The acid catalyzed hydration of 2-methylbut-2-ene is a reaction that involves the addition of water to the double bond in the presence of an acid catalyst. This reaction results in the formation of 2-methylbutan-2-ol.



The acid catalyzed hydration of 2-methylbut-2-ene proceeds through the following steps:

1. Protonation: The acid catalyst, typically sulfuric acid (H2SO4), protonates the double bond of 2-methylbut-2-ene, generating a carbocation intermediate. The protonation makes the double bond more susceptible to attack by the nucleophile (water).

2. Nucleophilic Attack: Water acts as a nucleophile and attacks the positively charged carbocation. This leads to the formation of a new bond between the carbon and the oxygen atom of water.

3. Deprotonation: The resulting intermediate is a protonated alcohol, which is then deprotonated by water or another water molecule present in the reaction mixture. This step regenerates the acid catalyst and forms the final product, 2-methylbutan-2-ol.



The correct answer is option 'C', (CH3)2C=CHCH3.

In this compound, the double bond is located between the second and third carbon atoms. When this compound undergoes acid catalyzed hydration, the double bond is protonated, and water adds across the double bond to form 2-methylbutan-2-ol. The resulting product has an additional methyl group attached to the second carbon atom.

The other options, CH3CH2CH=CH2 (option 'a') and CH3CH=CHCH2 (option 'b'), do not have the double bond located at the correct position for the acid catalyzed hydration to occur. Therefore, these compounds would not produce 2-methylbutan-2-ol upon acid catalyzed hydration.

Overall, the correct answer, option 'C', is the only compound among the given options that can yield 2-methylbutan-2-ol through acid catalyzed hydration.

Proceed backward i.e., remove a molecule of water to get the alkene.