Question

${\text{Y}}\xleftarrow{{Na}}{\text{X}}\xrightarrow{{{\text{conc}}{\text{.}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{,413K}}}}{{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}$
What are ${\text{X and Y}}$ in the above reactions?

Answer 1

Hint: The reactions in which symmetrical ethers are formed are mostly the results of reactions in which alcohols react with concentrated sulphuric acid. When alcohols react with any reactive metal the hydrogen atom is replaced by the metal and the hydrogen gas is released.

Complete step by step answer:
We will solve the reaction step by step. First we will find out ${\text{X}}$ and then ${\text{Y}}$ .
So,
 ${\text{X}}\xrightarrow{{{\text{conc}}{\text{.}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{,413K}}}}{{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}$
In this reaction ether is formed using the reagent concentrated sulphuric acid. The ether formed is diethyl ether in which the two moles of alcohol are used, so the only alcohol used for this reaction is ethanol.
Therefore the reaction is:
${\text{2}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}\xrightarrow{{{\text{conc}}{\text{.}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{,413K}}}}{{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O + }}{{\text{H}}_2}{\text{O}}$

Mechanism of the reaction:
First of all the concentrated sulphuric acid present at $413K$ attacks on the ethyl alcohol, one water molecule is released and ethyl hydrogen sulphate is produced as shown in the following reaction:
 ${\text{2}}{{\text{C}}_2}{{\text{H}}_5}{\text{OH + OH - S}}{{\text{O}}_3}{\text{H(}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}) \to {\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{O - S}}{{\text{O}}_3}{\text{H + }}{{\text{H}}_2}{\text{O}}$

Now this produced ethyl hydrogen sulphate reacts with one mole of ethyl alcohol and will form diethyl ether. In this reaction sulphuric acid is released. The reaction is as follows:
${{\text{C}}_2}{{\text{H}}_5}{\text{OH + C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{O - S}}{{\text{O}}_3}{\text{H}} \to {({{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O + }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}$
Hence we can conclude that the reactant ${\text{X}}$ is ethanol.

Now to find out ${\text{Y}}$ we have to place the ethanol in presence of sodium metal.
${\text{X}}\xrightarrow{{Na}}{\text{Y}}$
And we know that ${\text{X}}$ is ethanol.
Therefore,
 ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}\xrightarrow{{Na}}{{\text{C}}_2}{{\text{H}}_5}{\text{ONa}}$
Hence on reacting ethanol with sodium metal sodium ethoxide is produced.

Thus, the missing compounds in the reaction are ethanol and sodium ethoxide.
${\text{X = Ethanol}}$ and ${\text{Y = Sodium ethoxide}}$.


Note:
Ethers are basically carbon derivatives in which the oxygen atom is bounded by the alkyl or aryl groups. It is represented as $R - O - R'$ where ${\text{R and R'}}$ are the alkyl or aryl groups. Ethers are relatively unreactive so they are used as solvents for fats and oils.