Chapter 18, Problem 18.55P

Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism of a given reaction in the acidic medium is to be drawn and the major organic product is to be predicted.

Concept introduction:

When an aldehyde or ketone is treated with an alcohol under acidic conditions, the hemiacetal product is formed. By using an excess amount of alcohol under acidic conditions, the nucleophilic addition produces hemiacetal, which further forms an acetal. The acetal has two alkoxy groups that are bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. These types of reactions are carried forward by the proton transfer and nucleophilic addition on the carbonyl carbon. An acetal is produced under acidic conditions by a ketone or aldehyde but not under basic conditions because the nucleophilic substitution requires the leaving group to be ${\text{OH}}^{\text{-}}$ which is an unsuitable leaving group for an ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 18.55P

The complete, detailed mechanism of the given reaction in the acidic medium is shown below and an acetal is the major product.

Explanation of Solution

The given reaction is

This is an acetal formation reaction in which the reaction is catalyzed by sulfuric acid and the excess ethanol acts as the nucleophile.

First three steps are acid catalyzed nucleophilic addition reactions on the ketone or aldehyde. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes carbonyl carbon more electropositive for nucleophilic addition.

Next, the weak nucleophile, alcohol, attacks the activated electrophilic carbon by nucleophilic addition reaction.

In the next step, deprotonation produces the uncharged hemiacetal.

The remaining steps essentially make up ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in protonation step by generating a good ${\text{H}}_{\text{2}}\text{O}$ leaving group.

The ${\text{H}}_{\text{2}}\text{O}$ leaving group departs and forms resonance stabilized carbocation.

The resonance stabilized carbocation is further attacked by the ethyl alcohol nucleophile, which produces positively charged acetal.

In the last step, the deprotonation of charged acetal by alcohol results in uncharged acetal

formation. Acetal is the major product of the given aldehyde.

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is the major product.

Conclusion

The complete, detailed mechanism of the given reaction under acidic medium and excess alcohol is drawn.

Interpretation Introduction

(b)

The complete, detailed mechanism of a given reaction in the acidic medium is to be drawn and the major organic product is to be predicted.

Concept introduction:

When an aldehyde or ketone is treated with an alcohol under acidic conditions, the hemiacetal product is formed. By using an excess amount of alcohol under acidic conditions, the nucleophilic addition produces hemiacetal, which further forms an acetal. The acetal has two alkoxy groups that are bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reaction is carried forward by proton transfer and nucleophilic addition on the carbonyl carbon. An acetal is produced under acidic conditions by a ketone or aldehyde but not under basic conditions because the nucleophilic substitution requires the leaving group ${\text{OH}}^{\text{-}}$ which is an unsuitable leaving group for an ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 18.55P

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Explanation of Solution

The given reaction is

This is an acetal formation reaction in which the reaction is catalyzed by sulfuric acid and the excess alcohol ($\text{ethane-1, 2-diol}$) acts as the nucleophile.

First three steps are to the acid catalyze nucleophilic addition reactions on the ketone or aldehyde. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes carbonyl carbon more electropositive for nucleophilic addition.

Next, the weak nucleophile, $\text{ethane-1, 2-diol}$ attacks on the activated electrophilic carbon by nucleophilic addition reaction.

In the next step, deprotonation produces the uncharged hemiacetal.

The remaining steps essentially make up a ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in protonation step by generating a good ${\text{H}}_{\text{2}}\text{O}$ leaving group.

The ${\text{H}}_{\text{2}}\text{O}$ leaving group departs and form resonance stabilized carbocation.

The resonance stabilized carbocation further attacked by the $\text{ethane-1, 2-diol}$ nucleophile, which produced positively charged acetal.

In the last step, the deprotonation of charged acetal by alcohol results in the uncharged acetal formation. Acetal is the major product of the given aldehyde.

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Conclusion

The complete, detailed mechanism of given reaction under acidic medium and excess alcohol is drawn.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism of a given reaction in the acidic medium is to be drawn and major organic product is to be predicted.

Concept introduction:

When an aldehyde or ketone is treated with an alcohol under acidic conditions, the hemiacetal product is formed. By using an excess amount of alcohol under acidic conditions, the nucleophilic addition produces hemiacetal, which further forms an acetal. The acetal has two alkoxy groups are bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reactions carried forward by the proton transfer and nucleophilic addition on the carbonyl carbon. An acetal produced under acidic conditions by a ketone or aldehyde but not under basic conditions because the nucleophilic substitution that requires the leaving group to be ${\text{OH}}^{\text{-}}$. It is an unsuitable leaving group for an ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 18.55P

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Explanation of Solution

The given reaction is

This is an acetal formation reaction in which the reaction is catalyzed by sulfuric acid and the excess methanol acts as the nucleophile.

