We did a Williamson Ether Synthesis in lab last week and I am confused about one of the steps.
We reacted an o-cresol with KOH... the results being the conjugate base of the cresol to become soluble in H20, K+, H20, and an excess of KOH... chloroacetic acid was added, and an acid base reaction happens between the KOH and chloroacetic acid.... Sn2 reaction happens between conjugate cresol base and deprotonated chloroacetic acid... making the o-methylphenoxyacetic acid.. a final addition of HCl neutralizes the species for separation..
What would be the consequence of added only 1 mole equivalent of KOH instead of 4?
Is the excess KOH only needed to deprotonate the chloroacetic acid?
Why does the chloroacetic acid even need to be deprotonated?
Thanks for all your *delete me*