[Siro]

I have been teaching myself OC by studying various types of reactions online, so therefor have no teacher to go to.. I have learned a lot doing this, but I am at a roadblock where I can find very little information on quenching/working-up.

If I have an Organic Substance dissolved in Diethyl Ether with the reducing agent LAH(Lithium Aluminium Hydride) leftover in the solution from the reaction, I am told I should quench the reaction by working up with Na-K-Tartarate(Potassium Sodium Tartarate). This is supposed to "quench the reaction"... but what exactly does this mean? Does it safely eliminate the LAH by releasing the hydrogen and allowing the Lithium and Aluminum to be washed out in water? I have my doubts about this as a similar reaction would happen by reacting it with water, which would be very violent. How does this quenching assist in separating out the Diethyl Ether+Organics from the LAH+Na-K-Tartarate? What is happening?

From what I understand LAH is soluble in alcohol and reacts violently in water, and Na-K-Tartarate is soluble in water. The Diethyl Ether and the Organics need to be extracted by themselves. Does the NaKTartarate somehow make the LAH water soluable and nonreactive? How does this work?

I am only looking for a general explanation, but if a kind soul would like to go the extra step and help me understand the fundamental concept behind this, that would be greatly appreciated!

Edit: I realize if the Na-K-T can deactivate the LAH making all its components water soluble, then the solution can be washed with water a few times and then the aqueous separated leaving just the organics in the solvent. Is this what is supposed to happen when LAH is quenched?

[Archer]

Are you teaching yourself theoretical or practical organic chemistry?

This post reads a little like you are doing the latter.

[Siro]

Both I would suppose, for what's one without the other? I am going to school to Major in Chemical Engineering, though I don't have a college level chemistry class for over 1 more year... I own a small supplement/vitamin health business and realize I could expand my horizons and areas of profit by also producing various industry precursors, and hopefully eventually cut costs by being able to extract and isolate(or synthesis) vitamin concentrates myself so I won't have to get them from another manufacturer. For instance if I could produce my own Absorbic Acid(vitamin c) I could save up to a could thousand dollars a year since I won't be having to buy it from the chinese manufacturers.

I realize my goals are probably a fair ways bit away, knowledge wise, but seeing as I have nothing better to do with my time and am obsessively interested in knowing how to manipulate these various molecules to my financial advantage, then I figure why not start now.

At this point I believe I have figured out the LAH quenching for the most part, but I could be wrong, and would hope somebody could confirm the theory for me before I move on.

[magician4]

maybe this'll help you to understand the reason for of the use of Rochelle's salt:

http://curlyarrow.blogspot.de/2009/07/lithium-aluminium-hydride-reductions.html

regards

Ingo

[Babcock_Hall]

Does your quenching protocol include ethyl acetate or acetone?

[Siro]

I am not in the process of doing any particular experiment like this... but the only solvent used was Diethyl Ether.

I have no clue about the quenching protocol, or quenching much in general, hence why I posted here. From what I've read thus far the Na-K-Tartarate binds to the remaining LAH, allowing it to be washed out with a water solution isolating the Diethyl Ether and the Organic Material. Right? Wrong?

I will take a look into the above links. Thank you for the guidance!

[AlphaScent]

What happens when you react LAH with water? This is a quench.  You are stopping the reaction from proceeding any further by elimination any active reagent from being available to do chemistry. 

What happens when you add water to a pyridine and tosyl chloride solution.  This is a quench.

As a word of advice.  I would not work with LAH until you are taught properly in a lab atmosphere.

[Siro]

I realize the basic definition of the word quench, but LAH is alcohol soluble. When it is quenched, what happens to the molecule physically, that prevents it from being stuck in the alcohol/organic layer when the 2 are separated? What changes in the molecules behavior, in it's structure, etc?

[Babcock_Hall]

I have no clue about the quenching protocol, or quenching much in general, hence why I posted here. From what I've read thus far the Na-K-Tartarate binds to the remaining LAH, allowing it to be washed out with a water solution isolating the Diethyl Ether and the Organic Material. Right? Wrong?

Wrong.  Read through the link and see if you can determine what is happening.

