Question

Consider the following reaction and identify A and B: \[ CH_3Cl + NaI \xrightarrow{\text{dry acetone}} A + B \]

• A. A = CH\(_3\)I, B = NaCl
• B. A = CH\(_3\)OH, B = NaCl
• C. A = CH\(_3\)CHO, B = NaCl
• D. A = C\(_2\)H\(_6\), B = CH\(_3\)I

Hint:

{Finkelstein Reaction Shortcut:} \[ R{-}Cl/Br + NaI \xrightarrow{\text{acetone}} R{-}I \] Driven by precipitation of NaCl or NaBr.

Answer 1

The Correct Option is A

Concept: This reaction represents the
Finkelstein reaction, which involves halogen exchange in alkyl halides. It is an \( S_N2 \) reaction where a chlorine atom is replaced by iodine using sodium iodide in dry acetone. General reaction: \[ R{-}Cl + NaI \xrightarrow{\text{acetone}} R{-}I + NaCl \] The reaction proceeds forward because \( NaCl \) is insoluble in acetone and precipitates out, driving the reaction to completion.
Step 1: Identify the reaction type. Presence of NaI in dry acetone strongly indicates Finkelstein reaction.
Step 2: Apply halogen exchange. \[ CH_3Cl + NaI \rightarrow CH_3I + NaCl \]
Step 3: Identify products.

• A = CH\(_3\)I (methyl iodide)
• B = NaCl (precipitate)

\[ \therefore \text{Correct answer = (A)} \]