Why is 10% KOH(aq) used to extract the organic layer during the reaction work up for the reaction involving potassium tert-butoxide/tert-buthanol? Explain your answer. Hint: What is the pKa of tert-butanol?
10% KOH(aq) used to extract the organic layer during work up for the reaction involving potassium tert-butoxide/tert-buthanol because it makes the solution more basic.
pKa of teritiary butanol is around 18 and it's conjugate salt is potassium tert-butoxide.
When ter-butanol is treated base like KOH then it forms ter-butoxide, an alkoxide ion. Ter-butanol is easily gets oxidised and loses its acidic proton. In order to make it more basic. Potassium hydroxide is added to it.
ter-Bt-OH + KOH -------> ter-But- O- K+ + H2O
This shows that Potassium ter-butoxide formed from the tert-butanol and potassium abstracts the acidic protons from the substrates. This is strong and non nucleophilic base due to steric hinderance. This helps in salting out the main substituents in the organic layer
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