Chapter 18, Problem 18.56P

Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism of the 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, a hemiacetal product is formed. Use of an excess amount of alcohol under acidic conditions and after that the nucleophilic addition produces hemiacetals, which further form an acetal. The acetal has two alkoxy groups bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reactions 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 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 18.56P

The complete, detailed mechanism of the given reaction in the acidic medium is shown below; 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 an acid, and the excess diol acts as the nucleophile.

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

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

In the next step, deprotonation produces the uncharged hemiacetal.

The remaining steps essentially make up the ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in the 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 the resonance stabilized carbocation.

The resonance stabilized carbocation is further attacked by the diol nucleophile, which produces positively charged acetal. The carbonation step is intramolecular because it forms a five-membered ring, and it is stable.

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

The complete, detailed mechanism of the given reaction in the acidic medium is shown below; 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 the 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, a hemiacetal product is formed. Use of an excess amount of alcohol under acidic conditions and after that the nucleophilic addition produces hemiacetals, which further form an acetal. The acetal has two alkoxy groups bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reactions 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 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 18.56P

The complete, detailed mechanism of the given reaction in the acidic medium is shown below; 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 diols act as the nucleophile.

First three steps are that of the acid catalyzed nucleophilic addition reactions on ketone. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes the 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 an ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on the $\text{OH}$ group accept the proton in the 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 a resonance stabilized carbocation.

The resonance stabilized carbocation is further attacked by the diol nucleophile, which produces positively charged acetal. The carbonation step is intramolecular because it forms a five-membered ring, and it is stable.

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 the given reaction in the acidic medium is shown below; 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

(c)

Interpretation:

The complete, detailed mechanism of the 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, a hemiacetal product is formed. Use of an excess amount of alcohol under acidic conditions and after that the nucleophilic addition produces hemiacetals, which further form an acetal. The acetal has two alkoxy groups bonded to the same carbon. The formation of the acetal product is favored by using excess alcohol. This type of reactions 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 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 18.56P

The complete, detailed mechanism of the given reaction in the acidic medium is shown below; 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 diol acts as the nucleophile.

First three steps are that of the acid catalyzed nucleophilic addition reactions on the aldehyde. In the first step, the $\text{C=O}$ group is protonated in acidic condition, which makes the 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 an ${\text{S}}_{\text{N}}\text{1}$ reaction. The lone pairs on $\text{OH}$ group accept the proton in the 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 a resonance stabilized carbocation.

The resonance stabilized carbocation is further attacked by the diol nucleophile, which produced positively charged acetal. The carbonation step is intramolecular because it forms a six-membered ring, and it is stable.

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

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

Conclusion

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