Question

Complete the electron-pushing mechanism for the E1 reaction when 2-methylbutan-2-ol is treated with 20% sulfuric acid.

Use two curved arrows to show the fast protonation of the alcohol. Use a curved arrow to show the dissociation of the leaving group from the substrate. Draw the organic product of the E1 dehydration Use two curved arrow to show the deprotonation of the carbocation intermediate.

Answer 1

Concepts and reason

An organic reaction in which two substituents are removed from a molecule is termed as an elimination reaction. There are two elimination reaction mechanisms, namely, ${\rm{E1}}$ and ${\rm{E2}}$ .

Acid catalyzed dehydration of alcohols:

Alcohols undergo elimination reactions to form alkenes and water in the presence of strong acids, such as sulfuric acid. In this reaction, hydroxyl functional group in the alcohol abstracts a proton from the acid to form a protonated alcohol and which follows by expulsion of water to give alkenes as the final organic product.

Alkene formation takes place via E1 mechanism and then obeys Zaitsev’s rule.

Fundamentals

Two possible alkenes can be found by performing an elimination reaction on the given alkyl halide.

The general scheme for the reactions are depicted below:

Elimination reaction:

For an unsymmetrical molecule, the major product will be the more substituted alkene.

Dehydration of alcohols:

Zaitsev’s rule: This rule states that in an elimination reaction, the product that is more substituted is formed as a stable product.

Example for elimination reaction (E1):

Ans:

The curved arrow mechanism is as follows: