Question

The reagent 'X' used in the following reaction to obtain a good yield of the product is: 

• A. \( KI, H_2SO_4 \)
• B. \( KI, 95\% \ H_3PO_4 \)
• C. \( NaI, ZnCl_2 \)
• D. \( HI \)

Hint:

To convert alcohols to alkyl iodides efficiently, use \( KI \) with phosphoric acid (95\%) instead of sulfuric acid, as \( H_2SO_4 \) oxidizes iodide ions, reducing yield.

Answer 1

The Correct Option is B

Step 1: Understanding the Reaction Type 
- The reaction involves the conversion of an alcohol to an alkyl iodide.
- The preferred method for this transformation is using potassium iodide (\( KI \)) in the presence of a strong acid.
Step 2: Choosing the Suitable Acidic Medium 
1. Use of \( H_2SO_4 \) (Sulfuric Acid):
- Sulfuric acid oxidizes iodide ions (\( I^- \)) to molecular iodine (\( I_2 \)), reducing the availability of \( I^- \) needed for substitution.
- This results in a poor yield.
2. Use of \( H_3PO_4 \) (Phosphoric Acid, 95\% concentration):
- Unlike sulfuric acid, phosphoric acid does not oxidize \( I^- \), leading to a better yield of alkyl iodide.
- The reaction mechanism involves protonation of the hydroxyl group, making it a better leaving group, followed by nucleophilic substitution by \( I^- \).
Step 3: Evaluating the Given Options 
- Option (1): Incorrect, as sulfuric acid leads to poor yield due to oxidation of \( I^- \).
- Option (2): Correct, as \( KI, 95\% \ H_3PO_4 \) provides the best yield of alkyl iodide.
- Option (3): Incorrect, as sodium iodide (\( NaI \)) with zinc chloride (\( ZnCl_2 \)) is used for Lucas test, not for alkyl iodide preparation.
- Option (4): Incorrect, as pure hydriodic acid (\( HI \)) is not typically used due to instability and side reactions.
Thus, the correct answer is 

Option (2).