[TheAlbatross58]

I am having trouble with a homework problem assigned to me that is asking me to write the mechanism for the reaction of a two ring lactone with a grignard reagent.



reacts with CH₃MgBr to make the product:



All I have so far is that the CH₃^- adds to the carbon that is double bonded with the oxygen and the pi bond from the CO double bond moves to the oxygen.  Can someone help me or guide me in the right direction?

[discodermolide]

How will this lactone react with a Grignard reagent.
Try writing out the reaction with cyclic lactone  like tetrahydro-2H-pyran-2-one and you may see.

[orgopete]

Hint, the final product is an aldol condensation product of a methyl ketone and a cyclohexanone.

[TheAlbatross58]

Hint, the final product is an aldol condensation product of a methyl ketone and a cyclohexanone.

That helps a lot I just cant seem to find a way to get past this stage:



I added the CH₃^- to the C=O carbon but there just seem to be too many electrons in there and I am really confused.

[discodermolide]

Move the negative charge back towards the ring to re-form a carbonyl C=O, and at the same time break the bond linking the ring oxygen to the new formed carbonyl and see what you get.

[TheAlbatross58]

I got that far but I don't see how the newly formed O^- will turn into the other ketone with the double bond being right there. Maybe I'm just not thinking clearly.

[orgopete]

You can use water as it is present from the aldol condensation.

[discodermolide]

I got that far but I don't see how the newly formed O^- will turn into the other ketone with the double bond being right there. Maybe I'm just not thinking clearly.

When you do this electron movement you get an enol, which is nothing other than a carbonyl in disguise.
See picture.
Now you can perhaps understand Orgopete's first answer.