Chapter 18, Problem 18.57P

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

When a small amount of a strong acid is added to ketone or an aldehyde, the uncharged nucleophile ammonia (${\text{NH}}_{\text{3}}$) and amines ${\text{RNH}}_{\text{2}}$ can be added reversibly to the carbonyl group. The product is an imine, which has characteristic $\text{C=N}$ double bond. In imine formation, the nucleophile adds in the first step because ammonia is a stronger nuclophile than alcohol. The carbonyl carbon is not activated in mildly acidic conditions. In imine formation, N is deprotonated, therefore, it follows the $\text{E1}$ mechanism. Imine formation requires at least two hydrogen on the nucleophilic N atom.

Answer to Problem 18.57P

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 imine formation reaction in which the reaction is catalyzed under mildly acidic condition and the excess ammonia acts as the nucleophile. Ammonia attacks the $\text{C=O}$ group that has not been protonated.

First three steps are the nucleophilic addition reactions on the ketone. In the first step, nucleophile ${\text{NH}}_{\text{3}}$ attacks the $\text{C=O}$ electropositive carbon. The ${\text{NH}}_{\text{3}}$ nucleophile adds to the carbonyl carbon because ${\text{NH}}_{\text{3}}$ is a stronger nucleophile than alcohol.

In the next step, deprotonation takes place by ammonia ${\text{NH}}_{\text{3}}$ group under weakly acidic conditions.

Next, negativively charged oxygen ion abstracts the proton from the ammonium ion.

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.

Ammonia nucleophile abstracts the proton from the amino group and forms imine as the product.

The complete, detailed mechanism of the given reaction under mildly acidic medium is shown below and an imine is the major product.

Conclusion

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

Interpretation Introduction

(b)

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 a small amount of a strong acid is added to ketone or an aldehyde, the uncharged nucleophile ammonia (${\text{NH}}_{\text{3}}$) and amines ${\text{RNH}}_{\text{2}}$ can be added reversibly to the carbonyl group. The product is an imine, which has characteristic $\text{C=N}$ double bond. In imine formation, the nucleophile adds in the first step because ammonia is a stronger nuclophile than alcohol. The carbonyl carbon is not activated in mildly acidic conditions. In imine formation, N is deprotonated, therefore, it follows the $\text{E1}$ mechanism. Imine formation requires at least two hydrogen on the nucleophilic N atom.

Answer to Problem 18.57P

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 imine formation reaction in which the reaction is catalyzed under mildly acidic condition and excess methylamine acts as the nucleophile. ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ attacks the $\text{C=O}$ group that has not been protonated.

First three steps are nucleophilic addition reactions on ketone. In the first step, nucleophile ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ attacks the $\text{C=O}$ electropositive carbon. ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ nucleophile adds to the carbonyl carbon because ${\text{NH}}_{\text{3}}$ is a stronger nucleophile than alcohol.

In the next step, deprotonation takes place by the ammonia ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ group under weakly acidic conditions.

Next, negativively charged oxygen ion abstracts the proton from methylamino ion.

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 a resonance stabilized carbocation.

${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ nucleophile abstracts the proton from the amino group and forms imine as the product.

The complete, detailed mechanism of the given reaction under mildly acidic medium is shown below and an imine is the major product.

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

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

Concept introduction:

When a small amount of a strong acid is added to ketone or an aldehyde, the uncharged nucleophile ammonia (${\text{NH}}_{\text{3}}$) and amines ${\text{RNH}}_{\text{2}}$ can be added reversibly to carbonyl group. The product is an imine, which has characteristic $\text{C=N}$ double bond. In imine formation, nucleophile adds in the first step because ammonia is a stronger nuclophile than alcohol. The carbonyl carbon is not activated in the mildly acidic conditions. In imine formation, N is deprotonated, therefore it follows the $\text{E1}$ mechanism. In the hydrolysis of imine reaction, ketone is the major product followed by proton transfer and nucleophilic addition of water on the $\text{C=N}$ double bond.

Answer to Problem 18.57P

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

Explanation of Solution

The given reaction is

This is an imine hydrolysis reaction under strongly acidic condition to produce ketone as the major product.

In the first step, the $\text{C=N}$ group is protonated in acidic condition, which makes carbonyl carbon more electropositive for nucleophilic addition.