[Siro]

I did read the link, and I do understand it allows the layers to be separated... but I don't just want to know that it happens... I want to know exactly why it is happening? What happens to the LAH molecule when it is quenched by the Na-K-Tartarate?

[discodermolide]

When you quench LAH with water you get large quantities of insoluble aluminium oxides. They appear in your reaction flask as gels or fine solids which are very difficult to remove by filtration, very difficult. Using a solution of Na,K-tartrate alleviates this and you obtain a form of the aluminium salts which is supposed to be easier to remove by filtration, usually by stirring overnight. This is a salt of the aluminium with the tartaric acid.
My experience with this method has not always been successful. I would suggest that Li-borohydride would be easier to work with and the quench is the careful addition of dilute acetic acid, watching out for too vigerous  evolution of hydrogen. But it is addition controlled. The removal of the salts is done by simple extraction into water.
Something else, you do not want to use diethyl ether if you are contemplating doing this on a larger scale, as this is a solvent that is very dangerous. Use THF or a higher ether it's much safer.

[magician4]

first of all, NaK-tartrate doesn't quench your initial reaction product: it's just a "helper for later"

second, in an ideal situation, you don't have no more LiAlH₄ molecules around at all when you beginn quenching (as they'd been consumed by the very reaction you introduced them for in the first place - except of the usual small excess, that is)

third, the core of "quenching"  is the reaction of (usually) water with the (for example) aluminium alcoholates  you generated during reduction (here shown with a ketone):

R₂C=O + [AlH₄]^-  [ (R₂CH-O-)₄ Al ]^-
[ (R₂CH-O-)₄ Al ]^- + 4 H₂O  4 R₂CHOH + Al(OH)₃ + OH^-

in a related process, excess LiAlH₄ (and "partially used" LAH) would be hydrolyzed:
Li[AlH₄] + 4 H₂O  LiOH + Al(OH)₃ + 4 H₂

now, the " Al(OH)₃ " generated is a pretty slimy stuff  (in fact it is a bunch of substances, and most of them next to completely insoluble in your matrix), with a lot of disadvantages to your workup, including that it might encapsulate a lot of yet unhydrolyzed material, thereby preventing it from finishing hydrolysis.

this is not what chemists dream of

those of us not-so-faint-at-heart hence would perform the workup with, for example, more or less concentrated " aq. HCl on ice " , hence preventing the slime from forming at all:
"Al(OH)₃ " , insoluble + excess H⁺  Al³⁺_(aq.) , soluble + H₂O

unfortunately, even if you were the more adventurous type: not all your target substances would stand strong acid conditions combined with some heat evolving.
substances requiring neutral or even alkaline conditions (let's say: acetals) hence could be not worked up with those protocols ...
... and that's where Rochelle's salt comes into play:

tartrates will form (more or less soluble) complexes with aluminium ions even under alkaline conditions, and hence prevent those slimes from forming at all

the complexes formed will look something like this:


(from: link )

... and thus facilitate the workup big time


regards

Ingo

[Babcock_Hall]

The LAH needs to be used up, in order to be quenched.  That is what the acetone or ethyl acetate are for, to soak up reducing power.  Presumably you get isopropanol from acetone.

[Siro]

Ah, so I assume the leftover LAH/Aluminum sludge could be quenched with an aqueous solution of heavily saturated Na-K-Tartarate, though it would probably be best to do this addition with a pressure equalized addition funnel? I only assume so because I am not entirely how how violent the reaction between the LAH and water would be. This way the LAH can be quenched into it's insoluable aluminum oxides, and the Na-K-Tartarate would clean up the junk.

I'm really enjoying this thread so far... I've learned a lot.

[magician4]

The LAH needs to be used up, in order to be quenched. 

to the best of my knowledge, this is not true: quenching will work regardless
just so, my personal favourite "HCl on ice" will develop hydrogen, and hence needs some higher attention

That is what the acetone or ethyl acetate are for, to soak up reducing power.  Presumably you get isopropanol from acetone.

I always was afraid of an "after the show" Meerwein-Ponndorf-Verley with acetone here , as you would need to use some excess of it, and hence (partial) re-oxidation of my original compound, diminishing the yield. You happen to know anything about this?
Ethylacetate on the other hand was very slow in performance, if memory serves.


regards

Ingo