[Raine74]

Hello everyone!  This is the first of many posts I hope to make on this forum.  Here is a little background before I get to the meat and bones of my question.  I took organic chemistry (for professional fields) two years ago, without finishing the last class of the series.  I am now taking it, but my organic chemistry knowledge tree is dried up and untrimmed.  I remember some of the basics, and others come back to me every day.  I'm doing fairly well, but this quarter is a lab quarter and mechanisms are really important.  I'm learning the named reactions, but the professor has no qualms with throwing curve balls at the class.  I'm attempting to learn the underlying mechanisms to prepare myself for them, but acid-base reactions (I think I named them right) are a weak point in my comprehension.


I'll use this problem as the basis of my question, summarize how I believe I should answer it, then speculate on it further to understand how de/protonation works.



I have to find the product of this reaction.  It appears to be a regioselective alkylation of an asymmetrical ketone (I did some snooping!).  It's regioselective because it has (what appears to be) a kinetic reaction control (Lithium carbanion stabilizer; aprotic medium which will not de/protonate reactant, specific amounts of reactants; and a sterically hindering base, LDA).  Under these very specific conditions, the deprotonation will occur only at the alpha carbon to the right of the carbonyl carbon.  I'm not 100% on that, but I believe those are the more acidic hydrogens because the cyclopentane group is more (electron dense?, thus want's the left alpha carbon's hydrogen more than the methyl group of the right alpha carbon's hydrogen).  Also, when I draw the resulting enolate ion, I get a secondary carbanion when the right alpha carbon is deprotonated, instead of a tertiary carbanion for the left alpha carbon deprotonation.  A lot of this I just reviewed as of late, and could be awfully wrong

So, assuming that is right, what happens if I use a weaker base, like water or ammonia.  The reactant is a weak acid, and shouldn't be deprotonated by a weak base.  What is the final determination on the theoretical deprotonation?  Do I compare pKa values?  If there was an excess of LDA, would there be more deprotonations?  When and/or why would it stop if it could do multiple deprotonations.  I'm terribly sorry for the large background.  I want to show that I'm not using this post as a crutch, and allowing any answers to come at me in a frame I feel I have a good understanding of.  Thanks in advance for your *delete me*

[Doc Oc]

You've correctly sniffed out the main concepts.  You have two potential alpha carbons that can be deprotonated, and one will yield a more stable anion than the other.  This is the difference in attaining kinetic or thermodynamic products.

1) LDA is a bulky and very strong base, which means it will seek out the easiest access protons and not necessarily the most stable ones.
2) As far as weak bases go, you'll more likely see something like sodium methoxide (NaOMe) than water (too weak) or ammonia (amination side reaction will occur). 
3) pKas are certainly key, but other factors like the bulkiness of the base end up playing a role too.
4) In the presence of excess LDA, things start to get much more tricky after the first reaction with ethyl bromide.  Draw out the product to see why.