Question

In the above reaction product is

(A) \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{O}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}\]
(B) \[{{\rm{C}}_2}{{\rm{H}}_{\rm{5}}}{\rm{O}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}\]
(C) \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{O}}{{\rm{C}}_6}{{\rm{H}}_{\rm{5}}}\]
(D) \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{I}}\]

Answer 1

Hint: An ether is a functional group in which two alkyl groups are attached to the oxygen atom. In a chemical reaction, ether is written as R-O-R. There are many methods of preparation for ether. One such method is Williamson ether synthesis.

Complete Step by Step Solution:
Let's first understand what a Williamson ether synthesis is. In this reaction, deprotonated alcohol undergoes a reaction with an alkyl halide to give ether as a product.
Here, also the reaction is a Williamson ether synthesis. The first step of the reaction is the deprotonation of phenol to form a phenoxide ion.

Fig: Deprotonation of phenoxide ion

In the next step, the electrophilic carbon of the ethyl chloride is attacked by the phenoxide ion and the iodine separates.

Fig: Nucleophilic attack of phenoxide ion
The product formed is \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{O}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}\] and the hydrogen iodide separates as a by-product.
Therefore, option A is right.

Additional Information: Let's discuss some properties of ether. It is a liquid whose nature is volatile. It has no colour and a pleasant odour. The boiling of ether is comparable with alkanes but less than alcohols. The solubility of ether in water is because of the hydrogen bond formation by the oxygen of ether with the water molecules. But there is a decrease in solubility when the hydrocarbon chain increases.

Note: Ether is a very important compound in many industries. Its use in the manufacturing of hydrocarbons, gums, oil, dyes, resins, and paints is significant. They are also used in lubricating oils. The nature of ether is toxic.