Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):
$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$ $$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$
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?
Answer
The proposed reaction:
$$\ce{CH3OH(l) + NaOH(s) <<=> CH3O^-Na+(s) + H2O(l)}$$
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.
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