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I have read that ROH+ NaOH doesn't form alkoxide ions because water is a stronger acid than all alcohols except methanol. However while reading xanthate test I found that alcohol reacts with KOH to form alkoxide ions. Why is it so?

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The reaction of alcohol + NaOH producing $\ce{C2H5ONa}$ + water is an equilibrium which is usually strongly pushed to the left hand side. But in the presence of $\ce{CS2}$, the alkoxide $\ce{C2H5ONa}$ is transformed into xanthate. So the equilibrium is driven to the right hand side.

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  • $\begingroup$ And the same goes with the elimination reactions in which alcoholic KOH is used for alkene formation from alkyl halide? $\endgroup$ – Gaurav Kumar May 21 at 5:09

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