ORG.CHEM W/TEXT+SOLU.MANUAL
ORG.CHEM W/TEXT+SOLU.MANUAL
15th Edition
ISBN: 9780393252125
Author: KARTY
Publisher: W.W.NORTON+CO.
Question
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Chapter 21, Problem 21.39P
Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism and the products are to be drawn for the given reaction.

Concept introduction:

Carboxylic acid derivatives undergo acyl group substitution reactions when treated with appropriate nucleophiles. The reaction occurs via nucleophilic addition-elimination involving a tetrahedral intermediate. It may also involve proton transfer step(s), particularly when the nucleophile being added in the first step is not a strong nucleophile. The reaction occurs if the possible product is more stable than the reactant. If the two are of comparable stability, the reaction will occur reversibly. The order of increasing stability of acid derivatives is

acid chloride < acid anhydride < ester  carboxylic acid < amide < carboxylate anion

Expert Solution
Check Mark

Answer to Problem 21.39P

The complete mechanism is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  1

The product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  2

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  3

Alcohol is a weak nucleophile and adds to one of the carbonyl carbons of acetic anhydride to produce a protonated ether linkage. The carbonyl oxygen becomes negatively charged as a result of the π bond pair moving to it.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  4

In the next step, the positively charged oxygen is deprotonated by another molecule of the alcohol.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  5

The final step is nucleophilic elimination of acetate anion as a result of the lone pair of the negatively charged oxygen moving to reform the carbonyl group.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  6

Thus, the complete mechanism of the reaction can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  7

The product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  8

Conclusion

The product and mechanism of the given reaction are determined on the basis of nucleophilic addition-elimination mechanism.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism and the products are to be drawn for the given reaction.

Concept introduction:

Carboxylic acid derivatives undergo acyl group substitution reactions when treated with appropriate nucleophiles. The reaction occurs via nucleophilic addition-elimination involving a tetrahedral intermediate. It may also involve proton transfer step(s), particularly when the nucleophile being added in the first step is not a strong nucleophile. The reaction occurs if the possible product is more stable than the reactant. If the two are of comparable stability, the reaction will occur reversibly. The order of increasing stability of acid derivatives is

acid chloride < acid anhydride < ester  carboxylic acid < amide < carboxylate anion

Expert Solution
Check Mark

Answer to Problem 21.39P

The complete mechanism is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  9

The product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  10

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  11

The amine can act as both a nucleophile and a base.

In the first step, it acts as a nucleophile and adds to the carbonyl carbon of the acid chloride to produce a protonated amine linkage. The π bond pair of the carbonyl group moves to the oxygen, making it negatively charged.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  12

The protonated amine is then deprotonated by a second molecule of the amine.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  13

Finally, one lone pair on the negatively charged oxygen will move back to reform the carbonyl group, eliminating the chloride and forming the product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  14

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  15

And the product of the reaction will be

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  16

Conclusion

The product and mechanism of the given reaction are determined on the basis of nucleophilic addition-elimination mechanism.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products are to be drawn for the given reaction.

Concept introduction:

Carboxylic acid derivatives undergo acyl group substitution reactions when treated with appropriate nucleophiles. The reaction occurs via nucleophilic addition-elimination involving a tetrahedral intermediate. It may also involve proton transfer step(s), particularly when the nucleophile being added in the first step is not a strong nucleophile. The reaction occurs if the possible product is more stable than the reactant. If the two are of comparable stability, the reaction will occur reversibly. The order of increasing stability of acid derivatives is

acid chloride < acid anhydride < ester  carboxylic acid < amide < carboxylate anion

Expert Solution
Check Mark

Answer to Problem 21.39P

The complete mechanism of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  17

The product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  18

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  19

The amine will act as a nucleophile and add to the carbonyl carbon of the acid chloride in the first step to form a protonated amine linkage. The π bond pair of the carbonyl group moves to the oxygen to make it negatively charged.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  20

The protonated amine is deprotonated in the second step by the added base, pyridine.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  21

Finally, one lone pair on the negatively charged oxygen moves back to reform the carbonyl group, eliminating the leaving group chloride and forming the final product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  22

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  23

And the product of the reaction will be

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  24

Conclusion

The product and mechanism of the given reaction are determined on the basis of nucleophilic addition-elimination mechanism.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism and the products are to be drawn for the given reaction.

Concept introduction:

Carboxylic acid derivatives undergo acyl group substitution reactions when treated with appropriate nucleophiles. The reaction occurs via nucleophilic addition-elimination involving a tetrahedral intermediate. It may also involve proton transfer step(s), particularly when the nucleophile being added in the first step is not a strong nucleophile. The reaction occurs if the possible product is more stable than the reactant. If the two are of comparable stability, the reaction will occur reversibly. The order of increasing stability of acid derivatives is

acid chloride < acid anhydride < ester  carboxylic acid < amide < carboxylate anion

Expert Solution
Check Mark

Answer to Problem 21.39P

The complete mechanism of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  25

The product of the reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  26

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  27

The alcohol is a weak nucleophile. It will add to one of the carbonyl carbons in the anhydride to produce a protonated intermediate.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  28

In the seconds step, this intermediate will be deprotonated by a second molecule of the alcohol.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  29

Finally, one lone pair from the negatively charged oxygen will move back to reform the carbonyl, eliminating the leaving group benzoate in the process and forming the product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  30

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  31

And the product of the reaction will be

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.39P , additional homework tip  32

Conclusion

The product and mechanism of the given reaction are determined on the basis of nucleophilic addition-elimination mechanism.

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Chapter 21 Solutions

ORG.CHEM W/TEXT+SOLU.MANUAL

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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