Question

The Compound A on treatment with Na gives B, and with ${\text{PC}}{{\text{l}}_{\text{5}}}$ gives C. B and C react together to give diethyl ether. A, B and C are in the order.
A.${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl , }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{ , }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$
B.${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH , }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{ , }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}$
C.${\text{}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH ,}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa , }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}$
D.${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH ,}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl , }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$

Answer 1

To solve this question the knowledge of alcohol is required. Any alcohol would react with alkali metals to form salt of the form $\left( {{\text{ROM}}} \right)$ where R is the alkyl group, O is Oxygen of alcohol and M is the alkali metal. This salt when it reacts with ${\text{PC}}{{\text{l}}_{\text{5}}}$ , the $\left( { - {\text{ONa}}} \right)$ group is replaced by the chloride group.

Complete step by step answer:
The reaction given in the question is Williamson synthesis reaction. In the reaction, an alcohol reacts with an alkali metal to form a salt. The reaction goes as follows:
${\text{ROH}} + {\text{Na}} \to {\text{RONa}}$
As the R group given in the question is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$ , the reaction will be:
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}} + {\text{Na}} \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$
When an alcohol reacts with ${\text{PC}}{{\text{l}}_{\text{5}}}$ , the alcohol group is replaced by halide and it forms an alkyl halide. The reaction goes as follows:
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}} + {\text{PC}}{{\text{l}}_5} \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}$
The sodium salt of the ethyl alcohol and ethyl chloride reacts together to form diethyl ether.
Thus, A is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ , B is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$ and C is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}$ .

Hence, the correct option is option C.

Note:
Alcohols react with sodium metal or other alkali metals due to the difference in electronegativities between the alkyl group and Oxygen atom to form alkoxides, releasing Hydrogen in the process.
The alkoxide ion is a very good nucleophile and reacts easily with chlorinating agents like ${\text{PC}}{{\text{l}}_5}$ to form alkyl halide and this reaction is used in different organic transformations.
In the Williamson ether synthesis, an alkoxide, which is a nucleophile, reacts with an alkyl halide by an ${{\text{S}}_{\text{N}}}2$ substitution reaction to form the ether.