Question

Write mechanism of acid dehydration of ethanol to ethene.

Hint:

Acid dehydration: Protonation ? Loss of water ? Loss of proton ? Alkene. Acid is a catalyst (regenerated at the end).

Answer 1

Step 1: Concept
Acid-catalyzed dehydration of alcohols follows an $E1$ mechanism for primary alcohols involving three steps.

Step 2: Step-by-Step Mechanism

Step 1 -- Protonation of the hydroxyl group:
$CH_3CH_2OH + H^+ \rightarrow CH_3CH_2\overset{+}{O}H_2$ (oxonium ion)
The lone pair on oxygen accepts a proton from the strong acid, converting the poor leaving group ($OH^-$) into an excellent leaving group ($H_2O$).

Step 2 -- Loss of water (C--O bond breaks):
$CH_3CH_2\overset{+}{O}H_2 \rightarrow CH_3\overset{+}{C}H_2 + H_2O$
Water leaves, forming a primary carbocation (for ethanol, this step is driven by heat).

Step 3 -- Loss of proton ($E2$-like elimination):
$CH_3\overset{+}{C}H_2 + H_2O \rightarrow CH_2=CH_2 + H_3O^+$
A base (water molecule) abstracts an $\alpha$-hydrogen, and the electrons shift to form the $\pi$ bond, generating ethene.

Final Answer: Three steps: (1) Protonation of $-$OH by $H^+$. (2) Loss of $H_2O$ forming carbocation. (3) Loss of $H^+$ (by base) to form C=C double bond (ethene). Acid is regenerated.