Question

Conant-Finkelstein reaction for the preparation of alkyl iodide is based upon the fact that

• A. sodium iodide is soluble in methanol, while sodium chloride is insoluble in methanol
• B. sodium iodide is soluble in methanol, while $NaCl$ and $NaBr$ are insoluble in methanol
• C. sodium iodide is insoluble in methanol, while $NaCl$ and $NaBr$ are soluble
• D. the three halogens differ considerably in their electronegativity

Answer 1

Answer: (B)

Correct answer is (b) sodium iodide is soluble in methanol, while $NaCl$ and $NaBr$ are insoluble in methanol