Question

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

• A. $C _{2} H _{5} OH , C _{2} H _{5} ONa , C _{2} H _{5} Cl$
• B. $C _{2} H _{5} OH , C _{2} H _{6}, C _{2} H _{5} Cl$
• C. $C _{2} H _{5} Cl , C _{2} H _{6}, C _{2} H _{5} OH$
• D. $C _{2} H _{5} OH , C _{2} H _{5} Cl , C _{2} H _{5} ONa$

Answer 1

The Correct Option is A

To solve this question, let's analyze the reactions of compound A based on the given information and deduce the correct order of A, B, and C.

1. Compound A reacts with Na to form compound B. This typically indicates a reaction with an alcohol, as alcohol reacts with sodium to form an alkoxide and hydrogen gas. Therefore, A is likely ethanol, C_2H_5OH, and B would be sodium ethoxide, C_2H_5ONa.
2. Compound A also reacts with PCl_5 to form compound C. Reactions of PCl_5 with alcohols typically yield alkyl chlorides and phosphorus oxychloride. Thus, C would be ethyl chloride, C_2H_5Cl.
3. According to the information, compounds B and C react to produce diethyl ether, (C_2H_5)_2O. This reaction is consistent with the Williamson ether synthesis, where an alkoxide ion (from C_2H_5ONa) reacts with an alkyl halide (C_2H_5Cl) to form an ether.

From this analysis, we can conclude that the correct order of A, B, and C is: C_2H_5OH, C_2H_5ONa, C_2H_5Cl.

The correct answer is thus the option: C_2H_5OH, C_2H_5ONa, C_2H_5Cl.