Step 1: Understand the question.
We want the named reaction that turns an alkyl chloride into an alkyl iodide.
Step 2: Note what kind of change this is.
We are swapping one halogen (Cl) for another (I) on the same carbon. So this is a halogen exchange reaction.
Step 3: Recall the reagent for making iodides.
To put iodine in, we treat the chloride with sodium iodide (NaI) in dry acetone. The reaction is \[ \text{R-Cl} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaCl} \]
Step 4: See why it works.
NaI dissolves in dry acetone but NaCl does not, so NaCl drops out as a solid. This keeps pulling the reaction forward to make more alkyl iodide.
Step 5: Rule out the other names.
Swartz reaction makes alkyl fluorides (with AgF). Fittig and Friedel-Crafts are aromatic ring reactions, not simple halogen swaps.
Step 6: Pick the answer.
The Cl to I exchange is the Finkelstein reaction, option (B).
\[ \boxed{\text{Finkelstein reaction}} \]