hey everyone, im having a hard time trying to figure out the products from the following reaction:
You have benzyl chloride and treat it with Cl2 at 145°C. I know this is going to form Ph-CH-Cl2.
Then it says that you treat the previous product with NaOH 5% in acetone and water, and this is where im stuck, since i dont understand what type of reaction would happen there, i mean, i think -OH would act as a nucleophile and attack the CH-Cl2 Carbon via SN2, but that would give Ph-CH(OH)Cl as a product and i dont know if this is right, so i need you to please tell me what happens when you have a good nucleophile but it is going to attack a Carbon with 2 Cl´s attached to it. Is my prediction correct? Is it attacked twice and that produces a Diol?... i would also appreciate if you could explain a little bit about the solvent used in this reaction, since i dont really understand the low NaOH concentration and that acetone-water mixture they chose.