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Claisen condensation mechanism

Aren't two last steps superfluous? I just do not see how the fourth and last forms differ.

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    $\begingroup$ The deprotonation step is essentially irreversible due to the acidity of the proton. Up until that point, the steps are reversible and favor the left hand side of the equilibrium. The deprotonation is often called the driving force of the reaction because when the reaction manages to beat the odds and reach that step of the reaction it is prevented from returning to the starting material by this "irreversible" deprotonation. Acidic workup yields the product. If deprotonation could not occur, only the starting material would be recovered. $\endgroup$ – RobChem Jun 14 '15 at 12:34
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The Claisen reaction occurs in a strong base $(\ce{RO-})$, which is a conjugate base of an alcohol $(\text{p}K_a \approx 16-18)$; while for the $\beta$-dicarbonyl compound, $\text{p}K_a \approx 10$. The fourth molecule cannot help but be deprotonated in the strongly basic reaction mixture to produce the fifth molecule. Acid needs to be added to reform the desired product.

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You are correct that there is no difference between the last fourth and last molecules shown.

However, the thermodynamic driving force behind the Claisen condensation is the formation of the highly stable enolate of the $\beta$ ketoester, or in your case $\beta$ diketone.

enter image description here

In order to form this stable enolate the original ester must have at least two $\alpha$ hydrogens so that there is at least one left at the end to deprotonate. Therefore the Claisen condensation will not work with esters having only one $\alpha$ hydrogen (and obviously not with esters with no $\alpha$ hydrogens).

Once you have completed the reaction to yield the enolate product, it can be acidified as shown in your reaction scheme to yield the neutral molecule.

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The 4th and the last structures are same but the last two steps are not superfluous, because the 4th structure is consumed right after it is generate and will not persist while after addition of acid, the last structure persist.

Please pay attention to the condition of the last step (acid). All steps except the last step happened consequently once you mixed two starting materials with base. The last step happens after addition of acid.

Under basic condition (alkoxides), the beta-keto ester generated in step 3 will be deprotonated immediately to give the second 5th enolate structure which will be protonated after addition of proton source (acid in this case).

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