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

Question

12765 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-06T04:04:15

The mechanism for this is something like this:

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.
The negatively-charged oxygen grabs a proton from the acid that you add afterwards.
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.
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.
Repeat the curved arrows and electron flow in steps 1 and 2 with another equivalent of phenyl magnesium bromide.

Your final product is triphenylmethanol.