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

Concept: This reaction is known as the Finkelstein reaction. In this reaction, an alkyl chloride or alkyl bromide reacts with sodium iodide in dry acetone to form the corresponding alkyl iodide. The reaction proceeds because sodium chloride is insoluble in dry acetone and precipitates out.

Step 1: Writing the reaction involved. \[ CH_3Cl + NaI \rightarrow CH_3I + NaCl \]

Step 2: Understanding why the reaction proceeds forward. Dry acetone is used because \[ NaCl \] is insoluble in acetone. The precipitated sodium chloride shifts the equilibrium toward product formation.

Step 3: Identifying products A and B. Comparing with \[ CH_3Cl + NaI \rightarrow CH_3I + NaCl \] we get \[ A = CH_3I \] and \[ B = NaCl \] Therefore Option (A) is correct. \[ \boxed{\text{Option (A)}} \]