Question

Write the mechanism of acid dehydration of ethanol to yield ethene.

Hint:

At a lower temperature (\( 413 \, K \)), the same reaction yields ethoxyethane (ether) instead of ethene.

Answer 1

Step 1: Protonate the alcohol oxygen.
Concentrated $H_2SO_4$ provides a proton that the oxygen lone pair of ethanol readily accepts, converting the poor leaving group $-OH$ into a much better one, giving a protonated oxonium intermediate.
\[ CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2\overset{+}{O}H_2 \]
Step 2: Kick out water to form the carbocation, the slow step.
The carbon-oxygen bond then breaks heterolytically, water leaves as a neutral, stable molecule, and this generates an ethyl carbocation. Being the step that takes the most energy to get through, this is the rate determining step of the whole mechanism.
\[ CH_3CH_2\overset{+}{O}H_2 \xrightarrow{\text{slow}} CH_3\overset{+}{C}H_2 + H_2O \]
Step 3: Pull off a beta-hydrogen to complete the elimination.
A base in the medium, such as $HSO_4^-$ or another water molecule, plucks a proton off the carbon next to the positive centre. The electron pair left behind forms the new pi bond, giving ethene and regenerating the acid catalyst so it is free to act again.
\[ CH_3\overset{+}{C}H_2 \rightarrow CH_2=CH_2 + H^+ \]
\[ \boxed{\text{Ethanol} \xrightarrow{conc.\ H_2SO_4,\ 443\ K} \text{Ethene, via protonation, carbocation formation, then beta-elimination (E1)}} \]