The organic compound A of molecular formula $C_{2} H_{6} O$ liberates hydrogen gas with sodium metal. A gives B of formula $C_{4} H_{10} O$ , when it reacts with concentrated $H_{2} S O_{4}$ at 413K. At 443K with concentrated $H_{2} S O_{4}$ A gives compound C of formula $C_{2} H_{4}$ . This compound C decolourises bromine water. What are A, B, and C?

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Solution

$A : C H_{3} C H_{2} - O H$
$B : C H_{3} - C H_{2} - O - C H_{2} - C H_{3}$
$C : C H_{2} = C H_{2}$

$C H_{3} C H_{2} - O H + N a ⟶ C H_{3} C H_{2} - 0 N a + H_{2} ↑$

$H_{2} S O_{4}$
$C H_{8} C H_{2} - O H ⟶ C H_{3} C H_{2} - O - C H_{2} C H_{3}$
$413$ K

$H_{2} S O_{4}$
$C H_{3} C H_{2} - O H ⟶ C H_{2} = C H_{2}$
$443$ K

$B r_{2}$
$C H_{2} C H_{2} ⟶$ Bromine water decolourises.

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