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

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Sodium iodide is soluble in methanol, while NaCl and NaBr are insoluble in methanol

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Sodium iodide is insoluble in methanol, while NaCl and NaBr are soluble

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The three halogens differ considerably in their electronegativity

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Solution

The correct option is B
Sodium iodide is soluble in methanol, while NaCl and NaBr are insoluble in methanol

$R - X + N a I S o l u b l e i n (C H_{3} O H, M e_{2} C O) a c e t o n e - --- \to R - I + N a X ↓ I n s o l u b l e i n (C H_{3} O H . M e_{2} C O)$
(where X=Cl or Br)

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