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I was given the following mechanism to prove today post-lecture:

enter image description here

and my attempt is given below.

Now, when the negative charge on the alkene attacks the carbonyl, the electrons are pushed onto the oxygen - which is understandable. My question is: why do these electrons come back down, rather than just be protonated to give an alcohol?

my attempt

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    $\begingroup$ This is the Weinreb "amide". In your first rxn., Mg is bound to both oxygens. The complex is stable until aqueous workup causes decomposition to the ketone. Your proposal would cause the ketone to react with the Grignard reagent. Ketones are more reactive than amides. $\endgroup$ – user55119 Jun 11 at 2:14
  • $\begingroup$ Why don't ketones form geminal diols in water? colby.edu/chemistry/CH242/15-2.pdf $\endgroup$ – Karsten Theis Jun 11 at 22:50
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    $\begingroup$ @KarstenTheis perfect if you can add that as the answer then I can accept it as the answer! $\endgroup$ – vik1245 Jun 12 at 16:50
  • $\begingroup$ @Karsten Theis: Not at all. I appreciate your concern and thank you for doing the right thing. Cheers! :-) $\endgroup$ – Mathew Mahindaratne Jun 12 at 19:02
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Nucleophilic acetyl substitution is common among carboxylic acid, acid chlorides, anhydrides, esters, and amides.

Two very common examples are anhydride decomposition and esterification of carboxylic acids. For example, if you look at esterification, carboxylic acid and alcohol first make an addition, giving tetrahedral intermediates 1 and 2. Intermediate 2 then collapses and displaces the water molecule as a leaving group to give the corresponding ester (see the scheme).

Ester mechanism

Your substrate is an amide. Like all other carboxylic acid derivatives, this molecule also undergoes a nucleophilic acetyl substitution with Grignard reagent, which is a strong nucleophile. The polarized carbonyl group has partially positive carbon, which is a good electrophile.

The nucleophilic carbon on Grignard reagent attaches to carbonyl carbon and gives tetrahedral intermediate as a metal alkoxide complex. But, this alkoxide can collapse to reform strong C=O bond, displacing the amino portion of the amide as a leaving group, in the form of the amide ion, $\ce{R1R2N-}$. This produces the ketone as a final product, assuming you have used only one equivalent of Grignard reagent (Channel reactions of Grignard reagent with esters).

Why does the reaction continue after the tetrahedral carbon forms?

After nucleophilic attack on a carbonyl, the resulting tetrahedral species either gets protonated to form the final addition product, or it is an intermediate on the way to the substitution product. This depends on the type of reaction. For example, a geminal diol will usually dehydrate to form a ketone. There is a nice overview of the different scenarios, including some nitrogen chemistry, in this document.

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