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

Step 1: Identify the type of reaction.
Sodium iodide is dissolved in dry acetone and mixed with methyl chloride. This is a halogen exchange reaction, commonly called the Finkelstein reaction, used to swap a chlorine or bromine for an iodine.
Step 2: Balance the equation by atom count.
On the left side we have $\text{CH}_3\text{Cl}$ and $\text{NaI}$. Sodium must pair with the leaving chloride to give $\text{NaCl}$, and the methyl group must pair with iodide to give $\text{CH}_3\text{I}$. \[ \text{CH}_3\text{Cl} + \text{NaI} \rightarrow \text{CH}_3\text{I} + \text{NaCl} \]
Step 3: Explain why the reaction goes to completion.
Dry acetone dissolves $\text{NaI}$ well but barely dissolves $\text{NaCl}$. As $\text{NaCl}$ forms it drops out of solution as a solid, so there is nothing to pull the products back to reactants.
Step 4: Match to the given options.
Only $\text{A} = \text{CH}_3\text{I}$ and $\text{B} = \text{NaCl}$ matches this atom balance and mechanism; the other options introduce atoms that are not present anywhere in the reactants.
\[ \boxed{\text{A} = \text{CH}_3\text{I}, \ \text{B} = \text{NaCl}} \]