Explain the mechanism of the following reaction: CH3−CH2−OH + H+ → CH2=CH2 + H2O

The –OH group in the alcohol donates two electrons to H+ from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the nucleophile) then attacks the hydrogen adjacent to the carbocation and form a double bond. Primary alcohols undergo bimolecular elimination (E2 mechanism) while secondary and tertiary alcohols undergo unimolecular elimination (E1 mechanism). Oxygen donates two electrons to a proton from sulfuric acid H2SO4, forming an alkyloxonium ion. Then the nucleophile HSO−4 back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.