Question

Consider the following reaction and identify A and B :
\( \text{CH}_3\text{Cl} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{A} + \text{B} \)

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

Hint:

The precipitation of NaCl/NaBr in acetone provides the driving force for this reaction.
The order of nucleophilicity of halide ions in acetone is \( \text{I}^- \gt \text{Br}^- \gt \text{Cl}^- \).
This reaction is a classic example of an equilibrium being shifted by removal of a product.

Answer 1

The Correct Option is A

Concept:

• This reaction is specifically known as the Finkelstein reaction.

• It is a nucleophilic substitution \( (\text{S}_{\text{N}}2) \) reaction used for halogen exchange.

• It is the standard method for preparing alkyl iodides from alkyl chlorides or bromides.
Step 1: Nucleophilic attack mechanism
In this reaction, the iodide ion \( (\text{I}^-) \) serves as a strong nucleophile.
Since the substrate is a primary methyl halide \( (\text{CH}_3\text{Cl}) \), the reaction proceeds via an \( \text{S}_{\text{N}}2 \) mechanism.
The \( \text{I}^- \) attacks the carbon center from the back side while the \( \text{Cl}^- \) leaves simultaneously.

Step 2: Role of dry acetone and Le Chatelier's principle
The solvent, dry acetone, plays a critical role in driving the reaction to completion.
Sodium iodide \( (\text{NaI}) \) is soluble in acetone, but the byproduct sodium chloride \( (\text{NaCl}) \) is almost insoluble.
Consequently, NaCl precipitates out of the solution as it forms.
\[ \text{CH}_3\text{Cl} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{I} + \text{NaCl} \downarrow \]

Step 3: Identifying products A and B
Based on the exchange, product A is Methyl Iodide \( (\text{CH}_3\text{I}) \).
Product B is the precipitated Sodium Chloride \( (\text{NaCl}) \).