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

In the next step, deprotonation takes place by the nucleophile produced $\text{OH}$ group.

The lone pairs on $\text{NH}$ group accept the proton in the protonation step by generating a good ${\text{RNH}}_{\text{2}}$ leaving group.

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

The nucleophile water abstracts the proton in the resonance stabilized carbocation and produces ketone as the major product.

The complete, detailed mechanism of the given reaction under strongly acidic medium is shown below and ketone is the major product.

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

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

Concept introduction:

When a small amount of a strong acid is added to ketone or an aldehyde, the uncharged nucleophile ammonia (${\text{NH}}_{\text{3}}$) and amines ${\text{RNH}}_{\text{2}}$ can be added reversibly to the carbonyl group. The product is an imine, which has characteristic $\text{C=N}$ double bond. In imine formation, the nucleophile adds in the first step because ammonia is a stronger nuclophile than alcohol. The carbonyl carbon is not activated in mildly acidic conditions. An enamine underogoes reaction at the alpha carbon; it can be hydrolyzed back to the ketone or aldehyde when treated with large amounts of water under acidic conditions.

Answer to Problem 18.57P

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

Explanation of Solution

The given reaction is

This is an enamine formation reaction in which the reaction is catalyzed under mildly acidic conditions, and the secondary ${\text{R}}_{\text{2}}\text{NH}$ acts as the nucleophile. ${\text{R}}_{\text{2}}\text{NH}$ attacks the $\text{C=O}$ group that has not been protonated because the N atom in the amine does not have two hydrogen atoms. The second proton is lost from the carbon in the final step to produce enamine.

First three steps are nucleophilic addition reactions on ketone. In the first step, nucleophile ${\text{CH}}_{\text{3}}{\text{NHCH}}_{\text{3}}$ attacks $\text{C=O}$ electropositive carbon. ${\text{CH}}_{\text{3}}{\text{NHCH}}_{\text{3}}$ nucleophile adds to the carbonyl carbon because ${\text{R}}_{\text{2}}\text{NH}$ is a stronger nucleophile than alcohol.

In the next step, deprotonation takes place by ${\text{CH}}_{\text{3}}{\text{NHCH}}_{\text{3}}$ group under weakly acidic conditions.

Next, negatively charged oxygen ion abstracts the proton from dimethylamino ion.

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.

${\text{CH}}_{\text{3}}{\text{NHCH}}_{\text{3}}$ nucleophile abstracts the proton from the carbon ring and from enamine as the product.

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

Conclusion

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

Interpretation Introduction

(e)

Interpretation:

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

Concept introduction:

When a small amount of a strong acid is added to ketone or an aldehyde, the uncharged nucleophile ${\text{R}}_{\text{2}}\text{NH}$ can be added reversibly to the carbonyl group. The product is an enamine. The amino group is attached to an alkene carbon. In enamine formation, nucleophile ${\text{R}}_{\text{2}}\text{NH}$ adds in the first step because ${\text{R}}_{\text{2}}\text{NH}$ is a stronger nuclophile than alcohol. The carbonyl carbon is not activated in mildly acidic conditions. In the hydrolysis of enamine reaction, ketone is the major product followed by proton transfer and nucleophilic addition of water on ${\text{R}}_{\text{2}}\text{N-C}$ carbon atom.

Answer to Problem 18.57P

The complete, detailed mechanism of the given reaction under strongly acidic medium is shown below and ketone is the major product.

Explanation of Solution

The given reaction is

This is an enamine hydrolysis reaction under strongly acidic condition to produce ketone as the major product.

In the first step, $\text{C=C}$ pi electrons abstract the proton under strong acidic conditions which produces the strong electropositive $\text{C=N}$ carbon atom.

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

In the next step, deprotonation takes place by the nucleophile produced $\text{OH}$ group.

The lone pairs on $\text{N}$ atom accept the proton in protonation step by generating a good ${\text{R}}_{\text{2}}\text{NH}$ leaving group.

${\text{R}}_{\text{2}}\text{NH}$ leaving group departs and forms a resonance stabilized carbocation.

The nucleophile water abstracts the proton in the resonance stabilized carbocation and produces ketone as the major product.

The complete, detailed mechanism of the 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 strongly acidic medium and excess water is drawn.