Question

What is 'Y' in the following reaction?
        ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{O}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\,\text{+}\,\text{NaI}\,\text{+}\,\text{2HI}\,\,\to \,\,\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{I}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{O}$

Answer 1

Hint: If an alkyl halide is reacts with a less hindered substrate it follows ${{\text{S}}_{\text{N}}}\text{2}$ (substitution nucleophilic bimolecular) reaction, while steric ally hindered substrate follow the elimination reaction.

Complete Solution :
- Alkyl halide on reaction with sodium alkoxide gives ether, and ether on reaction with the excess of $\text{HI}$ produces two molecules of alkyl halide.
-When alkyl halide reacts with sodium alkoxide it gives ether, this reaction is known as Williamson's synthesis reaction of ether. This is the most common laboratory method for the production of ether. In this reaction ${{1}^{\circ }}$ halide is used to form ether. But if a sterically hindered (${{3}^{\circ }}$ halide) is used, elimination reaction takes place and alkene is formed as a major product.
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{I}\,\text{+}\,\text{NaO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\,\,\to \,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{O}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\,\text{+}\,\,\text{NaI}\,$
Where $\text{X}\,\,=\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{O}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$

- $\text{HI}$ Acts as a strong reducing agent. So when diethyl ether is a simple symmetrical ether, when it reacts with excess of $\text{HI}$ halogen goes with the alkyl group. Therefor two molecule of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{I}$ (ethyl iodide) and water molecule is formed.
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{O}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\,\text{+}\,\text{NaI}\,\text{+}\,\text{2HI}\,\,\to \,\,\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{I}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{O}$
Where $\text{Y}\,\text{=}\,\,\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{I}\,$

Here ether is solvent, so being a less polar compound it favours the ${{\text{S}}_{\text{N}}}\text{2}$ reaction and the nucleophile ${{\text{I}}^{\text{-}}}$ attacks the ${{1}^{\circ }}$ carbon of ethane.
(A) This reaction follows the nucleophile substitution reaction. So alkane will not form.
(B) $\text{HI}$ reduces the symmetrical ether molecule so alkyl halide will form as a final product.
(C) Alkene is formed as a product when the elimination reaction takes place. Elimination reaction is followed by sterically hindered substrate.
(D) Diethyl ether is reduced by $\text{HI}$ molecule.
So, the correct answer is “Option B”.

Note: When equimolar quantities of ether and $\text{HI}$ are reacted with each other, then one molecule of alkyl halide and one molecule of alcohol are formed as a product.
- While, when excess quantity of $\text{HI}$ is treated with less quantity of ether, two molecules of alkyl halide and water molecules are formed at the product side.