Question

Diethyl ether when refluxed with excess of $ {\text{HI}} $ gives two molecules of $ \left( {\text{i}} \right) $ . Ethers can be most commonly prepared by reaction of $ \left( {{\text{ii}}} \right) $ and $ \left( {{\text{iii}}} \right) $ . The method is called $ \left( {{\text{iv}}} \right) $ .
 $ \left( {\text{i}} \right){\text{,}}\,\left( {{\text{ii}}} \right){\text{,}}\,\left( {{\text{iii}}} \right) $ and $ \left( {{\text{iv}}} \right) $ respectively are?
(A) ethyl iodide, sodium alkoxide, alkyl halide, Williamson’s synthesis
(B) ethanol, alcohol, alkyl halide, substitution
(C) methyl iodide, Grignard’s reagent, alkyl halide, Williamson’s synthesis
(D) ethyl iodide, phenol, ethyl iodide, esterification

Answer 1

Hint: An ether in which the oxygen atom is linked to two ethyl groups is known as diethyl ether. It acts as an anesthetic for inhalation, a non-polar solvent, and a refrigerant. It's an ether and a volatile organic compound.

Complete answer:
Alcohols and alkyl halides can be formed when ethers are treated with strong acid in the presence of a nucleophile. If the ether is attached to a primary carbon, an $ {{\text{S}}_{\text{N}}}{\text{2}} $ pathway may be used.
When Diethyl ether $ \left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{O}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right) $ is refluxed with $ {\text{HI}} $ , it produces two molecules: \[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{I}}\,\,{\text{ + }}\,\,{\text{NaO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\]. In the presence of too much HI, the alkoxide undergoes a substitution reaction, yielding $ {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{I}} $ alkyl halide. As a result, $ \left( {\text{i}} \right)\,\,{\text{ = }}\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{I}} $ .
Williamson's synthesis, which involves a reaction between an alkyl halide and sodium alkoxide, is the most common way to make ethers. As a result, $ \left( {{\text{ii}}} \right) $ and $ \left( {{\text{iii}}} \right) $ are alkyl halide and sodium alkoxide, respectively. Williamson's synthesis is the name of the technique $ \left( {{\text{iv}}} \right) $ .
Ethyl iodide, sodium alkoxide, alkyl halide, Williamson’s synthesis are $ \left( {\text{i}} \right){\text{,}}\,\left( {{\text{ii}}} \right){\text{,}}\,\left( {{\text{iii}}} \right) $ and $ \left( {{\text{iv}}} \right) $ respectively.
Hence the correct option is (A).

Note:
The majority of diethyl ether is generated as a by-product of ethylene vapor-phase hydration to create ethanol. This technique involves solid-supported phosphoric acid catalysts and can be tweaked to produce more ether as required.