Complete the electron-pushing mechanism for the e1 reaction when 2-methylbutan-2-ol is treated with 20% sulfuric acid.

The final production of E1 elimination of 2-methylbutan-2-ol is 2-methyl-2-butene. For the electron pushing mechanism of E1 elimination reaction refer to the image attached.

Further explanation:

E1 elimination reaction:

E1 elimination reaction is a unimolecular reaction that removes an HX leaving group and produce a double bond. The mechanism of E1 elimination involves deprotonation of hydrogen from beta carbon and formation of carbocation intermediate. Finally, the removal of leaving group takes place.

The mechanism of E1 elimination of 2-methylbutan-2-ol is as follows:

Step 1: The alcohol group on 2-methylbutan-2-ol is hydrated by . This makes alcohol a better leaving group.

Step 2: The leaving group leaves and takes the electrons of bond pair with itself and forms a carbocation intermediate.

Step 3: Another molecule of comes and deprotonate the beta carbon and leaves an electron pair on beta carbon. The electron pair on beta carbon are donated to the carbocation carbon and formation of double bond takes place. The final product is 2-methyl-2-butene.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

2. Determine the product formed when 2-propanol react with NaH:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: E1 elimination reaction

Keywords: E1 elimination reaction, 2-methylbutan-2-ol, structure of alkenes, electron-pushing mechanism, and 20% sulfuric acid.

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