Why is LiAlH4 used instead of NaAlH4 ? NaAlH4 would be cheaperVosoryx - 14-12-2018 at 10:03
As with everything, it depends on reaction conditions and what you're aiming to accomplish. If i recall correctly, LiAlH4 is a more powerful reducing
agent than the sodium version, and is therefore more common. Cheaper than both is sodium borohydride, which is slightly less reducing again.
Sometimes, the reducing agent could be too powerful.
It really depends on the reaction conditions, reactants, expected products, and availability. Tsjerk - 14-12-2018 at 10:06
I guess the lithium salt is more soluble under common conditions, like less apolar solvent conditions. Nakhimov - 18-12-2018 at 06:33
The primary reason I've chosen LAH over the sodium species is solubility concerns, NaAlH4 isn't soluble in ether.
I also did a quick price check: LAH is actually cheaper at sigma than NaAlH4. Presumably due to economy of scale? I would suggest you check into the
industrial production methods for each and see if that gives a clue.Tsjerk - 18-12-2018 at 16:21
LAH is actually cheaper at sigma than NaAlH4. Presumably due to economy of scale?
Or maybe the production process requires a recrystallization, which would require less solvent in the case of a more soluble salt. clearly_not_atara - 18-12-2018 at 20:50
If I had to guess NaAlH4 is more reactive, which is probably annoying because LiAlH4 is already too reactive. To wit, NaAlH4 has a flash point of -22
and LiAlH4 has a flash point of +125, making the former compound a red 4 NFPA.
[Edited on 19-12-2018 by clearly_not_atara]Metacelsus - 18-12-2018 at 23:44
Doesn't flash point refer to the temperature at which the vapors are ignitable? I doubt either compound is very volatile.clearly_not_atara - 18-12-2018 at 23:56
If it didn't produce any vapor there wouldn't be a flash point:
"A sample of specified volume is introduced to the test cup which is maintained at the test temperature. After a specified time, a test flame is
applied and the presence or absence of a flash observe"
I imagine that dissociation to NaH (s) + AlH3 (g) gives the compound a nonzero flash point. The latter compound can sustain a flame at incredibly low
concentrations.
[Edited on 19-12-2018 by clearly_not_atara]zed - 7-1-2019 at 19:53
NaAlH4 is generally much cheaper than the Lithium Salt. And it can, in fact, be made straight-away with a Parr Pressure reactor, Na, Al, and H2.
Purported to be similar in reducing power to LiAlH4, but it is not very soluble in common solvents.
