This is taken from the patent "New method for the preparation of erlotinib

EP 2433931 A1". I have some question regarding one of the reaction in the patent, attached is the chemical reaction.

"In a 500 ml flask provided with a thermometer, Dean-Stark apparatus and a drip funnel there are introduced - in succession at 25°C and under nitrogen atmosphere - 30 g of 4-(3-aminophenyl)-2-methyl-3-butyn-2-ol of formula (IV), 2.2 g of caustic soda and 240 mL of Toluene.

The mixture is heated to the reflux temperature (105-110°C) for 4 hrs, collecting the distillate in the Dean-Stark apparatus. Fresh toluene is simultaneously added to the reaction mixture so as to maintain a minimum volume of 180 ml (6 vol) within the flask. This allows removing acetone from the reaction mixture. After the 4 hours, it is stirred at reflux until the reaction is completed, controlling the trend thereof by means of TLC.

Upon completion of the reaction it is cooled to 25 °C and a mixture made up of 20 mL of Toluene and 250 mg of dicalite is added. The reaction mixture is stirred for 15 min. The reaction mixture is filtered and the solid is washed using 15 mL of Toluene. The filtrates are combined and the resulting solution is concentrated at 40°C under vacuum up to residue. The product is obtained as oil (20.7 g) with quantitative yield"

Firstly, isn't NaOH insoluble in organic solvents like toluene?

Secondly: I understand acetone is released upon the cleavage. What will happen to NaOH in the toluene after reaction? Does it become a salt or remains as NaOH. I'm assuming the NaOH/toluene solution will still remain as it is without any chemical conversion.

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