Thursday, September 20, 2018

organic chemistry - 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 ($\ce{CH3CH2O^-Na+}$):


$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$ $$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$


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.




No comments:

Post a Comment

periodic trends - Comparing radii in lithium, beryllium, magnesium, aluminium and sodium ions

Apparently the of last four, $\ce{Mg^2+}$ is closest in radius to $\ce{Li+}$. Is this true, and if so, why would a whole larger shell ($\ce{...