How does ethyl alcohol react with the following? Write equations.
(i) Metallic $N a$
(ii) $C H_{3} C O O H$
(iii) $C H_{3} M g I$
(iv) $C o n c . H_{2} S O_{4}$ at $170^{\circ} C$

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Solution

(i) Metallic Na:
Ethyl alcohol react with metallic Na to form sodium ethoxide and hydrogen gas.
$C_{2} H_{5} - O H + N a \to C_{2} H_{5} - O N a + \frac{1}{2} H_{2} ↑$
(ii) $C H_{3} C O O H$ :
Ethyl alcohol react with acetic acid to form ethyl acetate.
$C_{2} H_{5} - O H + C H_{3} C O O H ⇌ H^{+} C H_{3} - C | | O O - C_{2} H_{5} + H_{2} O$
(iii) $C H_{3} M g I$ :
Ethyl alcohol react with methyl magnesium iodide (grignard reagent) to form methane
$C_{2} H_{5} - O H + C H_{3} M g I \to C H_{4} + C_{2} H_{5} - O M g I$
(iv) Conc. $H_{2} S O_{4}$ at 170 $^{o}$ C:
Ethyl alcohol react with conc. $H_{2} S O_{4}$ at 170 $^{o}$ C to form ethylene via dehydration
$C_{2} H_{5} - O H 95 % H_{2} S O_{4} - ------ \to 170^{o} C C H_{2} = C H_{2}$

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