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:

  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.