Answer:
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Explanation:
Alcohols are poor leaving groups.
To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.
Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.
After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.
