Write the overall reaction for reaction of methyl benzoate with excess phenyl magnesium brimide in ether, followed by H3O+. ????
1 Answer
Sep 6, 2015
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 about43 , so there is pretty much no chance (1 in10^27.5) 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.