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

To solve the given chemical reaction problem, we need to identify the products A and B formed from the reaction:

\[ CH_3Cl + NaI \xrightarrow{\text{dry acetone}} A + B \]

1. Understanding the Reaction: This chemical reaction describes a halide exchange reaction, specifically an example of the Finkelstein reaction. This reaction involves the exchange of halides using sodium iodide (NaI) in a solvent like dry acetone. It is classified as a nucleophilic substitution reaction.
2. Mechanism:
   • The chloride ion \((\text{Cl}^-)\) in \(\text{CH}_3\text{Cl}\) is substituted by the iodide ion \((\text{I}^-)\) from NaI.
   • The reaction proceeds because iodides are better leaving groups than chlorides, and NaI is soluble in dry acetone, thus favoring the reaction.
3. Result: The products of the reaction are iodomethane (CH₃I) and sodium chloride (NaCl).
4. Conclusion: Given this analysis, the correct identification of A and B in the reaction is:
   • A = CH\(_3\)I (iodomethane)
   • B = NaCl (sodium chloride)

Thus, the correct option is: A = CH\(_3\)I, B = NaCl.