Methanol (CHX3OH) and ethanol (CHX3CHX2OH) react with sodium metal (Na) to form sodium methoxide (CHX3OX−NaX+) and sodium ethoxide (CHX3CHX2OX−NaX+):
2CHX3OH+2Na⟶2CHX3OX−NaX++HX2
Do methanol and ethanol react with sodium hydroxide (NaOH) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?
Answer
The proposed reaction:
CHX3OH(l)+NaOH(s)−⇀↽−CHX3OX−NaX+(s)+HX2O(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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