locoghoul - 29-1-2008 at 08:12
Ok, it might seem a dumb question but,
Why does the LiAlH4 is a reducting agent while the NaH is just a base (not a reducing agent)?
My guess is that it has something to do with the fact that the NaH bond is totally ionic while the Al-H in the AlH4- ion is covalent. Still, im not
sure and some elaboration would be nice, thanks.
sparkgap - 29-1-2008 at 08:17
NaH can in fact be used as a reducing agent. Littered around the literature are examples where milder reducing agents just didn't cut it.
sparky (~_~)
locoghoul - 29-1-2008 at 08:31
Oh, i didn´t know that. Thanks.
Was I right about the reason behind their different behavior?
woelen - 30-1-2008 at 03:20
Actually, NaH is a very strong reductor. Crystals of NaH are flammable, and fine dust of this chemical is insanely inflammable, often autoigniting in
air. For this reason, NaH is not sold as the pure powder, but as a suspension in oil.
Drunkguy - 3-2-2008 at 05:23
How strong of a reducer is NaH?
For example, NaH is used in the Dieckman cyclization.
^Clearly if it was a really strong reducing agent then both esters would be reduced, so its obviously acting preferentially as a proton abstractor.
Perhaps the reason why LiAlH4 and NaBH4 are hydride donors is something to do with the mechanism of action?
I cant show u this visually at this time, but anybody who has studied organic theoretic chemistry will be familiar with it. What the curly arrows show
is the Al-[H] bond adding to the carbon, at the same time as C=[O] is adding to the aluminum.
For example, if deuterated reagents are used, it can be seen that following reduction of a ketone, the hydrogen on the alcohol comes from the quench,
whereas the deuterium is attached to the carbon.
Maybe, NaH and KH are unable to make and break intermolecular bonds simultaneously. from the curly arrows, i'd expect to see the hydride bond
adding to the acidic hydrogen at the same time as [H]-C is giving-up its electrons to form a carbanion. Then in a second step lasting pico/femto
seconds, the carbanion and electropositive sodium cation combine, after H2(g) has been removed.
The (ionically bonded) Na+ and Li+ in the hydride reducing agents must be honorably donating the electron in their outer-shell as part of a brave ploy
to stabilize the +ve charge on the Al or boron. The enhanced stability of these reagents means that they arent quite so desperate to
react. [This statement isnt quite correct in that LiAlH4 is clearly extremely reactive. The effect of Li donating its O.S.
electron has the effect of increasing the covalent character of the M-H hydride bonds, dictating the reaction pathway to one of preferential reduction
instead of simple proton abstraction.]
In cases where there is increased stability of reagents or where the reaction steps are reversible (ketal formation), the energetically more favorable
reaction products are formed preferentally.
Hence, proton abstraction is the *kinetically* controlled step, whereas carbonyl reduction is the THERMODYNAMIC reaction.
In the case of the simple hydrides (NaH, KH, CaH2), the metals are too electropositive/reactive to have the patience to hang around to accept the
negative charge from the oxygen.
Going back to the initial suggestion, I would agree that in the case of the hydride reducing agents, the partially
covalent M-H bonds is an important factor, whereas in the case of the simple metal-hydrides, the M-H bonds are completely ionic.
[Edited on 3-2-2008 by Drunkguy]
Nicodem - 3-2-2008 at 13:32
I've never seen a reduction of any organic substrate with NaH. I can't imagine how that could be since it is completely insoluble in every solvent I
know of. Its surface is however highly reactive and does reduce protons to H<sub>2</sub> of just about any acid, even such weak acids like
alcohols and amines (unlike CaH<sub>2</sub> which only very slowly reacts with alcohols unless heated). It also reduces oxygen and
halogens violently.
But obviously it can not be used for reductions of carbonyl groups or other such systems that require the nucleophilic addition of a hydride-like ion
on a double or triple bond. Clearly, for such an addition you need a hydride ion liganded to a Lewis acid since this is the only way for making it
soluble, nucleophilic and maintain its reducing ability at the same time. Complexation of hydride with BH<sub>3</sub> gives a stabile
complex even in protic solvents (the salts are called tetrahydroborates, NaBH<sub>4</sub> being the most known). Complexation with
AlH<sub>3</sub> gives the tetrahydroaluminates which are however too strong reducents and easily reduce protons, so only inert aprotic
solvents can be used (LiAlH<sub>4</sub> being the most known).