[Rutherford]

Maybe because in the first conversion you wrote, the intermediate is a hemiketal which are unstable, so the methoxy group leaves the molecule. Anyway, thanks.

[discodermolide]

I never mentioned a hemi-ketal You did.

[Sircodekill]

I'm sorry, but i prefer my hands to draw mechanisms. I don't know if this will help you, but here is the way my teachers taught me.

I put the charges on the grignard to notice the change of the reactivity ( + to -) of the carbon attached.

[discodermolide]

The Grignard does not attack the oxygen of the carbonyl. It attacks the carbonyl carbon, forming a C-C bond and a C-O^-MgBr⁺. The negative charge moves back in to reform the carbonyl and eliminate methoxy. Producing the vinyl ketone. Which then reacts with another mol of Grignard.

I apologise is you drew this but the scan did not come out too well and , for me was difficult to read.

[Rutherford]

Here is the mechanism in the answer. The arrows are confusing, but isn't the second compound a hemiketal?

[discodermolide]

A hemiketal would be the protonated form of your first intermediate. I suppose you could loosely say it is a hemiketal.
The arrows are confusing because they don't point to the correct bits. These arrows coming from the C=C suggest it is these electrons that are attacking, in reality it is the C=C^-.

[Sircodekill]

But it is necessary to form the ketone? I mean if it is possible to get the final product without passing through the ketone form in any step.

The mechanism seems good when you react the molecule with 1 mole of grignard and then another mole, but if you have a huge excess of grignard since beginning will follow the same path?
I don't know

[discodermolide]

I do not think that the benzylic cation is formed. That system's stability is better served when the ketone is an intermediate.

[Sircodekill]

I do not think that the benzylic cation is formed. That system's stability is better served when the ketone is an intermediate.

Ok, that sounds very logical. Never thought in that way. Thanks.

[Rutherford]

I guess that the methoxy is a good leaving group because of the electronegative oxygen. When the carboanion approaches the ester, the electrons are even more pushed to the CH₃O- group, so it leaves the molecule.
It would be then without a hemiketal, and as I look at your mechanism it is first S_N2 and then a nucleophilic addition followed by hydrolysis.