Step 1: Understanding the Question:
This question probes the acidic nature of the hydrogen atom attached to the oxygen in alcohols.
Active metals react with acids (even weak ones) to release gas. Step 2: Key Formula or Approach:
General reaction of alcohol with active metal:
\[ 2R-OH + 2Na \rightarrow 2R-ONa + H_2 \uparrow \] Step 3: Detailed Explanation:
Alcohols (\(R-OH\)) have a polar \(O-H\) bond. The hydrogen is weakly acidic.
Sodium is a highly electropositive metal. It displaces the hydrogen from the hydroxyl group.
The sodium atom loses an electron to become \(Na^+\), and the hydrogen atom accepts electrons to form \(H_2\) gas.
The resulting ionic compound \(R-O^-Na^+\) is called a sodium alkoxide.
Example with ethanol:
\[ 2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \]
This results in Sodium Ethoxide and Hydrogen gas. Step 4: Final Answer:
The reaction produces Sodium alkoxide and Hydrogen gas.