Write the overall reaction for reaction of methyl benzoate with excess phenyl magnesium brimide in ether, followed by H3O+. ????

The mechanism for this is something like this:

1. The magnesium bromide acts as a Lewis acid, turning the phenyl group into a nucleophile. Naturally, it backside-attacks the most susceptible electrophilic center, forming a fairly sterically-hindered tetrahedral intermediate. It is excess reactant so that it happens to a reasonable yield.
2. The negatively-charged oxygen grabs a proton from the acid that you add afterwards.
3. Electron conjugation from the newly-formed hydroxyl group's oxygen forces the methoxy leaving group to break off, a rapid intramolecular process. The pKa of methanol is about #15.5#, but the pKa of benzene is about #43#, so there is pretty much no chance (#1# in #10^27.5)# of the phenyl group breaking back off.
4. The reaction finishes when the methanolate grabs the proton off of the protonated final product. The methanolate does so rather than the water (#10^6.1# times as often), since the pKa of methanol (#~15.5#) is greater than the pKa of hydronium (#~9.4#), and acid-base equilibrium lies on the side of the weaker acid or stronger base.
5. Repeat the curved arrows and electron flow in steps 1 and 2 with another equivalent of phenyl magnesium bromide.

   Your final product is triphenylmethanol.