Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

Sagot :

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:.

View image almatheia
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.