Can you explain why is bimolecular dehydration not appropriate for the preparation of ethyl methyl ether?
Answer:
"Bimolecular dehydration" is a way of preparing diethyl ether from the dehydrating action of concentrated sulfuric acid on ethanol. The dehydrating agent, conc. H2SO4 abstracts a hydrogen from one ethanol and a hydroxyl (the -OH group) of a neighbor. Those radicals formed, join together. Sulfuric acid grabs a mole of water.
So this method works only with symmetrical ethers: di[alkyl] ethers. Methyl Ethyl ether is better prepared from Sodium ethoxide and methanol or Sodium methoxide and ethanol.
Random dehydration is unselective. You will also get MeOMe and EtOEt in statistical proportion less kinetic (steric) and thermodynamic effects.
The problem is if you put methanol and ethanol together, your going to get some methanol-methanol reactions and some ethanol-ethanol reactions. Not all of it is going to be the desired methanol-ethanol reaction. So it will be a low yield with lots of waste.
I guess it all depends on how you are preparing it!
If you preparing the ether by acid catalyzed intermolecular dehydration of an alcohol:
CH3CH2OH + H2SO4 --> CH3CH2OH2+
CH3CH2OH+ + CH3OH --> CH3CH2OCH3
then the mechanism WILL be bimolecular dehydration, as well as some unimolecular dehydration to form the intermediate carbocation and then reaction with either ethanol or methanol to form product.
If you are reacting by the Williamson ether synthesis
CH3CH2OH + Na --> CH3CH2O-Na+
CH3CH2O-Na+ CH3Br --> CH3CH2OCH3
Then there is NO dehydration and the ONLY product will be the ether.
lose water
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