fair, I know nothing about meth synthesis, so the experiences of those who dehydroxylate pseudoephedrine are lost on me.

Of course, what happens AFTER the Birch reduction of the ring, the cleavage of the benzylic carbon-oxygen bond, due to the electron-withdrawing characteristics of the oxygen, but I was hoping the OP would realize this conclusion on his own.

Your methods of pedagogy might differ, Nicodem.



[Converted unusable .svg to .png on 5/17/17 by PirateDocBrown]

BirchReductionScheme.png - 3kB


[Edited on 5/17/17 by PirateDocBrown]
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PirateDocBrown
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[*] posted on 17-5-2017 at 16:54


A little literature search finds an original article describing the secondary reaction.

There's no mention of amine derivatives, but that should only affect rate, not yield.

Li used in place of Na, but essentially the same thing.

Small, et al, JOC, (1975), 40(21), 3151.

http://pubs.acs.org/doi/pdf/10.1021/jo00909a036

[Edited on 5/18/17 by PirateDocBrown]
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MeshPL
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[*] posted on 17-5-2017 at 20:37


An interesting fact: lithium in ammonia will reduce almost any substituent on benzylic position to CH2 or CH3, be it amine, hydroxyl, ether, halogen, even acylated amine-that is amide (last one is not usually reduced by typical reduction making it a useful tool for debenzylating amides). This is useful in organic chemistry as not many things react wkth sodium/lithium in ammonia. Many groups containing acidic protons can be protected by reacting them with benzyl halide (by example benzyl chloride) and deprotected by reacting them with a suitable reducing agent, usually H2 and Pd/C, bur Li or Na in NH3 is also really effective.

Sodium and lithium are of simmilar reactivity in birch reduction, but lithium is more soluble in ammonia, hence it is often used.

The reason for this behaviour of benzylic position substituents is: mumble mumble benzylic position mumble mumble more reactive. That's what I know.
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PirateDocBrown
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[*] posted on 17-5-2017 at 23:06


Indeed that's true for reduceable groups at the benzylic position.

But for a reagent with groups elsewhere, it can be radically different. For groups directly substituent onto the aromatic ring, the outcome is highly dependent on the electron withdrawal or donation of that group to the ring. It's in these circumstances (or that of any event of no active groups at the benzylic position) that there is no further reaction, and the reduction is complete with the 1,4 cyclohexadiene product.

My intention was to get the OP to think about the energetics of the electron structure, and how the reactants change it. Hey, he might even start thinking about something other than shake-and-bake cookery.
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[*] posted on 18-5-2017 at 16:47


I've been doing amateur chemistry for at least a year now, and none of my other posts here have anything to do with meth. I was inspired to ask this because of a comment that NileRed made in his video on Birch reduction of benzene - he mentioned as an aside that Birch reduction is used to make meth, but that it is the hydroxyl group next to the aromatic ring that is reduced in that reaction. This got me thinking about why that might be the case.

My understanding of the theory is weak though, which is why I started this thread. I never took an organic chemistry class in college – I was a physics major rather than a chem major and only got interested in chemistry after graduating. In retrospect, I should have mentioned in the OP that this is a NileRed-inspired question, rather than leading with an “I swear I’m not making meth” paragraph, which made an already suspicious-seeming post seem even more suspicious. I was telling the truth, but it certainly didn’t come across that way.

Anyway, forget about ephedrine. I’ll start with what I think I know about normal Birch reductions – let me know if I’m wrong. Suppose we're reducing benzene. The solvated electron attacks the aromatic ring, giving it an unpaired electron on one side of the ring and a negative charge on the other. Then a proton donor (e.g. t-BuOH) comes along and protonates the carbon with negative charge. This still leaves a radical. Another solvated electron shows up and pairs with the unpaired one, turning it into a carbanion, which is then protonated. And its transformation to 1,4-cyclohexadiene is complete.

But I’m not sure what happens if you had 1-phenylethanol instead. I think the first electron attack starts in the aromatic ring like usual. From there, the presence of the oxygen, being really electronegative, changes what happens in ways I don’t quite understand. Does it somehow end up getting protonated, and then falling off because it is an excellent leaving group? And if instead it was beta-hydroxyphenethylamine, would the amine group affect anything?

I recently did a lithium/ethylenediamine Benkeser reduction of benzene to cyclohexene and cyclohexadiene. Li/ED is a stronger reducing agent than Li/NH3 and tends to reduce aromatics more than the normal Birch, so it likely reduced the majority of it all the way to cyclohexane. Nonetheless I was still able to smell that classic awful olefin smell and it rapidly decolorized bromine water, so I called it a success.

I just found 1-phenylethanol online and bought 30 mL. I will do a Benkeser on it and see if I get ethylbenzene, or whether the ring gets reduced instead. If the ring gets reduced, then the product should decolorize bromine water and might smell like olefins. If it doesn’t, then I should get ethylbenzene to distill out at its bp (~135 C, IIRC).
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PirateDocBrown
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[*] posted on 18-5-2017 at 21:23


Very good!

Here's some more food for thought:

I mentioned the activity of the amine only to see where your thinking went. Since the Birch environment is very Lewis basic, amines aren't going to be affected, playing only a role in reaction rate, not outcome. Likewise, a Lewis base environment will mean that a hydroxy anion can depart the reactant, once provided an electron to carry. But electrons cannot travel further up an aliphatic side chain any further than the benzylic position, and that only because of the great affinity for them that oxygen has.

Once you get the knack of noticing the possibilities in a reactant structure, you begin to gain the ability to deduce reaction outcomes from first principles, which is, of course, a large part of the synthetic organic chemist's stock in trade, usually taught to us in grad school Physical Organic.

I find it amusing that you smell the olefin odor even in the presence of stinky amines!

