Which of the following compounds is expected to have maximum enol content in ethanol?

(A) $\ce{CH3COCH2CO2CH3}$

(B) $\ce{CH3COCH2CO2C2H5}$ ✔(I think this is the correct answer)

(C) $\ce{CH3COCH2COCH3}$

(D) $\ce{CH3COCH(CH3)COCH3}$

I think the ketoesters will have greater enol content than the diketones because they have stronger $-I$ effect resulting in greater positive charge on the hydrogen atoms of the $\ce{CH2}$ group in between the ester and ketone group. This leads me to believe the hydrogen will be lost easily. Now between option one and two I think option (A) makes no sense because it will be converted to option (B) via trans esterification, so I have marked option (B). Please let me know where I am going wrong and the correct concept involved if what I think is not right.


Thermodynamically, the greater enol content can be associated with greater acidity. Thus, the question can be rephrased as ‘which is the strongest acid of these four?’ We can ignore the interconversion of the methyl ester A into the ethyl ester B — it will happen because we are in ethanolic solution, but extremely slowly as we do not have any strong bases present. (And even if we had strong bases, these would deprotonate A more further reducing the susceptibility of nucleophilic attack on the carboxy group.)

For reference and posteriority, these are the compounds we are talking about:

  • methyl acetoacetate (A)
  • ethyl acetoacetate (B)
  • acetoacetone (C)
  • 3-methylacetoacetone (D)

Simply by looking at the names we can see how highly related they are and how much subtle differences must matter. We can predict all four of them to be rather acidic since if the inter-carbonyl hydrogen is released and an enolate is formed, this enolate will always be in contact with an electron-withdrawing π system on the right-hand side: either another keto-group or an ester group.

That immediately gives us our first hint to compare the four: an ester group is less electron-withdrawing than a ketone, so we predict A and B to be less strong acids than C and D. This is convenient as it means we do not have to concern ourselves with the acidity difference of a methyl versus ethyl ester.

The difference in acidity between C and D is also subtle but more obvious. In both cases, the corresponding enolate anion would be planar. However, the methyl group is much larger than a hydrogen atom meaning that the enolate of D is subject to 1,3-allylic strain — which cannot be lessened without destroying planarity and thereby reducing conjugation. Thus, the enolate of C is more stable than the enolate of D.

This is backed by acidity data from the Bordwell tables — see table.

$$\textbf{Table 1: }\text{acidity data of the four compounds}\\ \begin{array}{lcccc}\hline \text{solvent} & \mathrm{p}K_\mathrm{a}(\textbf{A}) & \mathrm{p}K_\mathrm{a}(\textbf{B}) & \mathrm{p}K_\mathrm{a}(\textbf{C}) & \mathrm{p}K_\mathrm{a}(\textbf{D}) \\ \hline \ce{H2O} & -/- & 10.7 & \phantom{0}9.0 & -/-\\ \text{DMSO} & -/- & 14.2 & 13.3 & 15.1\\ \hline\end{array}$$

  • 1
    $\begingroup$ Thanks a lot . That was really helpful. I would like to ask you ,can't we explain on basis of inductive effect of methyl group ,the second part as to why c is chosen over d $\endgroup$
    – Ambarish
    Mar 1 '17 at 16:02
  • $\begingroup$ @Ambarish Not really. Remember that the anion — $\ce{Me-C(-O^-)=CMe-C(=O)-Me}$ is oxygen centred, thus pretty far away. In fact, I was initially inclined to assume a greater acidity for D since it leads to the higher-substituted double bond. $\endgroup$
    – Jan
    Mar 1 '17 at 16:07

I believe that C is the correct answer. If you draw out the enol structure from one of the ketones, you can see that the C-C double bonds is conjugated with the neighboring ketone which provides a stabilizing effect. Furthermore, either ketone can form the enol form, so there is double the chance for the enol to form.

You would not see the transesterification reaction happen here because it is thermodynamically unfavorable for ethanol to displace the methyl group without the presence of a catalyst. You are not wrong about the inductive effect of the ester, but since the solution we are in is ethanol losing the hydrogen would be unfavorable. In water, the ester might win out, as the enol could freely exchange its hydrogen.

  • $\begingroup$ If you draw out the enol of A , B and D, you should notice that they, too are conjugated to a pi system. Also, if you say C is strongest you should also consider why not D which is identical to C save an additional methyl group. $\endgroup$
    – Jan
    Mar 1 '17 at 15:24
  • $\begingroup$ I think Jan , the methyl group by its +I effect will decrease the positive charge density on the hydrogen atom ,thus making it difficult for Ethanol (already a weak base) to abstract it . Also you can think like this ,that the carbanion obtained after abstraction of hydrogen will be destabilized by the +I of methyl group . $\endgroup$
    – Ambarish
    Mar 1 '17 at 15:44
  • $\begingroup$ And regarding ester showing resonance ,I am still doubtful with the answer given . That is primarily the reason I thought the answer is b ,because resonance is present in the esters also , on addition the ester via its -I effect ,I think , creates a stronger active methylene group ,from which the hydrogen can be lost more easily ...... So that doubt still remains $\endgroup$
    – Ambarish
    Mar 1 '17 at 15:52

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