Write the mechanism of acid dehydration of ethanol to yield ethene.
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Solution and Explanation
• Dehydration of ethanol to ethene occurs in the presence of concentrated \( H_2SO_4 \) at \( 443 \, K \).
• It is an elimination reaction proceeding through a carbocation intermediate (\( E1 \) mechanism).
Step 1: Protonation of alcohol
The ethanol molecule acts as a base and accepts a proton from the acid to form an oxonium ion. \[ CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+ \]
Step 2: Formation of carbocation
This is the slowest step (RDS). The \( C-O \) bond breaks, and a water molecule is eliminated, leaving behind an ethyl carbocation. \[ CH_3CH_2OH_2^+ \xrightarrow{\text{slow}} CH_3CH_2^+ + H_2O \]
Step 3: Elimination of a proton
A proton is removed from the beta-carbon (the methyl group) by a base (like \( HSO_4^- \) or \( H_2O \)) to form the carbon-carbon double bond. \[ CH_3CH_2^+ \rightarrow CH_2=CH_2 + H^+ \] The acid catalyst is regenerated. The final answer is the three-step \( E1 \) mechanism.
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