Question

What happens when Ethyl iodide dissolved in dry ether is treated with sodium metal?

Answer 1

In this reaction, the iodine from the given compound (ethyl iodide) will be removed and that iodine will react with sodium to give sodium iodide. As iodine is removed, the only thing left in ethyl iodide will be ethyl, but it is not completely stable.

Complete answer:
When ethyl iodide dissolved in dry ether reacts with sodium metal, the product formed will be butane, and sodium iodide will be formed as a by-product.
Let us first see the chemical compositions of the given compounds.
${{C}_{2}}{{H}_{5}}I$ is the chemical formula of ethyl iodide, it reacts with sodium ( $Na$ ) to give butane and sodium iodide ( $NaI$ ).
When sodium attacks ethyl iodide, it removes the iodine atom from it and forms sodium iodide, but as iodine is removed from ethyl iodide, ethyl becomes unstable. As ethyl is unstable, it will react with another ethyl molecule to form a stable compound butane.
$2{{C}_{2}}{{H}_{5}}I+2Na\xrightarrow[ether]{Dry}{{C}_{4}}{{H}_{10}}+2NaI$
The reaction that will occur is shown above. As shown, two molecules of sodium will attack on two molecules of ethyl iodide to form two molecules of sodium iodide. Now, as ethyl will remain alone and unstable, it will get attached with another ethyl molecule.
${{C}_{2}}{{H}_{5}}-{{C}_{2}}{{H}_{5}}$
Thus, a stable compound butane will be formed, whose chemical formula is ${{C}_{4}}{{H}_{10}}$ .

Note:
In the reaction, two sodium atoms and two ethyl iodide molecules are used as two ethyl molecules will have to combine to form one butane. Dry ether works as a catalyst in this reaction, it does not have any role in the reaction other than being a catalyst.