Deprotonation of weak base?

[MedPR]

---

Members don't see this ad.

This is in regard to the Claisen Condensation reaction

TBR says"A B-ketoester product has a pair of hydrogens that are now conjugated to two carbonyl groups. As such, their acidity is enhanced, so the sodium ethoxide (the base used) deprotonates the B-ketoester product preferentially over the ester reactant (ethyl ethanoate)."

So what happens when the product (which doesn't have a positive charge) gets deprotonated? I don't understand why they would mention this fact. The first step in the reaction is deprotonation of the reactant.. So it seems like they are saying that as you produce more B-ketoester product, the reaction slows down not only because of product/reactant equilibrium, but also because the nucleophile will preferentially deprotonate the product instead of the reactant. ????

 

[MT Headed]

I don't quite follow what they are saying. I either see three hydrogens or one hydrogen, depending on how I squint at it.

If you can follow the mechanism in wikipedia: http://en.wikipedia.org/wiki/Claisen_condensation#Mechanism thenI wouldn't worry too much about it. Maybe make sure you can recognize the reaction and quickly predict the product, and you will be fine.

This reaction does consume a mole of base per 2 moles of reactants, which is important to know. The base is not a catalyst.

Two things worry me about your musings in the second paragraph.

One, you say "the nuclephile will preferentially deprotonate...". Nucleophiles attack and stick to a (partially?) positively charged electrophile like a carbon or carbocation. Bases deprotonate, that is their job. Nucleophiles don't deprotonate unless they are also bases and behaving like a base.

Two, if you really are pulling protons from your product (and again I honestly have no idea what in the hell TBR is trying to say here), you are removing product and creating some other negatively charged molecule instead. LeChatalier says if you remove product you will drive the reaction forward to create more product. (You will see this concept a lot more in biochemistry, where something as simple as sticking a phosphate on a glucose technically removes a glucose product and lechat causes more glucose to appear by whatever mechanism created that original glucose in the first place.)

 

[MedPR]

I don't quite follow what they are saying. I either see three hydrogens or one hydrogen, depending on how I squint at it.

If you can follow the mechanism in wikipedia: http://en.wikipedia.org/wiki/Claisen_condensation#Mechanism thenI wouldn't worry too much about it. Maybe make sure you can recognize the reaction and quickly predict the product, and you will be fine.

This reaction does consume a mole of base per 2 moles of reactants, which is important to know. The base is not a catalyst.

Two things worry me about your musings in the second paragraph.

One, you say "the nuclephile will preferentially deprotonate...". Nucleophiles attack and stick to a (partially?) positively charged electrophile like a carbon or carbocation. Bases deprotonate, that is their job. Nucleophiles don't deprotonate unless they are also bases and behaving like a base.

Two, if you really are pulling protons from your product (and again I honestly have no idea what in the hell TBR is trying to say here), you are removing product and creating some other negatively charged molecule instead. LeChatalier says if you remove product you will drive the reaction forward to create more product. (You will see this concept a lot more in biochemistry, where something as simple as sticking a phosphate on a glucose technically removes a glucose product and lechat causes more glucose to appear by whatever mechanism created that original glucose in the first place.)

Thanks, I understand the mechanism, but that one bit about the nucleophile (I meant base) preferentially deprotonating the product didn't make sense to me. I think maybe it's referring to the equilibrium that exists between the deprotonated form and the B-ketoester before workup? If you remove the base, I guess there is nothing left to deprotonate the product...?

 

[1stmeds]

Thanks, I understand the mechanism, but that one bit about the nucleophile (I meant base) preferentially deprotonating the product didn't make sense to me. I think maybe it's referring to the equilibrium that exists between the deprotonated form and the B-ketoester before workup? If you remove the base, I guess there is nothing left to deprotonate the product...?

It might have to do with the pKa of the product such that it is lower so more readily loses protons to the base being used.