(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms

(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:.

the answer for sodium is 0.186

the answer for rubidium is 0.251

wher is the rest of the question