**Title:** Determining the Major Product of an Alkene Hydration Reaction**Reaction:**The chemical reaction involves an alkene reacting with hydronium ion (\(H_3O^+\)) in the presence of water (\(H_2O\)) to yield an alcohol. Below is the depiction of the reaction:- **Reactant:** The alkene presented is 2-methylpropene.- **Reagents:** \(H_3O^+\) and \(H_2O\).This reaction is an example of acid-catalyzed hydration of an alkene, typically following Markovnikov's rule for the addition of water.**Possible Products:**There are four potential alcohol products shown below:1. 2-methyl-2-propanol2. 2-butanol3. 3-methyl-2-butanol4. 2-methyl-1-propanol **Explanation of Reaction and Markovnikov's Rule:**Markovnikov's rule states that in the addition of HX to an alkene, the hydrogen atom attaches to the carbon with more hydrogen atoms, and the halide (or in this case, the OH group from the water) attaches to the carbon with fewer hydrogen atoms.**Conclusion:**According to Markovnikov's rule, the expected major product of the given reaction is 2-methyl-2-propanol, which places the hydroxyl group on the more substituted carbon atom in the reactant alkene, resulting in a tertiary alcohol.

Given reaction: We have to find the product of the reaction.

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Select the major product of the following reaction. H3O+ H2O Тон OH ? OH HO.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter22: Organic And Biological Molecules

Section: Chapter Questions

Problem 170CP: Alcohols are very useful starting materials for the production of many different compounds. The...

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Concept explainers

Designing a Synthesis

Organic synthesis is a branch of chemical synthesis that involves the planned construction of organic compounds/molecules from smaller units that possess functional groups. Friedrich Wöhler was the first scientist who designed the synthesis of an organic compound, urea from inorganic starting materials silver cyanate with ammonium chloride via a simple, one-step synthesis.

Question

**Title:** Determining the Major Product of an Alkene Hydration Reaction

**Reaction:**

The chemical reaction involves an alkene reacting with hydronium ion (\(H_3O^+\)) in the presence of water (\(H_2O\)) to yield an alcohol. Below is the depiction of the reaction:

- **Reactant:** The alkene presented is 2-methylpropene.
- **Reagents:** \(H_3O^+\) and \(H_2O\).

This reaction is an example of acid-catalyzed hydration of an alkene, typically following Markovnikov's rule for the addition of water.

**Possible Products:**

There are four potential alcohol products shown below:

1. 2-methyl-2-propanol
2. 2-butanol
3. 3-methyl-2-butanol
4. 2-methyl-1-propanol

**Explanation of Reaction and Markovnikov's Rule:**

Markovnikov's rule states that in the addition of HX to an alkene, the hydrogen atom attaches to the carbon with more hydrogen atoms, and the halide (or in this case, the OH group from the water) attaches to the carbon with fewer hydrogen atoms.

**Conclusion:**

According to Markovnikov's rule, the expected major product of the given reaction is 2-methyl-2-propanol, which places the hydroxyl group on the more substituted carbon atom in the reactant alkene, resulting in a tertiary alcohol.

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