0 votes
19 views
ago in Organic Chemistry by (139k points)
What is the mechanism for the dehydration of 2-methyl-2-butanol using concentrated sulfuric acid, and what is the major product formed in the reaction?

1 Answer

0 votes
ago by (290 points)
The dehydration of 2-methyl-2-butanol using concentrated sulfuric acid follows an E1  Elimination Unimolecular  mechanism. Here's a step-by-step breakdown of the reaction:1. Protonation: The hydroxyl group  -OH  of 2-methyl-2-butanol acts as a nucleophile and attacks a proton  H+  from the concentrated sulfuric acid  H2SO4 . This results in the formation of a good leaving group, water  H2O , and a carbocation intermediate.2. Carbocation formation: The water molecule leaves, generating a carbocation at the tertiary carbon  the carbon attached to three other carbons  of the 2-methyl-2-butanol molecule. This carbocation is relatively stable due to the inductive effect and hyperconjugation from the three alkyl groups attached to the positively charged carbon.3. Elimination: A base  HSO4-  from the sulfuric acid abstracts a proton  H+  from one of the adjacent carbons to the carbocation, forming a double bond between the carbocation carbon and the adjacent carbon. This step results in the formation of an alkene.The major product formed in this reaction is 2-methyl-2-butene. Since the reaction follows Zaitsev's rule, the more substituted alkene is the major product. In this case, the double bond is formed between the tertiary and secondary carbons, resulting in a more substituted and stable alkene.
Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...