Question

Metallic sodium reacts with methanol and ethanol to give:
(A) \[{{\text{O}}_{\text{2}}}\]
(B) \[{{\text{H}}_2}{\text{O}}\]
(C) \[{\text{C}}{{\text{O}}_2}\]
(D) \[{{\text{H}}_2}\]

Answer 1

Hint: Sodium is a highly electropositive element and is a strong reducing agent. It readily donates electrons to the hydroxyl group of the alcohol molecule. The gas obtained is as a result of the reaction of the hydroxyl group of alcohol molecules with electrons donated by sodium metal.

Complete step by step answer:
Methanol and ethanol belong to the homologous series of alkanes. The structural formula of methanol and ethanol are \[{\text{C}}{{\text{H}}_3}{\text{OH}}\] and \[{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}\] respectively.
Thus, the functional group present in methanol and ethanol is the hydroxyl functional group. The hydroxyl functional group is represented as the \[ - {\text{OH}}\] group. In the hydroxyl functional group, the oxygen atom has partial negative charge and the hydrogen atom has partial positive charge.
When a small piece of sodium metal is added to either methanol or ethanol, a steady reaction occurs to form sodium methoxide or sodium ethoxide. The byproduct is the hydrogen gas which is evolved during the reaction in the form of bubbles.
Write balanced chemical equations for the reaction of sodium metal with methanol and ethanol:
\[{\text{2C}}{{\text{H}}_3}{\text{OH + 2Na }} \to {\text{ 2C}}{{\text{H}}_3}{\text{ONa + }}{{\text{H}}_2}\]
\[{\text{2C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH + 2Na }} \to {\text{ 2C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{ONa + }}{{\text{H}}_2}\]
In the above reactions, each sodium metal loses an electron to form sodium cation. These electrons are gained by protons of hydroxyl groups of methanol or ethanol to form hydrogen molecules.
Metallic sodium reacts with methanol and ethanol to give \[{{\text{H}}_2}\] gas.

Hence, the option (D) \[{{\text{H}}_2}\] is the correct option.

Note: In this reaction, sodium atom loses an electron. Hence, sodium is oxidized and acts as a reducing agent. Hydrogen of the hydroxyl group gains electrons. Hence, hydrogen is reduced. Thus, the reaction of sodium metal with methanol or ethanol is a redox reaction.