Question

Organic chem!
I did this excercise, but I'm not sure how to explain the following:
RC(O)CH2C(O)R' + R3N --> RC(O)CHC(O)R' + R3NH+
The equilibrium position goes to the right, because the proton is donated to R3N, so R3NH+, though I don't really know how or why?
pKa value for RC(O)CH2C(O)R' is 9, and R3NH+ is 12. How do I know those are the indicated pKa values whose difference will result in the equilibrium constant, Keq, and why do I have to expess it like= 10^+3 (12-9=3)?
Is RC(O)CH2C(O)R' a Bronsted acid? It donates protons.
Please help me :D, I'm having a really, really hard time :'(

Answer 1

\(R_3N\) is a the general form for a tertiary amine base, look at the attachment i drew the mechanism for the reaction.

pKa is the negative logarithm of Ka, so when you convert to Ka, you take the inverse logarithm which is

\(\sf\large pKa=-log(Ka)\rightarrow Ka=10^{-pKa}\)

When you wanna find the K for a reaction of two K's combine then you multiply those values.

\(\sf \large K_{final}=K_1*K_2\)

For this reaction one of them is actually acting as a base, you have the use the Kb instead of the Ka:

\(\sf \large K_b=\dfrac{1}{Ka}\)

\(\sf \large K_{final}=K_1*\dfrac{1}{K_2}=10^{-9}*\dfrac{1}{10^{-12}}=10^3=1000\)