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While reviewing my old materials before an external exam which also covers these topic I stopped by one of the reactions - Grignard w/esters:

Grignard rx

From what I remember, and what my notes are also telling me, Grignard is performed by creating Grignard reagent containing alkane & MgX. Then, such complex attacks the carbonyl leaving oxygen with a negative charge, which easily deprotonates water creating 2° alcohol.

However, when done with ester, which contains a good leaving group the reaction produces ketone. Is this only because no aqueous workup is performed after the first round?

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    $\begingroup$ Your picture is correct. The second paragraph is incorrect. Even with one equivalent of ester and Grignard reagent there is overreaction because the ketone that is formed in situ is more reactive than the ester. $\endgroup$ – user55119 Jun 19 '20 at 18:42
  • $\begingroup$ So if I understand correctly ketone while being an intermediate cannot be readily isolated and we should assume the major product of ester+G.reagent will be a 3° alcohol? Unfortunately I didn't have a chance to perform this rx in practice. $\endgroup$ – kiler129 Jun 19 '20 at 19:04
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    $\begingroup$ If you want to make a ketone from a carboxylate group don't use an ester. The standard reagent is a Weinreb amide. $\endgroup$ – Waylander Jun 19 '20 at 19:10
  • $\begingroup$ True, you get the alcohol. Follow @Waylander's lead. Alternatively, the carboxylic acid, six-carbon or greater, will react with 2 equiv. of RLi to give the lithium dialkoxide of the the ketone which, upon aqueous workup, gives the ketone. $\endgroup$ – user55119 Jun 19 '20 at 19:42

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