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The compound A on treatment with Na gives B, and with $PCl_5$ gives C. B and C react together to give diethyl ether. A, B and C are in the order
Explanation
Sodium ethoxide + ethyl chloride gives diethyl ether by $S_N2$.
Detailed Solution
$C_2H_5OH\ (A) \xrightarrow{Na} C_2H_5O^-Na^+\ (B)$
$C_2H_5OH\ (A) \xrightarrow{PCl_5} C_2H_5Cl\ (C)$
$C_2H_5O^-Na^+\ (B) + C_2H_5Cl\ (C) \xrightarrow{S_N2} C_2H_5OC_2H_5$
So A, B and C are $C_2H_5OH$, $C_2H_5ONa$ and $C_2H_5Cl$.
$C_2H_5OH\ (A) \xrightarrow{PCl_5} C_2H_5Cl\ (C)$
$C_2H_5O^-Na^+\ (B) + C_2H_5Cl\ (C) \xrightarrow{S_N2} C_2H_5OC_2H_5$
So A, B and C are $C_2H_5OH$, $C_2H_5ONa$ and $C_2H_5Cl$.