Write the mechanism of dehydration of ethyl alcohol with conc. H$_2$SO$_4$ at 413 K.
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Solution and Explanation
1. Protonation of ethanol (CH$_3$CH$_2$OH) to form an ethyl oxonium ion (CH$_3$CH$_2$OH$_2^+$).
2. Loss of water (H$_2$O) from the ethyl oxonium ion to form a carbocation (CH$_3$CH$_2^+$).
3. Rearrangement, if necessary, to stabilize the carbocation (in this case, no rearrangement is needed for ethyl alcohol).
4. The carbocation then loses a proton (H$^+$) to form ethene (CH$_2$=CH$_2$).
Thus, the overall reaction is:
\[ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 413 \, \text{K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \]
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