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 21.40P
Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism of a given compound synthesized from an acid chloride and water is to be drawn.

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.

Expert Solution
Check Mark

Answer to Problem 21.40P

The complete mechanism is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  1

Explanation of Solution

The given molecule is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  2

The carboxylic acid formation requires water as a nucleophile in the nucleophilic addition step. In this reaction, Cl- acts as the leaving group and water can act as a base in the proton transfer step.

In the first step, water acts as a nucleophile and adds to the carbonyl carbon of the acid chloride to produce a protonated intermediate ion. 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 21.40P , additional homework tip  3

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

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  4

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 21.40P , additional homework tip  5

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  6

Conclusion

The product and mechanism of the given reaction were determined based on nucleophilic addition-elimination mechanism.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism of a given compound synthesized from an acid chloride and water is to be drawn.

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.

Expert Solution
Check Mark

Answer to Problem 21.40P

The complete mechanism is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  7

Explanation of Solution

The given molecule is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  8

The ester formation requires isopropanol as a nucleophile in the nucleophilic addition step. In this reaction, Cl- acts as the leaving group and alcohol can act as a base in the proton transfer step.

In the first step, isopropanol acts as a nucleophile and adds to the carbonyl carbon of the acid chloride to produce a protonated intermediate ion. 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 21.40P , additional homework tip  9

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

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  10

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 21.40P , additional homework tip  11

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  12

Conclusion

The product and mechanism of the given reaction were determined based on nucleophilic addition-elimination mechanism.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism of a given compound synthesized from an acid chloride and water is to be drawn.

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.

Expert Solution
Check Mark

Answer to Problem 21.40P

The complete mechanism is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  13

Explanation of Solution

The given molecule is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  14

The amide formation requires amine as a nucleophile in the nucleophilic addition step. In this reaction, Cl- acts as the leaving group and alcohol can act as a base in the proton transfer step.

In the first step, amine acts as a nucleophile and adds to the carbonyl carbon of the acid chloride to produce a protonated intermediate ion. 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 21.40P , additional homework tip  15

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

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  16

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 21.40P , additional homework tip  17

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  18

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism of given compound synthesized from an acid chloride and water is to be drawn.

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.

Expert Solution
Check Mark

Answer to Problem 21.40P

The complete mechanism is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  19

Explanation of Solution

The given molecule is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  20

The amide formation requires amine as a nucleophile in the nucleophilic addition step. In this reaction, Cl- acts as the leaving group and pyridine can act as a base in the proton transfer step.

In the first step, pyridine acts as a nucleophile and adds to the carbonyl carbon of the acid chloride to produce a protonated intermediate ion. 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 21.40P , additional homework tip  21

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

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  22

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 21.40P , additional homework tip  23

Thus, the complete mechanism can be drawn as

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 21, Problem 21.40P , additional homework tip  24

Conclusion

The product and mechanism of the given reaction were determined based on 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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