Chapter 12, Problem 12.52P

Interpretation Introduction

(a)

Interpretation:

The mechanism and major organic product(s) of the given reaction are to be drawn.

Concept introduction:

In the hydroboration reaction of an alkene, a molecule of water adds to the carbon-carbon double bond of the alkene in an anti-Markovnikov manner. The boron atom in borane is less electronegative than hydrogen and acts as the electron-poor atom. Therefore, the $\text{π}$ bond pair of the alkene attacks and forms a partial bond with the boron atom. Boron is bonded to the fewer substituted carbon. The hydrogen atom forms a partial bond with the carbocation. The addition is therefore against Markovnikov’s rule. A cyclic transition state is formed, leading to a syn addition. The remaining two hydrogen atoms of the $\text{B-H}$ bond are similarly replaced by the addition of another two molecules of alkenes to form a trialkylborane. Trialkylborane then reacts with the anion hydrogen peroxide (${}^{\text{-}}\text{O-O-H}$) to form a trialkylborate ester. The trialkylborate ester, on base hydrolysis, gives the final product, alcohol.

Answer to Problem 12.52P

The mechanism for the given reaction is

The major product for the given reaction is

Explanation of Solution

The given reaction is

The structure of cyclopentylethene is

In the first step, borane adds across the carbon-carbon double bond of the alkene via a four-membered cyclic transition state. The addition is anti-Markovnikov as the boron atom adds to the fewer substituted (fewer sterically hindered) carbon, and the hydrogen adds to the most substituted carbon. The remaining two hydrogen atoms from borane are replaced in a similar way by two more alkene molecules to form a trialkylborane.

A trialkylborane is then oxidized by the hydroperoxide anion (${}^{-}\text{OOH}$) to form a trialkylborate ester. Tri-substituted boron is an electron-deficient atom and acts as a Lewis acid. The hydroperoxide anion, formed by deprotonation of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ by the strong base NaOH acts as a Lewis base and forms a bond with the boron atom. Then, a borate ester is formed via a 1, 2-alkyl shift shown below. Repetition of this step two times forms a reactive trialkylborate ester.

Then the trialkylborate ester undergoes base hydrolysis to three molecules of the alcohol product and boric acid.

The ${}^{\text{-}}\text{OH}$ ion is electron-rich and coordinates with the electron-poor boron. Heterolysis of this adduct liberates the alkoxide, which extracts a proton from the water molecule to form the product alcohol. Repetition of these two steps hydrolyzes the remaining alkoxide groups from the ester.

Thus, the major product of the given reaction is

Conclusion

The mechanism and the major organic product(s) for the given reaction are drawn on the basis of an anti-Markovnikov syn addition of borane across the double bond.

Interpretation Introduction

(b)

Interpretation:

The mechanism and the major organic product(s) of the given reaction are to be drawn.

Concept introduction:

In the hydroboration reaction of an alkene, a molecule of water adds to the carbon-carbon double bond of the alkene in an anti-Markovnikov manner. The boron atom in borane is less electronegative than hydrogen and acts as the electron-poor atom. Therefore, the $\text{π}$ bond pair of the alkene attacks and forms a partial bond with the boron atom. Boron is bonded to the fewer substituted carbon. The hydrogen atom forms a partial bond with the carbocation. The addition is therefore against Markovnikov’s rule. A cyclic transition state is formed, leading to a syn addition. The remaining two hydrogen atoms of the $\text{B-H}$ bond are similarly replaced by the addition of another two molecules of alkenes to form a trialkylborane. Trialkylborane then reacts with the anion hydrogen peroxide (${}^{\text{-}}\text{O-O-H}$) to form a trialkylborate ester. The trialkylborate ester, on base hydrolysis, gives the final product, alcohol.

Answer to Problem 12.52P

The mechanism for the given reaction is

The major product of the given reaction is

Explanation of Solution

The given reaction is

In the first step, borane adds across the carbon-carbon double bond of the alkene via a four-membered cyclic transition state. The addition is anti-Markovnikov as the boron atom adds to the fewer substituted (fewer sterically hindered) carbon, and the hydrogen adds to the most substituted carbon. The remaining two hydrogen atoms from borane are replaced in a similar way by two more alkene molecules to form a trialkylborane.

