Fundamentals of Organic ChemistryChemistryAlcohols Phenols and Ethers
Anisole is treated with HI under two different conditions to yield products A , B , C , and D as shown below: C + D HI ( g ) C ₆ H ₅ OCH ₃ Conc. HI A + B Identify the products A , B , C , and D .
Options
- AA and B are CH ₃ I and C ₆ H ₅ OH , while there is no reaction in the second case.
- BA and B are CH ₃ I and C ₆ H ₅ OH , while C and D are CH ₃ OH and C ₆ H ₅ I .
- CBoth sets of products ( A and B , as well as C and D ) are CH ₃ I and C ₆ H ₅ OH .
- DA and B are CH ₃ OH and C ₆ H ₅ I , while C and D are CH ₃ I and C ₆ H ₅ OH .
Correct answer
C. Both sets of products ( A and B , as well as C and D ) are CH ₃ I and C ₆ H ₅ OH .
Step-by-step solution
The reaction of anisole ( C ₆ H ₅ OCH ₃ ) with HI proceeds via the protonation of the ether oxygen to form an oxonium ion, C ₆ H ₅- O ⁺( H )- CH ₃ . The O - C ₆ H ₅ bond has partial double bond character due to the resonance between the lone pair of electrons on the oxygen atom and the benzene ring, making it stronger and difficult to break. In contrast, the O - CH ₃ bond is a pure single bond. Therefore, the nucleophile I ⁻ attacks the less sterically hindered CH ₃ group via an S _ N 2 mechanism, cleaving the O -