Physics, 03.10.2019 13:50 deeoki3384

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

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E1 reaction works in the mechanism that the removal of an HX substituent results in the formation of a double bond. The E1 reaction for 2-methylbutan-2-ol is shown in the figure. This reaction is called acid-catalyzed dehydration of a tertiary alcohol.

The mechanism works in three major steps:
1. The OH group of the main reactant is hydrated by H2SO4 so it becomes H2O.
2. The H2O leaves taking electrons with it. This results to a carbocation intermediate on the carbon atom where it was attached.
3. Another H2O protonates the beta carbon. This is the carbon atom next to the carbocation. It will donate its electrons to the neighboring C-C bond, as indicated by the arrow. The carbons are rehybridized from sp3 to sp2, which is a pi bond. As a result, a double bond forms.

The product is 2-methyl-2-butene.

Complete the electron-pushing mechanism for the e1 reaction when 2-methylbutan-2-ol is treated with

Complete the electron-pushing mechanism for the e1 reaction when 2-methylbutan-2-ol is treated with

Answer from: bossboybaker

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 ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ . 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 ${{\text{H}}_{\text{2}}}{\text{O}}$ 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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