Question

Match the following: A. Swarts reaction, B. Finkelstein reaction, C. Wurtz-Fittig reaction, D. Sandmeyer reaction.

• A. A-I, B-III, C-II, D-V
• B. A-III, B-I, C-II, D-IV
• C. A-III, B-IV, C-V, D-II
• D. A-IV, B-I, C-III, D-V

Hint:

Swarts = Fluorine; Finkelstein = Iodine!

Answer 1

The Correct Option is B

Step 1: Recall Swarts reaction.
Swarts converts an alkyl chloride or bromide into an alkyl fluoride using metal fluorides like $AgF$. So it gives $C_2H_5F$, which is item III. A is III.

Step 2: Recall Finkelstein reaction.
Finkelstein swaps a chloride or bromide for iodide using NaI in acetone. So it gives $C_2H_5I$, which is item I. B is I.

Step 3: Recall Wurtz-Fittig reaction.
This joins an aryl halide and an alkyl halide using sodium to make an alkyl benzene. So the product is an alkylbenzene, item II. C is II.

Step 4: Recall Sandmeyer reaction.
Sandmeyer turns a diazonium salt into an aryl halide using copper halides. Here it gives bromobenzene, item IV. D is IV.

Step 5: Put the matches together.
A-III, B-I, C-II, D-IV.

Step 6: Conclusion.
So the answer is A-III, B-I, C-II, D-IV. \[ \boxed{\text{A-III, B-I, C-II, D-IV}} \]