Explanation: When sodium (Na) reacts with ethyl iodide (C₂H₅I), it undergoes a type of reaction called the Wurtz reaction, which is a coupling reaction. In this reaction, two molecules of ethyl iodide react with sodium metal, resulting in the formation of ethane (C₂H₆) and the byproduct sodium iodide (NaI).
The Wurtz Reaction:

The general form of the Wurtz reaction is:
2 R-X+2 Na→R-R+2 NaX
2R-X+2Na→R-R+2NaX

Where R-X is an alkyl halide, and Na is sodium metal.

In the case of ethyl iodide (C₂H₅I), the reaction would be:
2 C2H5I+2 Na→C2H6+2 NaI
2C2​H5​I+2Na→C2​H6​+2NaI
How It Works:

Nucleophilic Substitution: The sodium metal donates electrons to the ethyl iodide molecules, causing the iodine ion (I⁻) to be replaced by a sodium ion (Na⁺). This results in the formation of ethyl radicals (C₂H₅•).

Radical Coupling: Two ethyl radicals (C₂H₅•) generated by this process combine to form a new bond between the two ethyl groups, producing ethane (C₂H₆).

Formation of Sodium Iodide: The iodine atoms (I⁻) from the ethyl iodide molecules bond with sodium ions (Na⁺) to form sodium iodide (NaI).

Why This Happens:

The Wurtz reaction is a way to form carbon-carbon (C-C) bonds by using alkyl halides and sodium metal. The reaction is particularly useful for creating alkanes (saturated hydrocarbons) like ethane (C₂H₆) from alkyl halides.

Summary:

In summary, when sodium (Na) reacts with ethyl iodide (C₂H₅I), the Wurtz reaction occurs, leading to the formation of ethane (C₂H₆), which is a simple alkane, and the byproduct sodium iodide (NaI). The key feature of this reaction is the coupling of two ethyl groups, which forms the new C-C bond in ethane.