Question

Explain the mechanism of the following reaction:
$C{H}_3-C{H}_2-OH\underset{443K}{\stackrel{H^{+}}{}}C{H}_2=C{H}_2+H_2^{-}O$

Answer 1

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 $(E^2$ mechanism) while secondary and tertiary alcohols undergo unimolecular elimination ($E^1$ mechanism). Oxygen donates two electrons to a proton from sulfuric acid $H_{2}S{O}_4$, forming an alkyloxonium ion. Then the nucleophile $HS{O}_4^{–}$ back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond.