Sciencemadness Discussion Board

More than BuLi ?

tatapouette - 18-11-2012 at 03:41

Hi all !

I posted a question in another post, which maybe was not the best one for.
Here it is :
Industrially, n-BuLi seems to be made following this reaction :

n-BuX + Li -> n-BuLi + LiX
with X = Cl or Br

As LiX precipitates, a simple filtration should be enough to recover the base.

If it is this easy, how come other compounds such as n-BuNa or n-BuK or even t-BuK (which must be really badass :D) etc, aren't commercially produced ?
(even if the potassium species can be made in situ by Schlosser's method)

Hexavalent - 18-11-2012 at 03:59

An extract from; http://forums.xkcd.com/viewtopic.php?t=39927&p=1598395

"Cotton and Wilkinson has a brief section on organo-sodium and organo-potassium compounds. As would be expected from the decreasing electronegativity down the group they are more ionic than organo-lithium. In contrast to organo-lithium compounds, organo-sodium and organo-potassium compounds are highly dissociated into ions, and therefore insoluble in non-polar solvents. To quote C&W, they are also 'exceedingly reactive' and are very difficult isolate - presumably many of the polar solvents that will solvate them will also react. Perhaps this is the reason they aren't used? But this isn't really my area of chemistry, so I might be barking up the wrong tree."

tatapouette - 18-11-2012 at 04:31

Thanks !
This gives some interesting starting points.
So :
- for these compounds, the stronger the base, the shorter half-life before reacting with solvent. Even when used in THF or diethyl ether.
- but a polar solvent could make them easier to get.

If a polar solvent is needed, which couldn't be reactive with these species, what about good ol' HMPA ?
The phosphorous atom would probably not been attacked (too much hindrance) and I guess the dimethylamino groups couldn't be deprotonated.

Problem solved ?

kristofvagyok - 18-11-2012 at 05:29

Quote: Originally posted by tatapouette  
If a polar solvent is needed, which couldn't be reactive with these species, what about good ol' HMPA ?
The phosphorous atom would probably not been attacked (too much hindrance) and I guess the dimethylamino groups couldn't be deprotonated.

Problem solved ?

No.

There is a P=O bond what could easily react with those stong bases. This is why they use TMEDA (tetramethyl-ethylenediammine) for dissolving the unsoluble mass at living polymerisation of styrene initiated with BuLi.

Also, I have never ever used stronger bases than BuLi, but EtNa, and MeNa could easily explode, and I have red long ago from organoceasium reagents like C6H6Cs6 what is a black solid, prepared from benzene and metallic Cs. It explodes on contact almost anything.

tatapouette - 18-11-2012 at 06:14

Quote: Originally posted by kristofvagyok  

There is a P=O bond what could easily react with those stong bases.

Really ?
I thought that :
- as the phosphorous atom is in a tetrahedron geometry ;
- and because of the (nearly charged) oxygen proximity,
the only possible attack would have been a SN2-type one, opposite to the oxygene.
But with the three NMe2 that hinder the phosphorous, I thought this couldn't happen.
Quote: Originally posted by kristofvagyok  

This is why they use TMEDA (tetramethyl-ethylenediammine) for dissolving the unsoluble mass at living polymerisation of styrene initiated with BuLi.


In this case, doesn't TMEDA act both as a polar solvent which cannot be (or at least, in these conditions, is not) attacked by the BuLi and as a lithium complexant (which then can increase the activity of the carbanion) ?
Quote: Originally posted by kristofvagyok  

Also, I have never ever used stronger bases than BuLi, but EtNa, and MeNa could easily explode, and I have red long ago from organoceasium reagents like C6H6Cs6 what is a black solid, prepared from benzene and metallic Cs. It explodes on contact almost anything.


Small molecules with big reactivity are often a problem ; in a certain way, bigger molecules can 'calm down' this extreme reactivity.
For example (and even if I would prefer to avoid this :)), I wouldn't make it a big deal of making some phenyl azide (I actually made some substituted-phenyl azide derivatives on a few grams scale). Some caution is needed of course, but nothing impossible.

But I would certainly not make even 100mg of alkyl azide such as methyl azide or ethyl azide !!

I agree with you that methylsodium, if it exists, would probably be very unstable. I guess bigger molecules, with more carbons, may be more stable. Let's say t-butyl ? Or even bigger ones ?

And C6H6Cs6 : wow !! :o :o
THAT is incredible !!
I guess it is even far more sensitive than diazomethane (which needs special glassware and stuff) !!

ThatchemistKid - 18-11-2012 at 14:24

actually, tatapouette, If I remember correctly if one uses n-chlorobutane for the reaction Lithium chloride precipitates out and can be removed but if n-bromobutane is used then the LiBr associates with the BuLi and does not precipitate. This is actually why I obtained n-chlorobutane for my BuLi synthing purposes :D. here is the wiki exert that talks about this!

If the lithium used for this reaction contains 1–3% sodium, the reaction proceeds more quickly than if pure lithium is used. Solvents used for this preparation include benzene, cyclohexane, and diethyl ether. When BuBr is the precursor, the product is a homogeneous solution, consisting of a mixed cluster containing both LiBr and BuLi, together with a small amount of octane. BuLi forms a weaker complex with LiCl, so that the reaction of BuCl with Li produces a precipitate of LiCl.