1 mole of cyclohexyl ethyl ether reacts with 1 mole of HI, what are the 2 compounds formed in the highest amounts? a) ethyl iodide and cyclohexyl iodide b) ethanol and cyclohexyl iodide c) ethanol and cyclohexanol d) ethyl iodide and cyclohexanol
The correct answer and explanation is:
The correct answer is d) ethyl iodide and cyclohexanol.
When 1 mole of cyclohexyl ethyl ether reacts with 1 mole of hydrogen iodide (HI), the ether undergoes an acid-catalyzed cleavage. Ethers react with HI by breaking the carbon-oxygen bond, producing an alkyl iodide and an alcohol. The reaction mechanism involves protonation of the ether oxygen to form a better leaving group, followed by nucleophilic attack by iodide ion (I⁻).
Cyclohexyl ethyl ether consists of two alkyl groups attached to the oxygen: a cyclohexyl group and an ethyl group. The cleavage can occur on either side, but the cleavage favors the formation of the more stable carbocation or the less hindered alkyl iodide. In this case, the cleavage happens preferentially at the bond between oxygen and the ethyl group because ethyl iodide is a primary alkyl iodide and forms more readily than cyclohexyl iodide, which would require breaking the bond on the cyclohexyl side (a secondary carbon).
The products formed are ethyl iodide (from the ethyl group) and cyclohexanol (from the cyclohexyl side), because the cyclohexyl group remains attached to the oxygen but is protonated and converted into cyclohexanol.
To summarize:
- The ether oxygen is protonated by HI, making it a good leaving group.
- Iodide ion attacks the ethyl carbon (less hindered side), producing ethyl iodide.
- The cyclohexyl group is converted into cyclohexanol.
This reaction pathway is typical for ethers reacting with HI, producing an alkyl iodide from the less hindered side and an alcohol from the more hindered side.
Thus, the main products are ethyl iodide and cyclohexanol.