Question

chcch2ch2oh reacted with with NaH and ch3i both excess? what is the mechanism and the expected products ?

Answer 1

Given below is a possible mechanism for the reaction of alkynol with base NaH and electrophilic reagent CH3I. the first step is deprotonation of terminal alkyne and alcoholic -OH by base NaH. 2 moles of H2 gas is formed. The carbanion and oxoanion thus formed then attacks on the electrophilic end of CH3I, to form a dialkylated product shown below. It is important to remember here that although both the deprotonation has been shown in one step, they may not actually occur at the same time in the real experimental conditions. However the product would remain to be the same.