In the presence of acid, if a Grignard reagent reacts with a keto-ester, which carbonyl carbon would be targeted? Does it matter?

1 Answer
Jun 19, 2015

The grignard reagent is a great nucleophile. Once you make it, it'll look for the most electrophilic site it can find. Let's say you were working with... ethyl acetoacetate. That's a keto-ester molecule.

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Now, I notice you've basically said "in the presence of acid" (with some protons being the acidic content).

  • Notice how if you protonate the carbonyl on the left (step 1), the mechanism basically stops after the grignard reagent attacks there. The molecule is stuck as a tetrahedral structure centered on the carbon of the left carbonyl.
  • Water CAN transfer the proton to the other carbonyl (step 2, then step 3, separately), or another acid protonates the other carbonyl, because it's available. Either way...
  • It makes way for a nucleophilic attack on the OTHER carbonyl (step 4),
  • a two-step proton transfer (steps 5 and 6, separately),
  • and the ethanol leaving (step 7).

Now it's no longer the left-carbonyl being attacked, though, and you can see that naturally the carbonyl on the right would be targeted... eventually. So why not target that earlier, right? Minimize the amount of time and energy it takes, because molecules love being as low in energy as possible.


  • Protonate the one on the right (step 1).
  • Then, water comes in, grabs the proton from the protonated carbonyl on the right (step 2),
  • transfers to the ethoxy group (step 3),
  • and ethanol breaks off as a leaving group after #O^-# transfers its electrons downwards to reform the double bond (step 4).

The gist of it is, there's a pointless mechanistic pathway, and a useful one. Both happen, but the more useful one is better for learning. The idea is that the ester group is supposed to leave, giving you a diketone.

In the end, you'd get a mixture of products that result from the grignard attacking the left carbonyl, the right carbonyl, AND both. A bit hard to tell which is most prominent, but they all occur. I would guess that attacking the right carbonyl first, though, will be not only the most educationally useful, but the fastest one (because it's the least number of steps, minimizing the amount of energy needed to overcome the activation complex), and so the most common result.