Question

Explain the mechanism of acidic dehydration of ethyl alcohol to form ethene.

Hint:

Acidic dehydration of alcohols is an elimination reaction. Higher temperature favors alkene formation over ether formation.

Answer 1

Step 1: Protonation of alcohol Ethyl alcohol reacts with concentrated \(\text{H}_2\text{SO}_4\) at about \(443\,\text{K}\), and the hydroxyl group gets protonated, forming an oxonium ion: \[ \text{CH}_3\text{CH}_2\text{OH} + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{OH}_2^+ \]
Step 2: Formation of carbocation The protonated alcohol loses a water molecule to form an ethyl carbocation: \[ \text{CH}_3\text{CH}_2\text{OH}_2^+ \rightarrow \text{CH}_3\text{CH}_2^+ + \text{H}_2\text{O} \]
Step 3: Elimination of proton The carbocation loses a proton to form ethene: \[ \text{CH}_3\text{CH}_2^+ \rightarrow \text{CH}_2 = \text{CH}_2 + \text{H}^+ \]
Step 4: Regeneration of acid The proton released in the last step regenerates the acid catalyst.