8.4: Hydration of Alkenes- Acid-Catalyzed Hydration

Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane.

What Is Electrophilic Hydration?

Electrophilic hydration is the reverse dehydration of alcohols and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:

The phrase "electrophilic" literally means "electron-loving" (whereas "nucleophilic" means "nucleus-loving"). Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see S_N2 for more details).

Temperatures for Types of Alcohol Synthesis

Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the formed product.

• Primary Alcohol: Less than 170ºC
• Secondary Alcohol: Less than 100ºC
• Tertiary Alcohol: Less than 25ºC

How Does Electrophilic Hydration Work?

Mechanism for 3º Alcohol (1º and 2º mechanisms are similar):

The reaction takes place through the treatment of the alkene with water and a strong acid catalyst, such as H₂SO₄, by a mechanism similar to that of HX addition. Thus, as shown in Figure \(\PageIndex{1}\), protonation of an alkene double bond yields a carbocation intermediate, which reacts with water to yield a protonated alcohol product, ROH₂⁺. Loss of H⁺ from this protonated alcohol gives the neutral alcohol and regenerates the acid catalyst.

Figure \(\PageIndex{1}\): Mechanism of the acid-catalyzed hydration of an alkene to yield an alcohol. Protonation of the alkene gives a carbocation intermediate, which reacts with water. The initial product is then deprotonated.

Now that the reaction is complete, the non-nucleophilic strong acid is regenerated as a catalyst, and an alcohol forms on the most substituted carbon of the current alkane. At lower temperatures, more alcohol product can be formed.

What is Regiochemistry and How Does It Apply?

In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. 

In the mechanism for a 3º alcohol shown above, the green H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugation—electrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double-bonded molecules can cause a completely different reaction.

If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products:

• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.

• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift

The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.

 

Is this a Reversible Synthesis?

Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To shift the equilibrium one way or the other, the temperature or the concentration of the non-nucleophilic strong acid can be adjusted. For example:

• Less sulfuric or phosphoric acid and excess water facilitate the synthesis of more alcohol products.
• Lower temperatures facilitate the synthesis of more alcohol products.

 

Exercises

1) Predict the product of each reaction.

 

 

 

2) Predict the product of each reaction. How does the cyclopropane group affect the reaction?

 

 

 

3) Predict the product of each reaction. What is different about this problem?

 

 

 

4) Predict the product of each reaction. Consider stereochemistry.

 

 

 

5) Indicate any shifts as well as the major product:

 

 

 

 

 

 

 

Answers

1) This is a basic electrophilic hydration.

2) The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).

3) A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed.

4) This reaction will have poor yields due to a very unstable intermediate. For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp² hybridization, and when the water is added on, the carbons change their hybridization to sp³. This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic.

5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes.

References

1. Vollhardt and Schore. Organic Chemistry, Structure and Function- Fifth Edition. New York: W. H. Freeman and Company, 2007.
2. Krow, Grant. "Sulfuric Acid." Encyclopedia of Reagents for Organic Synthesis. Philadelphia, Pennsylvania: John Wiley & Sons, 2001.