Chapter 21, Problem 21.60P

Interpretation Introduction

(a)

Interpretation:

The mechanism and the major organic product for the given reaction are to be drawn.

Concept introduction:

${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ and $\text{MCPBA}$ are acids and will react with a ketone in a Baeyer-Villiger oxidation. Baeyer–Villiger oxidation produces carboxylic acids from aldehydes, and esters from ketones. In these reactions, a hydrogen or alkyl group departs from the carbonyl carbon, facilitated by the breaking of the weak $\text{O-O}$ bond from a peroxyacid $\left({\text{RCO}}_{\text{3}}\text{H}\right)$. In an unsymmetric ketone or aldehyde, the major product depends on which group leaves from carbonyl C. The ability of a group to do so is called its migratory aptitude.

Migratory Aptitude in a Baeyer–Villiger Oxidation:

$\text{Methyl group < 1° Alkyl group < 2° Alkyl group }\approx \text{ Aryl group < 3° Alkyl group < H}$

Answer to Problem 21.60P

The mechanism and the major organic product for the given reaction are:

Explanation of Solution

The given reaction is:

The ketone, substrate in the reaction, is asymmetrical ketone. carbonyl C is bonded to a primary alkyl group and an aryl group. According to migratory aptitude, the aryl group has greater migratory aptitude, so its bond will prefentially break.

In this reaction, an O atom from the acid is inserted between carbonyl C and phenyl group, initially bonded to the carbonyl C. C=O is activated by ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ acid. The weak peroxyacid $\text{MCPBA}$ attacks C=O in a nucleophilic addition step. A proton is removed by the conjugate base ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}{}^{\text{-}}$ in the step $\text{3}$. Step $\text{4}$ is a nucleophilic elimination step in which the phenyl group leaves at the same time as it forms a bond to the O of $\text{MCPBA}$. The weak bond O-O bond of MCPBA is simultaneously broken in step $\text{4}$. The final step is a proton transfer that yields the uncharged ester as the product.

The complete mechanism and the ester formed as a product for the reaction are:

Conclusion

The mechanism and the product for the reaction are drawn on the basis of the given reaction conditions.

Interpretation Introduction

(b)

Interpretation:

The mechanism and the major organic product for the given reaction are to be drawn.

Concept introduction:

${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ and $\text{MCPBA}$ are acids and will react with a ketone in a Baeyer-Villiger oxidation. The Baeyer–Villiger oxidation produces carboxylic acids from aldehydes, and esters from ketones. In these reactions, a hydrogen or alkyl group departs from the carbonyl carbon, facilitated by the breaking of the weak $\text{O-O}$ bond from a peroxyacid $\left({\text{RCO}}_{\text{3}}\text{H}\right)$. In an unsymmetric ketone or aldehyde, the major product depends on which group leaves from the carbonyl C. The ability of a group to do so is called its migratory aptitude.

Migratory Aptitude in a Baeyer–Villiger Oxidation:

$\text{Methyl group < 1° Alkyl group < 2° Alkyl group }\approx \text{ Aryl group < 3° Alkyl group < H}$

Answer to Problem 21.60P

The mechanism and the major organic product for the given reaction are:

Explanation of Solution

The given reaction is:

The ketone, substrate in the reaction, is asymmetrical ketone. carbonyl C is bonded to a H atom and an aryl group. According to the migratory aptitude, the H atom has greater migratory aptitude, so its bond will prefentially break.

In this reaction, an O atom from the acid is inserted between carbonyl C and the H atom, initially bonded to carbonyl C. C=O is activated by ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ acid. The weak peroxyacid $\text{MCPBA}$ attacks the C=O in a nucleophilic addition step. A proton is removed by the conjugate base ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}{}^{\text{-}}$, in the step $\text{3}$. Step $\text{4}$ is a nucleophilic elimination step in which the phenyl group leaves at the same time as it forms a bond to the O of $\text{MCPBA}$. The weak O-O bond of MCPBA is simultaneously broken in step $\text{4}$. The final step is a proton transfers to yield uncharged ester as the product.

The complete mechanism and the ester formed as a product for the reaction are:

Conclusion

The mechanism and the product for the reaction are drawn on the basis of given reaction conditions.

Interpretation Introduction

(c)

Interpretation:

The mechanism and the major organic product for the given reaction are to be drawn.

Concept introduction:

${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ and $\text{MCPBA}$ are acids and will react with a ketone in a Baeyer-Villiger oxidation. The Baeyer–Villiger oxidation produces carboxylic acids from aldehydes, and esters from ketones. In these reactions, a hydrogen or alkyl group departs from the carbonyl carbon, facilitated by the breaking of the weak $\text{O-O}$ bond from a peroxyacid $\left({\text{RCO}}_{\text{3}}\text{H}\right)$. In an unsymmetric ketone or aldehyde, the major product depends on which group leaves from the carbonyl C. The ability of a group to do so is called its migratory aptitude.

Migratory Aptitude in a Baeyer–Villiger Oxidation:

$\text{Methyl group < 1° Alkyl group < 2° Alkyl group }\approx \text{ Aryl group < 3° Alkyl group < H}$

Answer to Problem 21.60P

The mechanism and the major organic product for the given reaction are:

Explanation of Solution

The given reaction is:

The ketone, substrate in the reaction, is asymmetrical ketone. carbonyl C is bonded to a primary alkyl group and tertiary alkyl group. According to migratory aptitude, the tertiary alkyl group has greater migratory aptitude, so its bond will prefentially break.

In this reaction an O atom from the acid is inserted between carbonyl C and the tertiary alkyl group, initially bonded to carbonyl C. C=O is activated by ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ acid. The weak peroxyacid $\text{MCPBA}$ attacks C=O in a nucleophilic addition step. A proton is removed by the conjugate base ${\text{CF}}_{\text{3}}{\text{CO}}_{\text{3}}{}^{\text{-}}$, in the step $\text{3}$. Step $\text{4}$ is a nucleophilic elimination step in which the phenyl group leaves at the same time as it forms a bond to the O of $\text{MCPBA}$. The weak O-O bond of MCPBA is simultaneously broken in step $\text{4}$. The final step is a proton transfers to yields the uncharged ester as product.

The complete mechanism and the ester formed as a product for the reaction are:

Conclusion

The mechanism and the product for the reaction are drawn on the basis of given reaction conditions.