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

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

Tardigrade Admin · Accepted Answer

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

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

sodium iodide is soluble in methanol, while sodium chloride is insoluble in methanol

sodium iodide is soluble in methanol, while $N a Cl$ and $N a B r$ are insoluble in methanol

sodium iodide is insoluble in methanol, while $N a Cl$ and $N a B r$ are soluble

the three halogens differ considerably in their electronegativity

Correct answer is (b) sodium iodide is soluble in methanol, while $N a Cl$ and $N a B r$ are insoluble in methanol