To a solution of dimethyl isophthalate B2 (60 g, 0.31 mol) in 500 mL MeOH was added a solution of NaOH (12.4 g, 0.31 mol) in 200 mL MeOH. The mixture was stirred at room temperature overnight. It was concentrated and the residue was dissolved in 500 mL of water and extracted with Et2O. The aqueous solution was acidified with concentrated HCl. The white precipitate formed in the HCl solution at pH = 2 was collected and dried under vacuum to give 46 g of the crude compound monomethyl isophthalate as a white solid.
