Question

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

• A. \(A = CH_3I,\; B = NaCl\)
• B. \(A = CH_3OH,\; B = NaCl\)
• C. \(A = CH_3CHO,\; B = NaCl\)
• D. \(A = C_2H_6,\; B = CH_3I\)

Hint:

Finkelstein reaction: \[ RCl + NaI \xrightarrow{\text{dry acetone}} RI + NaCl \] Dry acetone is used because NaCl precipitates out, driving the reaction forward.

Answer 1

The Correct Option is A

Step 1: Recognize the Finkelstein reaction.
Treating an alkyl chloride or bromide with sodium iodide in dry acetone produces the corresponding alkyl iodide. This halide-exchange reaction is called the Finkelstein reaction.
Step 2: Understand why the equilibrium shifts forward.
$NaI$ is soluble in dry acetone, but the $NaCl$ formed is insoluble and precipitates immediately. Removal of $NaCl$ from solution drives the equilibrium toward the product $CH_3I$, making the reaction essentially irreversible.
Step 3: Identify A and B.
\[ CH_3Cl + NaI \xrightarrow{\text{dry acetone}} CH_3I + NaCl \] Comparing: $A = CH_3I$ and $B = NaCl$.
\[ \boxed{A = CH_3I,\; B = NaCl} \]