Chapter 21, Problem 21.14P

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

The ester is hydrolyzed to the carboxylic acid under acidic conditions, so no strong base should appear in the mechanism. Water is the nucleophile, and the leaving group is a molecule of ethanol. Proton transfer steps are incorporated to avoid the appearance of strong bases.

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the ester’s carbonyl group, and in Step 2, water attacks the carbonyl $\text{C}$ atom yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the $\text{O}$ atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded $\text{O}$ atom of the original ester gains a positive charge, which increases its leaving group ability. In Step 5, the leaving group departs and the $\text{C=O}$ bond is reformed, and in Step 6, the carbonyl $\text{O}$ is deprotonated yielding the overall uncharged product.

Conclusion

The complete, detailed mechanism and the overall product for the given reaction is drawn by acid-catalyzed ester hydrolysis.

Interpretation Introduction

(b)

Interpretation:

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

Concept introduction:

This is an acid-catalyzed transesterification, and a new ester is formed. $\text{Cyclohexanol}$ acts as the nucleophile in this acid-catalyzed nucleophilic addition–elimination mechanism, and the leaving group is a molecule of methanol. Proton transfer steps are incorporated to avoid the appearance of strong bases.

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the ester’s carbonyl group, and in Step 2, $\text{Cyclohexanol}$ attacks the carbonyl $\text{C}$ atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the $\text{O}$ atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded $\text{O}$ atom of the original ester gains a positive charge, which increases its leaving group ability. In Step 5, the leaving group departs and the $\text{C=O}$ bond is reformed, and in Step 6, the carbonyl $\text{O}$ is deprotonated, yielding the overall uncharged product.

Conclusion

The complete, detailed mechanism and the overall product for the given reaction is drawn by acid-catalyzed transesterification.

Interpretation Introduction

(c)

Interpretation:

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

Concept introduction:

This is an acid catalyzed amide hydrolysis which produces a carboxylic acid. Water acts as the nucleophile in this acid-catalyzed nucleophilic addition–elimination mechanism, and the leaving group is a molecule of amine. Proton transfer steps are incorporated to avoid the appearance of strong bases.

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the ester’s carbonyl group, and in Step 2, water attacks the carbonyl $\text{C}$ atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the $\text{O}$ atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded $\text{N}$ atom of the original ester gains a positive charge, which increases its leaving group ability. In Step 5, the leaving group departs and the $\text{C=O}$ bond is reformed, and in Step 6, the carbonyl $\text{O}$ is deprotonated, yielding the overall uncharged product.

Conclusion

The complete, detailed mechanism and the overall product for the given reaction is drawn by acid-catalyzed amide hydrolysis.

Interpretation Introduction

(d)

Interpretation:

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

Concept introduction:

This is an acid catalyzed ester hydrolysis which produces a carboxylic acid. Water acts as the nucleophile in this acid-catalyzed nucleophilic addition–elimination mechanism, and the leaving group is a molecule of alcohol. Proton transfer steps are incorporated to avoid the appearance of strong bases.

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the ester’s carbonyl group, and in Step 2, water attacks the carbonyl $\text{C}$ atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the $\text{O}$ atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded $\text{O}$ atom of the original ester gains a positive charge, which increases its leaving group ability. In Step 5, the leaving group departs and the $\text{C=O}$ bond is reformed, and in Step 6, the carbonyl $\text{O}$ is deprotonated, yielding the overall uncharged product.

Conclusion

The complete, detailed mechanism and the overall product for the given reaction is drawn by acid-catalyzed ester hydrolysis.

Interpretation Introduction

(e)

Interpretation:

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

Concept introduction:

This is Fischer esterification reaction. $\text{Ethanol}$ acts as the nucleophile in this acid-catalyzed nucleophilic addition–elimination mechanism, and the leaving group is a water molecule. Proton transfer steps are incorporated to avoid the appearance of strong bases.

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the acid’s carbonyl group, and in Step 2, $\text{Ethanol}$ attacks the carbonyl $\text{C}$ atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the $\text{O}$ atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded $\text{O}$ atom of the original ester gains a positive charge, which increases its leaving group ability. In Step 5, the leaving group departs and the $\text{C=O}$ bond is reformed, and in Step 6, the carbonyl $\text{O}$ is deprotonated, yielding the overall uncharged product.

Conclusion

The complete, detailed mechanism and the overall product for the given reaction is drawn by Fischer esterification reaction.

Interpretation Introduction

(f)

Interpretation:

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

Concept introduction:

This is an acid catalyzed amide hydrolysis which produces a carboxylic acid. Water acts as the nucleophile in this acid-catalyzed nucleophilic addition–elimination mechanism, and the leaving group is a molecule of amine. Proton transfer steps are incorporated to avoid the appearance of strong bases.

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the ester’s carbonyl group, and in Step 2, water attacks the carbonyl $\text{C}$ atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the $\text{O}$ atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded $\text{N}$ atom of the original ester gains a positive charge, which increases its leaving group ability. In Step 5, the leaving group departs and the $\text{C=O}$ bond is reformed, and in Step 6, the carbonyl $\text{O}$ is deprotonated, yielding the overall uncharged product.

Conclusion

The complete, detailed mechanism and the overall product for the given reaction is drawn by acid-catalyzed amide hydrolysis.