I'm gratified that you are leaving precursor chemicals alone. Seeing bright minds lured away from real science can be disheartening.

I'd be happy to see your results when posted. Post pics of your apparatus, too!

[Edited on 5/19/17 by PirateDocBrown]
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[*] posted on 18-5-2017 at 22:20


It was pretty easy to get rid of the amine smell. First I converted the ED to a salt with an acid - I used phosphoric that time, but will probably use HCl this time. That was impressively exothermic but got rid of the amine stink pretty fast! I then did a liquid-liquid extraction with DCM, then distilled off the DCM to get the crude product. I did it on a really small scale and only got a couple mL of this, which may have contained side-products too because I didn't try to separate anything out after the DCM was boiled off. Still, it was impressively olefin-smelling and cleared bromine water very quickly, so it must have had a fair amount of cyclohexene and a little cyclohexadiene as well.

So then the major factor is that it's in an extreme Lewis basic environment, and as soon as it is provided an electron, an OH- can just break off despite being a poor leaving group under more normal circumstances? I also take it that doing this on 2-phenylethanol wouldn't work, or at best would be very slow - the electron can't travel that far. I can see why the amine wouldn't do anything - it's quite happy as R-NH2 in a superbasic environment. I just wasn't sure if anything weird would happen with the solvated electrons given that ED (or NH3) is being used as the solvent. I'm still not sure why amines can dissolve alkali metals and create solvated electrons while nothing else can.

One other thing that I have run across in the literature is that calcium can sometimes work as well to produce milder conditions and make the reduction more selective. I did try that as well on benzene but didn't get anywhere - the Ca only barely dissolved at all even after several days. I do have a few grams of barium metal and might give that a shot, and maybe sodium as well for good measure, once the 1-phenylethanol gets here.
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PirateDocBrown
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[*] posted on 19-5-2017 at 03:03


I would love to see photos!
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[*] posted on 19-5-2017 at 05:25


Mechanism of Ephedrine to methamphetamine by birch reduction
US20040049079

mechanism.png - 24kB
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[*] posted on 13-6-2017 at 07:30


Quote: Originally posted by Stibnut  
But I’m not sure what happens if you had 1-phenylethanol instead. I think the first electron attack starts in the aromatic ring like usual. From there, the presence of the oxygen, being really electronegative, changes what happens in ways I don’t quite understand.


The exact mechanism for these reductions isn't actually known; however, the one Waffles posted is more or less how they are generally thought to work. I drew the following mechanism for a similar thread in the past, but I'll repost it here as well since it shows a few steps the one posted above does not. The mechanism below is for the reduction of 1-phenylethanol under Birch-like conditions, but it also applies to other benzyl alcohols as well:

mech Li.png - 23kB

The key to this reduction is the electron-withdrawing nature of the hydroxyl group and the relative stability of benzylic radicals. As you correctly guessed, the reduction initially begins like any other Birch reduction, with an electron being transferred into the phenyl ring to give a radical anion. But because of the location of the C-OH bond, the electron pair at the ipso-position ends up cleaving the adjacent hydroxyl group (which most likely leaves as lithium hydroxide) through a sort of E1cB-type elimination, forming an exocyclic double bond with the benzylic carbon instead. The resulting radical species then "rearranges" to a more favorable benzylic radical, and a second electron is transferred to give a carbanion intermediate. The carbanion then abstracts a proton from the alcohol to complete the reduction.

[Edited on 6-13-2017 by Darkstar]
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AJKOER
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[*] posted on 17-6-2017 at 17:41


Assuming that the creation of solvated electrons reflect the main route in this synthesis, some may wish to experiment with newer routes.

In particular, as an example, see "The O2-assisted Al/CO2 electrochemical cell: A system for CO2 capture/conversion and electric power generation", by Wajdi I. Al Sadat and Lynden A. Archer. Link: http://advances.sciencemag.org/content/2/7/e1600968.full

To quote some comments and possible implications:

"On this basis, we demonstrate that an electrochemical cell that uses metallic aluminum as anode and a carbon dioxide/oxygen gas mixture as the active material in the cathode provides a path toward electrochemical generation of a valuable (C2) species and electrical energy. Specifically, we show that the cell first reduces O2 at the cathode to form superoxide intermediates."

Implying the creation and a reaction with solvated electrons:

O2 + e- = .O2-

"Aluminum is an attractive anode material for electrochemical capture and conversion of CO2 because of its relatively low cost and lower reactivity, in comparison to Li and Na, which makes electrochemical systems involving Al inherently safer and potentially easier to manufacture."

Which suggests an an alternative to employing sodium metal.

"In Al electrochemical systems, room-temperature ionic liquids present attractive alternatives to alkaline and saline (aqueous and nonaqueous) electrolytes, which are associated with parasitic corrosion and hydrogen evolution problems (17–20). The ionic liquid/salt melt of 1-ethyl-3-methylimidazolium chloride ([EMIm]Cl)/aluminum chloride (AlCl3) is particularly important because of its thermal and electrochemical properties (21). "

Note, I am not suggesting a carboxylation reaction, just the use of a galvanic cell employing ionic RT liquid/salt melt plus pseudoephedrine to possibly effect the desired product via solvated electrons.
----------------------------------------

For those wishing to experiment with more common reagents see: "Magnesium in methanol: substitute for sodium amalgam in desulfonylation reactions", by Alan C. Brown and Louis A. Carpino, in The Journal of Organic Chemistry, Vol. 50: Issue. 10: Pages. 1749-1750, May 1985, DOI: 10.1021/jo00210a035, link: http://pubs.acs.org/doi/abs/10.1021/jo00210a035 .

[Edited on 18-6-2017 by AJKOER]
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