Question

The compound $A$ on treatment with $Na$ gives $B$ and with $PC{l}_5$ gives $C.$ $B$ and $C$ react together to give diethyl ether. $A,B$ and $C$ are in the order:

• A. $C_2{H}_{5}Cl,C_2{H}_6,C_2{H}_{5}OH$
  
• B. $C_2{H}_{5}OH,C_2{H}_6,C_2{H}_{5}Cl$
  
• C. $C_2{H}_{5}OH,C_2{H}_{5}ONa,C_2{H}_{5}Cl$
  
• D. $C_2{H}_{5}OH,C_2{H}_{5}Cl,C_2{H}_{5}ONa$
  

Answer 1

The correct option is C $C_2{H}_{5}OH,C_2{H}_{5}ONa,C_2{H}_{5}Cl$


$\underset{A}{C_2{H}_{5}OH}+Na\to \underset{B}{C_2{H}_{5}ONa}+\frac{1}{2}{H}_2$

$\underset{A}{C_2{H}_{5}OH}+PC{l}_5\to \underset{C}{C_2{H}_{5}Cl}+POC{l}_3+HCl$

$\underset{C}{C_2{H}_{5}Cl}+\underset{B}{C_2{H}_{5}ONa}\to \underset{\text{diethyl ether}}{C_2{H}_{5}O{C}_2{H}_5}+NaCl$

Hence, option (c) is the correct answer.