Chapter 11, Problem 11.11P

Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn and the product is to be predicted.

Concept introduction:

A weak acid can add to an alkene in the presence of a strong acid. Because of the leveling effect, the protonated form of a weak acid is the strongest acid that can exist, so the strong acid protonates the weak acid. This protonated form of the weak acid is a good electrophile because of positive charge. The double bond in the alkene is an electron-rich region and behaves as a nucleophile. The $\text{π}$ electron pair from the double bond breaks heterolytically extracting the proton from the protonated form of the weak acid. The product of this step is the most stable carbocation. In the next step, the oxygen from the weak acid behaves as a nucleophile and forms a bond with the carbocation using one of its lone pairs. This forms a protonated form of the addition product. A final deprotonation step with another molecule of the weak acid gives the final product.

Answer to Problem 11.11P

The complete mechanism of the given addition reaction is

The product of the reaction is

Explanation of Solution

The given addition reaction is

Water is a weak acid and does not add to the alkene in neutral conditions. In the presence of the strong acid HCl, water is protonated to ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$, a strong electrophile because of the positive charge. The $\text{π}$ bond of cyclohexene breaks heterolytically and attacks the electrophile ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$. The proton is extracted and adds to one carbon while the other carbon from the double bond is converted to a carbocation.

This is the electrophilic addition step. In the next step, a molecule of water acts as a nucleophile and attacks the carbocation to form protonated alcohol.

Final deprotonation by another molecule of water gives the final product, cyclohexanol.

Thus, the complete mechanism for the reaction can be drawn as

And the product of the reaction is cyclohexanol

Conclusion

Weak Bronsted acids can add to an alkene in the presence of a strong acid.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn and the product is to be predicted.

Concept introduction:

A weak acid can add to an alkene in the presence of a strong acid. Because of the leveling effect, the protonated form of a weak acid is the strongest acid that can exist, so the strong acid protonates the weak acid. This protonated form of the weak acid is a good electrophile because of the positive charge. The double bond in the alkene is an electron-rich region and behaves as a nucleophile. The $\text{π}$ electron pair from the double bond breaks heterolytically, extracting the proton from the protonated form of the weak acid. The product of this step is the most stable carbocation. In the next step, the oxygen from the weak acid behaves as a nucleophile and forms a bond with the carbocation using one of its lone pairs. This forms a protonated form of the addition product. A final deprotonation step with another molecule of the weak acid gives the final product.

Answer to Problem 11.11P

The complete mechanism for the given addition reaction is

The product of the reaction is

Explanation of Solution

The given addition reaction is

In the presence of HCl, the weak acid ${\text{H}}_{\text{2}}\text{O}$ is converted to the strong electrophile ${\text{H}}_3{\text{O}}^{+}$. The $\text{π}$ bonds of the benzene rings are too stable to act as a nucleophile. However, the alkene double bond is also electron-rich and can behave as a nucleophile. It uses the $\text{π}$ bond pair to extract a proton from ${\text{H}}_3{\text{O}}^{+}$ in an electrophile addition step. The result is a carbocation. Because of the symmetry of the substrate, there is no preferred carbocation.

In the next step, a molecule of water acts as a nucleophile and adds to the carbocation to form a protonated alcohol.

In the final step, another molecule of water deprotonates to give the final product, $\text{1, 2-diphenylethan-1-ol}$.

Thus, the complete mechanism for this addition reaction can be drawn as

And the product of the reaction is

Conclusion

Weak Bronsted acids can add to an alkene in the presence of a strong acid.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn and the product is to be predicted.

Concept introduction:

A weak acid can add to an alkene in the presence of a strong acid. Because of the leveling effect, the protonated form of a weak acid is the strongest acid that can exist, so the strong acid protonates the weak acid. This protonated form of the weak acid is a good electrophile because of the positive charge. The double bond in the alkene is an electron-rich region and behaves as a nucleophile. The $\text{π}$ electron pair from the double bond breaks heterolytically, extracting the proton from the protonated form of the weak acid. The product of this step is the most stable carbocation. In the next step, the oxygen from the weak acid behaves as a nucleophile and forms a bond with the carbocation using one of its lone pairs. This forms a protonated form of the addition product. A final deprotonation step with another molecule of the weak acid gives the final product.

Answer to Problem 11.11P

The complete mechanism for the given addition reaction is

The product of the reaction is

Explanation of Solution

The given reaction is

In the presence of a strong acid ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ethanol is protonated to strongly electrophilic ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{\text{+}}{\text{H}}_{\text{2}}$. The double bond in the alkene is an electron-rich region and acts as a nucleophile. The $\text{π}$ bond pair is used to extract a proton from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{\text{+}}{\text{H}}_{\text{2}}$ in the electrophile addition reaction. The result is the formation of a carbocation. The carbocation has a positive charge on the tertiary carbon as that is more stable than the alternative secondary carbocation.

In the next step, a molecule of ethanol will act as a nucleophile and form a bond with the carbocation, using a lone pair on oxygen.

This results in the formation of protonated ether, which is deprotonated by another molecule of ethanol in the final step.

Thus, the complete mechanism for this addition reaction can be drawn as

And the product of the reaction is $\text{2-ethoxy-2-methylbutane}$

Conclusion

Weak Bronsted acids can add to an alkene in the presence of a strong acid.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism for the given reaction is to be drawn and the product is to be predicted.

Concept introduction:

A weak acid can add to an alkene in the presence of a strong acid. Because of the leveling effect, the protonated form of a weak acid is the strongest acid that can exist, so the strong acid protonates the weak acid. This protonated form of the weak acid is a good electrophile because of the positive charge. The double bond in the alkene is an electron-rich region and behaves as a nucleophile. The $\text{π}$ electron pair from the double bond breaks heterolytically, extracting the proton from the protonated form of the weak acid. The product of this step is the most stable carbocation. In the next step, the oxygen from the weak acid behaves as a nucleophile and forms a bond with the carbocation using one of its lone pairs. This forms a protonated form of the addition product. A final deprotonation step with another molecule of the weak acid gives the final product.

Answer to Problem 11.11P

The complete mechanism for the reaction is

The product of the reaction is

Explanation of Solution

The given reaction is

In the presence of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, acetic acid is protonated to strongly electrophilic species ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{+}}{\text{H}}_{\text{2}}$. The double bond in the substrate is an electron-rich region and acts as a nucleophile. The $\text{π}$ bond pair deprotonates ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{+}}{\text{H}}_{\text{2}}$ and forms a carbocation. The proton will preferentially add to the less substituted carbon from the double bond so that the more stable tertiary carbocation is formed.

In the next step, a molecule of acetic acid will act as a nucleophile using a lone pair on OH oxygen to form a bond with the carbocation. The result is a protonated form of the ester product.

Final deprotonation by another molecule of acetic acid will give the product.

Thus, the complete mechanism for the reaction can be drawn as

The product of the reaction is $\text{1-methylcyclopentyl acetate}$.

Conclusion

Weak Bronsted acids can add to an alkene in the presence of a strong acid.