Q59E - Organic Chemistry (Mcmurry)

Question

Meerwein’s reagent, triethyloxoniumtetrafluoroborate, is a powerful ethylating agent that converts alcohols into ethyl ethers at neutral pH. Show the reaction of Meerwein’s reagent with cyclohexanol, and account for the fact that trialkyloxonium salts are much more reactive alkylating agents than alkyl iodides.

${\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}\mathit{B}{{\mathbf{F}}_{\mathbf{4}}}^{\mathbf{-}}$ Meerwein’s reagent

Step-by-Step Solution

Answer

Here, trialkyloxonium ion is used because of its higher reactivity than alkyl iodides. Moreover, neutral ether can leave easily than the iodide ions. 

Step 1: Alcohol attacks on triethyloxoniumcation

In cyclohexanol, the hydroxyl group or alcohol group attacks on the triethyloxoniumcation. Thus electron shift occurs, diethyl ether is lost and cyclohexyl ethyl oxidanium ion is formed. 

Step 2: Attack of base and loss of proton

When base is attacked at the hydroxyl group, a proton is lost from it. Thus it forms ethoxycylohexane. This reaction is a reversible reaction. 

The whole mechanism is shown below: