Lithium aluminum hydride is one of the most prodigous reducing agents you find in organic synthesis. In organic chemistry, reduction almost always means the addition of hydrogens - the "hydride" part is the business end of LAH. It will reduce just about anything but an olefin or an aromatic ring.
LAH will go absolutely nuts in the presence of water, evolving a great deal of hydrogen and heat. Sodium does the same thing, and you may have seen people blow it up by throwing chunks of it into water. LAH is a powder, and that huge surface area means a very, very, quick reaction. "Quenching" a LAH reaction (destroying the unused material) involves addition of water at some point, which has to be done pretty carefully and slowly (often, on ice).
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please tell me how to kill/quenching of the reagents like LiAlH4, NaBH4, Raney-Ni, Zn dust
How is it manufactured? I take it a lot of energy is involved?
The hydride is not the only powerful ingridient of LAH. I once dissolved my teflon coated stirring bar in LAH (filled the LAH in my reaction vessel before adding solvent because of another round of vacuum/N2). O sweet lattice energy of LiF. After a quite exothermic reaction under nitrogen atmosphere I got pure carbon and my magnet. I don't want to think about the reaction products (HF etc.).
Worst accidents that happened to me by now....beware of the power of the almighty LAH!
actually, i'm not sure there is. it is made (at least one way it's made) by the reaction of LiH with AlCl3 in ether, which puts an upper limit of ~40 deg C on the temperature during its formation. so thats not a lot of spare energy floating about. and its deltaHf0 is only 116 kJ/mol
I would have used the phrase "It will reduce just about anything but a double bond an alkene or aromatic ring" since carbonyls, imines, etc. contain double bonds and LAH reduces those just fine.
That's just me being picky though.
CBr4 and CPh4 form a 1:1 co-crystal in which each para-hydrogen interacts with a bromine end-on into a porous dimandoid network.
J. Am. Chem. Soc. 118 4090 (1996)
[AlH4]- is a fierce little tetrahedral ion. Would it co-crystallize with CPh4 via hydrogen-hydride bonding... and (reversibly) evolve hydrogen into a 3-D organoaluminum matrix (dense fuel storage for the H*Y*D*R*O*G*E*N* car)?
Russ- Fair enough, fixed, thanks.
I'm just amazed to run across ol' Uncle Al! Man, I haven't heard from him since I hung around sci.chem about 10 years ago. Glad to see the fumes haven't gotten to you just yet!
It's deltaHf0 may be low, but manufacturing elemental lithium and aluminum involves lots of energy.
I find it useful to quench my LithAl reactions with a concentrated solution of Sodium Sulfite, it is not nearly as exothermic as doing it just with water.
Sodium sulfate. Using sulfite could be awful in so many ways. There are many quench recipes for avoiding aluminum slimes - sulfate, tartrate, citrate, EDTA... either grabbing the aluminum or encouraging formation of something granular. Large scale LAH reductions are not happy things.
Yes what you call aluminum slime is a bit tiring, but I have as of yet tried a large scale, only about 0.5 moles.
Next time I will try out one of the others.
Thank you
LiAlH4 is one of those reagents we used to use as a last resort. Reduction tends to be indiscriminate and it will reduce practically everything with a double bond (except aromatics). Can be extremely vigorous and can suffer from latency, so you add some .. nothing happens, you add more .. still nothing .. / keep stirring .. add more .. suddeny woosh it's out of the flask.
Try not to use diethyl ether as solvent because it's easily evaporated. There have been accidents with LiAlH4 and Et2O.
Can you use LAH in dichloromethane or any other chlorinated solvent to reduce stuff?
u can use chlorinated solvents for LAH reactions. But be sure that ur solvent should be anhydrous.