Question

Two compounds X and Y have the same formula \[{{\text{C}}_2}{{\text{H}}_4}{{\text{O}}_2}\]. One of them reacts with sodium metal to liberate \[{{\text{H}}_2}\] and \[{\text{C}}{{\text{O}}_2}\] with \[{\text{NaHC}}{{\text{O}}_3}\]. Second one does not reacts with Na metal and \[{\text{NaHC}}{{\text{O}}_3}\] but undergoes hydrolysis with \[{\text{NaOH}}\] to form salts of carboxylic acid and compound Z which is called wood spirit. Identify X, Y and Z and write chemical equations for the reaction involved.

Answer 1

Hint: Compound having two oxygen atoms will have to be either carboxylic acid or ester. There is only one combination possible for the given general formula for both the compounds X and Y.

Complete step by step solution:
- There is a very common test that is used for the identification of the carboxylic acid in functional group analysis. Carboxylic acid reacts with sodium bicarbonate or sodium hydrogen carbonate that is \[{\text{NaHC}}{{\text{O}}_3}\] to produce brisk effervescence and the carbon dioxide releases out. The possible acid with the given general formula is \[{\text{C}}{{\text{H}}_3}{\text{COOH}}\], that is, ethanoic acid the reaction occurs as:
\[{\text{2C}}{{\text{H}}_3}{\text{COOH}} + 2{\text{NaHC}}{{\text{O}}_3} \to {\text{2C}}{{\text{H}}_3}{\text{COONa}} + {\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\]

As we know acids react with metals to liberate hydrogen gas. Carboxylic acid being an acid reacts with sodium to form hydrogen gas and sodium acetate. The reaction occurs as:
\[{\text{2C}}{{\text{H}}_3}{\text{COOH}} + 2{\text{Na}} \to {\text{2C}}{{\text{H}}_3}{\text{COONa}} + {{\text{H}}_2}\]

Now the other compound that neither react with sodium bicarbonate and sodium metal is methyl methanoate, the molecular formula is \[{\text{HCOOC}}{{\text{H}}_3}\], that is, ester molecule. The possible ester with the given number of carbon is \[{\text{HCOOC}}{{\text{H}}_3}\]. It reacts with sodium hydroxide to form two compounds that are sodium salt of carboxylic acid and methanol. The reaction occurs as:
\[{\text{HCOOC}}{{\text{H}}_3} + {\text{NaOH}} \to {\text{HCOONa}} + {\text{C}}{{\text{H}}_3}{\text{OH}}\]
The above reaction is known as the saponification in which an ester reacts with an inorganic base to form alcohol and soap that is sodium salt of fatty acids.
Methanol is known as wood spirit.

Hence, the compound X is \[{\text{C}}{{\text{H}}_3}{\text{COOH}}\]
The compound Y is \[{\text{HCOOC}}{{\text{H}}_3}\]
The compound Z is \[{\text{C}}{{\text{H}}_3}{\text{OH}}\]

Note: Methanol is known as wood spirit because it was produced by the destructive distillation of wood. Now a day’s methanol is mainly prepared by the hydrogenation of carbon monoxide.