[Ciara]

The pka of methyltriphenylphosphonium bromide is 30. Could you generate an ylid from this salt using sodium methoxide as base? Please explain the answer, im sure its really easy but im just not getting it 

[KritikalMass]

This is the Wittig reaction I assume? I am going to say that no, the methoxide is not a strong enough base since it has a pKa of 15.5. In order to deprotonate the methyltriphenylphosphonium bromide it would have to have a pKa > 30.

If you look up "Wittig reaction" you will see that it requires the use of very strong bases such as n-butyllithium (pKa=51  ) or phenyllithium (pKa=44).

[luiee]

how about t-BuOK?

[Dan]

t-BuOH has a pKa of about 18, so no. As KritikalMass says, you need a base whose conjugate acid has a pKa>30, which is LDA/NaH/BuLi sort of territory.

[movies]

If you want to completely deprotonate the Wittig reagent, yes, you need a very strong base, but this is not always necessary.  Potassium t-butoxide is, in fact, a rather common base for Wittig reactions.

Anyone want to hazard a guess as to why?

[stewie griffin]

re: movies
Do you know why? B/c I don't know the "official" answer and have always been curious, though I do have a guess. I always assumed it was just to avoid any possible problems with the aldehyde or ketone which was undergoing olefination. Possible problems with BuLi include 1,2-addition and also enolization of the carbonyl compound. However with KO^tBu this isn't really an issue.

[movies]

Sure, there are good reasons to prefer one set of conditions over the other, as you pointed out.  You could say that t-butoxide might be worse in some cases because it might promote some other side reaction like an aldol if your substrate is enolizable.  It will definitely depend on the substrate you are dealing with.  Also if you are trying to get selective olefination of one carbonyl over another you might want to have the fully formed reagent so you control the addition rate.

What I was getting at was why you could use a weaker base like t-butoxide and still get the reaction to go despite the pK_a difference.

[stewie griffin]

Ahh, in that case I would say Le Chalier's principle (spelling??). Although the equilibrium is poor for the deprotonation to make the ylide, once an ylide molecule reacts with the carbonyl and forms the olefin there's no way that particular ylide molecule can be reformed. In other words, as olefination proceeds it pulls away more and more ylide thus pulling the deprotonation reaction more and more to the right.

[movies]

Interesting – I guess I hadn't thought of it quite in those terms, but it seems to work.

I go in for the Curtin–Hammett explanation: you only form a small amount of the ylide at any given time, but it is very reactive and goes on to the product rapidly.  It is hard to say the exact pK_a difference since it is highly dependent on solvent.  In DMSO, the phosphonium has a lower pK_a than an alcohol (22.4 vs ~30)!  I am not sure what solvent the phosphonium value above (~30) is measured in – I would assume that it is water, but water is not a good solvent for the Wittig reaction as far as I know.

[stewie griffin]

I don't think  Curtain-Hammett is appropriate description here b/c Curtain-Hammett describes the case where we have two rapidly interconverting species, each of which can irreversibly go on to products. From the delta G's invovled we can then predict the product ratio.
However in this case only one of the species (the ylide) involved can go on irreversibly to form products (since the KO^tBu can't irreversibly form a product with the cabonyl). So in the end it comes down to Le Chatlier's principle b/c we're only applying a stress to one side of the chemical equation.
In the end though, I believe I understand what your saying and I don't mean to be splitting hairs  . I'm just curious as to what you think about my take on the Curtain-Hammett.

[movies]

You have certainly described the classic Curtin–Hammett situation.  Perhaps I am using the concept too liberally, but I think of it as most any situation where the reactive intermediate is not the predominant species in solution.  In this case, you might consider an aldol reaction as a competing pathway that is not observed, although that really also involves another equilibrium as well.