Now as H2SO4 will dehydrate Ethanol:
2 C2H5OH ====> C2H5OC2H5 + H2O
140 deg
upon substituting NaHSO4 for H2SO4 assuming its similarly dehydrates as:
NaHSO4 + H2O <===> NaOH + H2SO4
then:
2 C2H5OH ====> C2H5OC2H5 + H2O
x deg
or diethyl ether which is colorless and boils at 34.6 C. This is my best guess as I am not much of an organic chemist.
Per wiki:
"Ethanol is mixed with a strong acid, typically sulfuric acid, H2SO4. The acid dissociates in the aqueous environment producing hydronium ions, H3O+.
A hydrogen ion protonates the electronegative oxygen atom of the ethanol, giving the ethanol molecule a positive charge:
CH3CH2OH + H3O+ → CH3CH2OH2+ + H2O
A nucleophilic oxygen atom of unprotonated ethanol displaces a water molecule from the protonated (electrophilic) ethanol molecule, producing water, a
hydrogen ion and diethyl ether.
CH3CH2OH2+ + CH3CH2OH → H2O + H+ + CH3CH2OCH2CH3
This reaction must be carried out at temperatures lower than 150 °C in order to ensure that an elimination product (ethylene) is not a product of the
reaction. At higher temperatures, ethanol will dehydrate to form ethylene. The reaction to make diethyl ether is reversible, so eventually an
equilibrium between reactants and products is achieved. Getting a good yield of ether requires that ether be distilled out of the reaction mixture
before it reverts to ethanol, taking advantage of Le Chatelier's principle."
[Edited on 2-2-2012 by AJKOER] |