The trialkylborane is then oxidized by the hydroperoxide anion (${}^{-}\text{OOH}$) to form a trialkylborate ester. Tri-substituted boron is an electron-deficient atom and acts as a Lewis acid. The hydroperoxide anion, formed by deprotonation of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ by the strong base NaOH, acts as a Lewis base and forms a bond to with the boron atom. Then, a borate ester is formed via a 1, 2-alkyl shift shown below. Repetition of this step two times forms a reactive trialkylborate ester.

Then the trialkylborate ester undergoes base hydrolysis to three molecules of the alcohol product and boric acid.

The ${}^{\text{-}}\text{OH}$ ion is electron-rich and coordinates with the electron-poor boron. Heterolysis of this adduct liberates the alkoxide, which extracts a proton the from water molecule to form the product alcohol. Repetition of these two steps hydrolyzes the remaining alkoxide groups from the ester.

Borane can add across the double bond from either above the plane of the ring or from below the plane of the ring. This will lead to the formation of a racemic mixture as the carbon bonded to the OH group is chiral.

Thus, the major product for the given reaction is

Conclusion

The mechanism and the major organic product(s) for the given reaction are drawn on the basis of an anti-Markovnikov syn addition of borane across the double bond.

Interpretation Introduction

(c)

Interpretation:

The mechanism and the major organic product(s) of the given reaction are to be drawn.

Concept introduction:

Like alkenes, alkynes also undergo hydroboration reaction but give a different product.

The addition of one molecule of borane produces an alkene, and if it is a terminal alkene, another molecule of borane can add to it. This leads to a mixture of products.

This problem can be avoided by the use of a bulky dialkylborane like disiamylborane, ${{\text{(C}}_{\text{5}}{\text{H}}_{\text{11}}\text{)}}_{\text{2}}\text{BH}$. The bulky nature of the alkyl groups prevents the addition of a second molecule. The product is an enol, which tautomerizes to a more stable aldehyde.

Answer to Problem 12.52P

The mechanism for the given reaction is

The major product for the given reaction is

Explanation of Solution

The given reaction is

The structure of disiamylborane is

It is a sterically hindered alkyl borane, having only one $\text{B-H}$ bond.

So the first step of the above reaction is a syn anti-Markovnikov electrophilic addition of boron and hydrogen to the carbon-carbon triple bond of the substrate.

Oxidation of this adduct by basic ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ produces an enol. The enol then tautomerizes to a more stable aldehyde.

Thus, the major product of the given reaction is

Conclusion

The mechanism and the major organic product(s) of the given reaction are drawn on the basis of the reaction conditions.

Interpretation Introduction

(d)

Interpretation:

The mechanism and the major organic product(s) of the given reaction are to be drawn.

Concept introduction:

Like alkenes, alkynes also undergo hydroboration reaction, but the product is different than alcohol. The oxidation of alkynes is carried out by using bulky dialkylborane like disiamylborane ${{\text{(C}}_{\text{5}}{\text{H}}_{\text{11}}\text{)}}_{\text{2}}\text{BH}$ instead of simple borane (${\text{BH}}_{\text{3}}$) to avoid the formation of a mixture of products. The product of the reaction is an enol, which tautomerizes to a more stable keto form.

Answer to Problem 12.52P

The mechanism for the given reaction is

The major product for the given reaction is

Explanation of Solution

The given reaction is

The structure of disiamylborane is

It is a sterically hindered alkyl borane having only one $\text{B-H}$ bond.

The first step of the reaction is a syn anti-Markovnikov electrophilic addition of $\text{B-H}$ bond to the carbon-carbon triple bond of the substrate. Oxidation with basic ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ produces an enol, which tautomerizes to the more stable keto form.

Thus, the major product for the given reaction is

Conclusion

The mechanism and major organic product(s) for the given reaction are drawn on the basis of the reaction conditions.