First three steps are to the acid catalyze nucleophilic addition reactions on the ketone or aldehyde. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes carbonyl carbon more electropositive for nucleophilic addition.

Next, the weak nucleophile, alcohol attacks on the activated electrophilic carbon by nucleophilic addition reaction.

In the next step, deprotonation produces the uncharged hemiacetal.

The remaining steps essentially make up a ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in protonation step by generating a good ${\text{H}}_{\text{2}}\text{O}$ leaving group.

The ${\text{H}}_{\text{2}}\text{O}$ leaving group departs and form resonance stabilized carbocation.

The resonance stabilized carbocation further attacked by the methyl alcohol nucleophile, which produced positively charged acetal.

In the last step, the deprotonation of charged acetal by alcohol results in the uncharged acetal formation. Acetal is the major product of the given aldehyde.

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Conclusion

The complete, detailed mechanism of given reaction under acidic medium and excess alcohol is drawn.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism of a given reaction in the acidic medium is to be drawn and major organic product is to be predicted.

Concept introduction:

When an aldehyde or ketone is treated with an alcohol under acidic conditions, the hemiacetal product is formed. By using an excess amount of alcohol under acidic conditions, the nucleophilic addition produces hemiacetal, which further forms an acetal. The acetal has two alkoxy groups are bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reactions carried forward by the proton transfer and nucleophilic addition on the carbonyl carbon. An acetal produced under acidic conditions by a ketone or aldehyde but not under basic conditions because the nucleophilic substitution that requires the leaving group to be ${\text{OH}}^{\text{-}}$. It is an unsuitable leaving group for an ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 18.55P

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Explanation of Solution

The given reaction is

This is an acetal formation reaction in which the reaction is catalyzed by sulfuric acid and the excess $\text{3-sulfanylpropan-1-ol}$ acts as the nucleophile.

First three steps are to the acid catalyze nucleophilic addition reactions on the ketone or aldehyde. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes carbonyl carbon more electropositive for nucleophilic addition.

Next, the weak nucleophile, thiols attacks on the activated electrophilic carbon by nucleophilic addition reaction.

In the next step, deprotonation produces the uncharged hemiacetal.

The remaining steps essentially make up a ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in protonation step by generating a good ${\text{H}}_{\text{2}}\text{O}$ leaving group.

The ${\text{H}}_{\text{2}}\text{O}$ leaving group departs and form resonance stabilized carbocation.

The resonance stabilized carbocation further attacked by the alcohol nucleophile, which produced positively charged acetal.

In the last step, the deprotonation of charged acetal by alcohol results in the uncharged acetal formation. Acetal is the major product of the given aldehyde.

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Conclusion

The complete, detailed mechanism of given reaction under acidic medium and excess alcohol is drawn.

Interpretation Introduction

(e)

Interpretation:

The complete, detailed mechanism of a given reaction in the acidic medium is to be drawn and major organic product is to be predicted.

Concept introduction:

When an aldehyde or ketone is treated with an alcohol under acidic conditions, the hemiacetal product is formed. By using an excess amount of alcohol under acidic conditions, the nucleophilic addition produces hemiacetal, which further forms an acetal. The acetal has two alkoxy groups are bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reactions carried forward by the proton transfer and nucleophilic addition on the carbonyl carbon. An acetal produced under acidic conditions by a ketone or aldehyde but not under basic conditions because the nucleophilic substitution that requires the leaving group to be ${\text{OH}}^{\text{-}}$. It is an unsuitable leaving group for an ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 18.55P

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Explanation of Solution

The given reaction is

This is an acetal formation reaction in which the reaction is catalyzed by sulfuric acid and the excess $\text{propane-1, 3-diol}$ acts as the nucleophile.

First three steps are to the acid catalyze nucleophilic addition reactions on the ketone or aldehyde. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes carbonyl carbon more electropositive for nucleophilic addition.

Next, the weak nucleophile, $\text{propane-1, 3-diol}$ attacks on the activated electrophilic carbon by nucleophilic addition reaction.

In the next step, deprotonation produces the uncharged hemiacetal.

The remaining steps essentially make up a ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in protonation step by generating a good ${\text{H}}_{\text{2}}\text{O}$ leaving group.

The ${\text{H}}_{\text{2}}\text{O}$ leaving group departs and form resonance stabilized carbocation.

The resonance stabilized carbocation further attacked by the alcohol nucleophile, which produced positively charged acetal.

In the last step, the deprotonation of charged acetal by alcohol results in the uncharged acetal formation. Acetal is the major product of the given aldehyde.

The complete, detailed mechanism of a given reaction in the acidic medium is shown below and an acetal is a major product.

Conclusion

The complete, detailed mechanism of given reaction under acidic medium and excess alcohol is drawn.