Reaction of NaOH with methanol and ethanol?
The proposed reaction:

$$\ce{CH3OH(l) + NaOH(s) &lt;&lt;=&gt; CH3O^-Na+(s) + H2O(l)}$$

According to wikipedia:


  The solid hydrolyzes in water to give methanol and sodium hydroxide


The equilibrium is biased to the left, as @vapid has mentioned in the comments. To quote him:


  In the large excess of alcohol, the reaction will be considerably shifted towards alkoxide.