[Squeak1107]

For the reaction attaached i was thinking of attacking the carbon with a Gringyard R'-MgCl and the forcing the NMe2 group off. However I know that NMe2 is a very poor leaving group so I'm not sure how to go about this reaction

[spirochete]

I'm not 100% sure how this works. But it's possible it forms a tetrahedral intermediate that is trapped with a negative charge on oxygen. Then in acid work up it collapses to make the ketone. This would also explain the selectivity for only forming ketone and not alcohol with a second addition of grignard reagent.

NMe2 anion can be a leaving group in extreme cases, though, during the elimination of anionic tetrahedral intermediates.

Careful you are not proposing Sn2 at an sp2 carbon, by the way. I'm pretty sure you aren't, but I wanted to be clear on that.

[AWK]

Term: leaving group concerns substitution reactions.  Your mechanism is addition followed by reaction with water and the elimination of amine.

[Squeak1107]

So if i understand i get to this intermediate and then react with water. Would the nitrogen donate a pair of electrons to become NHMe₂⁺ and then when the oxygen reforms the double bond is forced off?

[OCSaviour]

Did you try other methods? Hint: Acylation. Now, think your way back. Keep acylation in mind, and think of the products preceding the acylated product.

[spirochete]

So if i understand i get to this intermediate and then react with water. Would the nitrogen donate a pair of electrons to become NHMe₂⁺ and then when the oxygen reforms the double bond is forced off?

In dilute acid/water both O- and nitrogen could be protonated, then the tetrahedral intermediate collapses (eliminates) to make C=O, then loss of H+ from carbonyl. In neutral water it's harder to say what the exact sequence of protonations would be.

AWK were you addressing me? You are right I could have thrown in the term "addition" there.