EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393630817
Author: KARTY
Publisher: W.W.NORTON+CO. (CC)
Question
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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.

Expert Solution
Check Mark

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 21, Problem 21.14P , additional homework tip  1

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 C atom yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the O atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded 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 C=O bond is reformed, and in Step 6, the carbonyl 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. 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.

Expert Solution
Check Mark

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 21, Problem 21.14P , additional homework tip  2

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the ester’s carbonyl group, and in Step 2, Cyclohexanol attacks the carbonyl C atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the O atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded 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 C=O bond is reformed, and in Step 6, the carbonyl 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.

Expert Solution
Check Mark

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 21, Problem 21.14P , additional homework tip  3

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 C atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the O atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded 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 C=O bond is reformed, and in Step 6, the carbonyl 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.

Expert Solution
Check Mark

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 21, Problem 21.14P , additional homework tip  4

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 C atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the O atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded 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 C=O bond is reformed, and in Step 6, the carbonyl 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. 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.

Expert Solution
Check Mark

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 21, Problem 21.14P , additional homework tip  5

Explanation of Solution

In Step 1 of the mechanism, the acid protonates the acid’s carbonyl group, and in Step 2, Ethanol attacks the carbonyl C atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the O atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded 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 C=O bond is reformed, and in Step 6, the carbonyl 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.

Expert Solution
Check Mark

Answer to Problem 21.14P

The mechanism and product for the given reaction is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 21, Problem 21.14P , additional homework tip  6

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 C atom, yielding a tetrahedral intermediate. Steps 3 and 4 are proton transfers. In Step 3, the O atom from the original nucleophile becomes uncharged, and thus is stabilized. In Step 4, the singly bonded 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 C=O bond is reformed, and in Step 6, the carbonyl 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.

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Chapter 21 Solutions

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M

Ch. 21 - Prob. 21.11PCh. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Prob. 21.21PCh. 21 - Prob. 21.22PCh. 21 - Prob. 21.23PCh. 21 - Prob. 21.24PCh. 21 - Prob. 21.25PCh. 21 - Prob. 21.26PCh. 21 - Prob. 21.27PCh. 21 - Prob. 21.28PCh. 21 - Prob. 21.29PCh. 21 - Prob. 21.30PCh. 21 - Prob. 21.31PCh. 21 - Prob. 21.32PCh. 21 - Prob. 21.33PCh. 21 - Prob. 21.34PCh. 21 - Prob. 21.35PCh. 21 - Prob. 21.36PCh. 21 - Prob. 21.37PCh. 21 - Prob. 21.38PCh. 21 - Prob. 21.39PCh. 21 - Prob. 21.40PCh. 21 - Prob. 21.41PCh. 21 - Prob. 21.42PCh. 21 - Prob. 21.43PCh. 21 - Prob. 21.44PCh. 21 - Prob. 21.45PCh. 21 - Prob. 21.46PCh. 21 - Prob. 21.47PCh. 21 - Prob. 21.48PCh. 21 - Prob. 21.49PCh. 21 - Prob. 21.50PCh. 21 - Prob. 21.51PCh. 21 - Prob. 21.52PCh. 21 - Prob. 21.53PCh. 21 - Prob. 21.54PCh. 21 - Prob. 21.55PCh. 21 - Prob. 21.56PCh. 21 - Prob. 21.57PCh. 21 - Prob. 21.58PCh. 21 - Prob. 21.59PCh. 21 - Prob. 21.60PCh. 21 - Prob. 21.61PCh. 21 - Prob. 21.62PCh. 21 - Prob. 21.63PCh. 21 - Prob. 21.64PCh. 21 - Prob. 21.65PCh. 21 - Prob. 21.66PCh. 21 - Prob. 21.67PCh. 21 - Prob. 21.68PCh. 21 - Prob. 21.69PCh. 21 - Prob. 21.70PCh. 21 - Prob. 21.71PCh. 21 - Prob. 21.72PCh. 21 - Prob. 21.73PCh. 21 - Prob. 21.74PCh. 21 - Prob. 21.75PCh. 21 - Prob. 21.76PCh. 21 - Prob. 21.77PCh. 21 - Prob. 21.78PCh. 21 - Prob. 21.79PCh. 21 - Prob. 21.80PCh. 21 - Prob. 21.81PCh. 21 - Prob. 21.82PCh. 21 - Prob. 21.83PCh. 21 - Prob. 21.84PCh. 21 - Prob. 21.85PCh. 21 - Prob. 21.86PCh. 21 - Prob. 21.87PCh. 21 - Prob. 21.88PCh. 21 - Prob. 21.89PCh. 21 - Prob. 21.90PCh. 21 - Prob. 21.91PCh. 21 - Prob. 21.92PCh. 21 - Prob. 21.93PCh. 21 - Prob. 21.94PCh. 21 - Prob. 21.95PCh. 21 - Prob. 21.96PCh. 21 - Prob. 21.97PCh. 21 - Prob. 21.98PCh. 21 - Prob. 21.1YTCh. 21 - Prob. 21.2YTCh. 21 - Prob. 21.3YTCh. 21 - Prob. 21.4YTCh. 21 - Prob. 21.5YTCh. 21 - Prob. 21.6YTCh. 21 - Prob. 21.7YTCh. 21 - Prob. 21.8YTCh. 21 - Prob. 21.9YTCh. 21 - Prob. 21.10YTCh. 21 - Prob. 21.11YTCh. 21 - Prob. 21.12YT
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