The mechanism for the dehydration of ethanol (CH3CH2OH) to ethene (CH2=CH2) under acidic conditions involves the following steps:
Step 1: Formation of protonated alcohol (Fast step)
Ethanol (CH3CH2OH) reacts with the proton (H+) to form the protonated alcohol (ethyl oxonium ion).
CH3CH2OH + H+ → CH3CH2OH+
Step 2: Formation of carbocation (Slow step)
The protonated alcohol undergoes the loss of water to form a carbocation, which is the rate-determining step.
CH3CH2OH+ → CH3C+ + H2O
Step 3: Formation of ethene by elimination of a proton (Fast step)
The carbocation formed in Step 2 loses a proton to form ethene.
Overall reaction:
CH3C+CH2 → C2H4 (Ethene) + H+