Write the mechanism of the acid dehydration of ethanol to yield ethene.
(N/A) The dehydration of ethanol to ethene proceeds via a $\beta$-elimination mechanism in three steps:
Step-$1$: Protonation of ethanol to form ethyl oxonium ion.
$CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+$
Step-$2$: Formation of carbocation. This is the slow,rate-determining step where the $C-O$ bond breaks and water is eliminated.
$CH_3CH_2OH_2^+ \rightleftharpoons CH_3CH_2^+ + H_2O$
Step-$3$: Formation of ethene by elimination of a proton. The carbocation loses a $\beta$-proton to form ethene.
$CH_3CH_2^+ \rightleftharpoons CH_2=CH_2 + H^+$
Step-$1$: Protonation of ethanol to form ethyl oxonium ion.
$CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+$
Step-$2$: Formation of carbocation. This is the slow,rate-determining step where the $C-O$ bond breaks and water is eliminated.
$CH_3CH_2OH_2^+ \rightleftharpoons CH_3CH_2^+ + H_2O$
Step-$3$: Formation of ethene by elimination of a proton. The carbocation loses a $\beta$-proton to form ethene.
$CH_3CH_2^+ \rightleftharpoons CH_2=CH_2 + H^+$
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