Explain the mechanism of the following reaction.
$C_{2} H_{5} O H C o n c . H_{2} S O_{4} - ------- \to 443 K C H_{2} = C H_{2} + H_{2} O$

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Solution

$C_{2} H_{5} - O H Ethanol Conc. H_{2} S O_{4} - ------- \to 443 K C H_{2} = C H_{2} Ethene + H_{2} O$
The given reaction is a dehydration reaction which turns ethanol $(C H_{3} C H_{2} O H)$ into ethene $(C_{2} H_{4})$ . The concentrated $H_{2} S O_{4}$ is added as a catalyst. The acidic environment enables the alcohol group $(- O H)$ to be more easily removed from the ethanol to form ethene.

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2 C H_{3} - C H_{2} - O H H^{+} - -- \to 413 K C H_{3} - C H_{2} - O - C H_{2} - C H_{3} + H_{2} O

C H_{3} - C H_{2} - O H H^{+} - -- \to 443 K C H_{2} = C H_{2} + H_{2} O

C H_{2} = C H_{2} H_{3} O^{+} - -- \to C H_{3} - C H_{2}^{+} + H_{2} O

C H_{3} - C H_{2} - O H H^{+} - - \to 443 K C H_{2} = C H_{2} + H_{2}^{-} O

C H_{2} = C H_{2} + H B r \to C H_{3} - C H_{2} - B r

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