Question

Question 1: The pKa of ethyl acetoacetate is 10.7. What is the Keq for the reaction shown below? +HO - + H20 OCH2CH3 OCH2CH3 Question 2: Why is it important that the alcohol solvent consist mostly of ethanol? Explain why it would not be a good idea to perform this reaction in 95% methanol. You should provide any appropriate mechanisms to support your answer. Question 3: In the reading in the lab manual, the original procedure by Robinson is described. Why was important for Robinson to add a lithium enolate to the enone that he used? That is, what would be the concern with mixing the ketone and enone together and then adding base?

Answer 1

Answer No. 1: The given reaction is

Solvent : Ethyl alcohal so base would be ethoxide ion.

Look up the pKa values of both and write the equilibrium reaction seperatly:

Equation i for EAA:

Ka_(i) = [EAA^-][H₃O⁺]/  [EAA] = 10^-10.7 (since pKa = - logKa, Ka = 10^-pKa andpKa(ethyl acetoacetate) = 10.7)

Similarly equation (ii) for CH₃CH₂OH:

Ka_(ii)= [CH₃CH₂^-][H₃O⁺]/  [CH₃CH₂OH] = 10^-15.9 pKa(CH₃CH₂OH) = 15.9

by reversing the equation ii)

Ka_(iii)= = 1/ 10^-15.9 (Equation iii)

On combining reaction (i) and (iii) , We have

K_eq = [EAA^-][CH₃CH₂OH] / [EAA] [CH₃CH₂^-]

= Ka_(i) * Ka_(iii) =10^-10.7*(1/10^-15.9)^₌10^5.2

Answer No. 2:  Methanol solvent (MeOH) will generate methoxide ion (OMe^-), which will react with EAA (via SN² ) and form the Methyl derivative.

Transesterification.

Answer No. 3: what would be the concern with mixing the ketone and enone together and then adding base?

In this way, there is a possibility that the enone itself forms the enolate and attack the other "enone molecule" via michael addition to form a sideproduct.