The compound $A$ on treatment with $N a$ gives $B$ , and with $P C l_{5}$ gives $C . B$ and $C$ react together to give diethyl ether. $A, B$ and $C$ are in the order.

C_{2} H_{5} C l, C_{2} H_{6}, C_{2} H_{5} O H

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C_{2} H_{5} O H, C_{2} H_{6}, C_{2} H_{5} C l

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C_{2} H_{5} O H, C_{2} H_{5} O N a, C_{2} H_{5} C l

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C_{2} H_{5} O H, C_{2} H_{5} C l, C_{2} H_{5} O N a

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Solution

The correct option is C $C_{2} H_{5} O H, C_{2} H_{5} O N a, C_{2} H_{5} C l$
The compound A is ethanol $C_{2} H_{5} O H$ . The compound A on treatment with $N a$ gives sodium ethoxide $(C_{2} H_{5} O N a)$ which is compound B.

$C_{2} H_{5} O H N a - - \to - H_{2} C_{2} H_{5} O N a$

The compound A on treatment with $P C l_{5}$ gives ethyl chloride $(C_{2} H_{5} - C l)$ which is compound C.

$C_{2} H_{5} O H P C l_{5} - -- \to C_{2} H_{5} - C l + H C l + P O C l_{3}$

B and C react together to give diethyl ether.

$C_{2} H_{5} O N a + C_{2} H_{5} C l \to C_{2} H_{5} O C_{2} H_{5} + N a C l$
A,B and C are in the order $C_{2} H_{5} O H, C_{2} H_{5} O N a, C_{2} H_{5} C l$ .

Option C is correct.

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