Question

Anisole is treated with HI under two different conditions. \[ C_6H_5OCH_3 + D \xrightarrow{\text{conc. HI}} A + B \] The nature of A and B will be

• A. A and B are \( \text{CH}_3 \) and \( \text{C}_6\text{H}_5\text{OH} \), while C and D are \( \text{CH}_3 \text{I} \) and \( \text{C}_6\text{H}_5\text{I} \)
• B. A and B are \( \text{C}_6\text{H}_5\text{OH} \) and \( \text{CH}_3 \text{I} \), while C and D are \( \text{CH}_3\text{I} \) and \( \text{C}_6\text{H}_5\text{OH} \)
• C. A and B are \( \text{C}_6\text{H}_5\text{OH} \) and \( \text{CH}_3 \), while C and D are \( \text{CH}_3\text{I} \) and \( \text{C}_6\text{H}_5\text{I} \)
• D. Both A and B as well as both C and D are \( \text{CH}_3\text{I} \) and \( \text{C}_6\text{H}_5\text{OH} \)

Hint:

In the presence of HI, anisole undergoes nucleophilic substitution, replacing the methoxy group with iodine to form phenol and methyl iodide.

Answer 1

The Correct Option is A

Step 1: Nucleophilic Substitution with HI.
In this reaction, anisole undergoes nucleophilic substitution with HI. The methoxy group (\( \text{-OCH}_3 \)) is substituted by the iodine atom, forming \( \text{C}_6\text{H}_5\text{OH} \) (phenol) and \( \text{CH}_3 \text{I} \) (methyl iodide).
Step 2: Conclusion.
The correct answer is (A), as the products are phenol and methyl iodide.