Treatment of Cyclohexanol with Sodium Metal to Form Cyclohexyl Ethyl Ether

The reaction between cyclohexanol and sodium metal leads to the formation of an alkoxide ion, which subsequently reacts with iodoethane to yield cyclohexyl ethyl ether. Here’s a step-by-step breakdown of this reaction:

Step 1: Formation of Alkoxide Ion

When cyclohexanol (C₆H₁₂O) is treated with sodium (Na), sodium donates an electron to the hydroxyl group (-OH) of cyclohexanol. This results in the loss of a hydrogen atom and the formation of the cyclohexoxide ion (C₆H₁₁O⁻) and sodium ion (Na⁺):

C6H11OH + Na → C6H11O−Na+ + 1/2 H2

Step 2: Nucleophilic Attack on Iodoethane

Next, the alkoxide ion generated in the previous step acts as a nucleophile. It attacks the electrophilic carbon in iodoethane (C₂H₅I), resulting in the displacement of the iodide ion (I⁻):

C6H11O− + C2H5I → C6H11O-C2H5 + I−

Step 3: Formation of Cyclohexyl Ethyl Ether

The nucleophilic attack leads to the formation of cyclohexyl ethyl ether (C₈H₁₈O) as the final product:

C6H11O-C2H5

In summary, this series of reactions effectively illustrates how cyclohexanol can be converted into cyclohexyl ethyl ether through the action of sodium and iodoethane. The main steps involve the generation of an alkoxide ion followed by a nucleophilic substitution reaction with iodoethane.