Reaction of NaOH with methanol and ethanol? Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):

$$\ce{2CH3OH + 2Na -&gt; 2CH3O^-Na+ + H2}$$
$$\ce{2CH3CH2OH + 2Na -&gt; 2CH3CH2O^-Na+ + H2}$$

Do methanol and ethanol react with sodium hydroxide ($\ce{NaOH}$) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?
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.