Get Ready for Organic Chemistry
Get Ready for Organic Chemistry
2nd Edition
ISBN: 9780321774125
Author: KARTY, Joel
Publisher: PEARSON
Question
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Chapter 17, Problem 17.60P
Interpretation Introduction

(a)

Interpretation:

For the given reaction, the major product and reaction mechanism is to be drawn.

Concept introduction:

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

Expert Solution
Check Mark

Answer to Problem 17.60P

The nucleophile adds reversibly and undergoes a 1, 4-addition reaction.

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  1

Explanation of Solution

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon.

In this reaction, both carbonyl carbon as well as beta carbon are electrophilic. Here the nucleophile adds reversibly; therefore, this reaction undergoes 1, 4-addition. In this reaction, O atom forms in the nucleophilic addition step. It is then protonated in a subsequent proton transfer reaction to form uncharged alcohol. The reaction mechanism is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  2

Here the enol tautomerizes back to the ketone via back to back proton transfer steps. It is represented as below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  3

Conclusion

From the location of the electrophilic carbon, the major product for the given reaction is predicted, and the reaction mechanism is drawn.

Interpretation Introduction

(b)

Interpretation:

For the given reaction, the major product and the reaction mechanism is to be drawn.

Concept introduction:

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

Expert Solution
Check Mark

Answer to Problem 17.60P

The nucleophile adds irreversibly and undergoes a 1, 2-addition reaction.

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  4

Explanation of Solution

A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

In this reaction, the nucleophile adds irreversibly; hence this reaction undergoes direction addition (1, 2-addition). In this reaction, O atom forms in the nucleophilic addition step. It is then protonated in a subsequent proton transfer reaction to form uncharged alcohol. The reaction mechanism is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  5

Conclusion

From the location of the electrophilic carbon, the major product for the given reaction is predicted, and the reaction mechanism is drawn.

Interpretation Introduction

(c)

Interpretation:

For the given reaction, the major product and the reaction mechanism is to be drawn.

Concept introduction:

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

Expert Solution
Check Mark

Answer to Problem 17.60P

The nucleophile adds irreversibly and undergoes a 1, 2-addition reaction.

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  6

Explanation of Solution

A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

In this reaction, the nucleophile adds irreversibly; hence this reaction undergoes direct addition (1, 2-addition). Here O atom forms in the nucleophilic addition step. The O^- is then protonated in a subsequent proton transfer reaction to form uncharged alcohol. The reaction mechanism is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  7

Conclusion

From the location of the electrophilic carbon, the major product for the given reaction is predicted, and the reaction mechanism is drawn.

Interpretation Introduction

(d)

Interpretation:

For the given reaction, the major product and the reaction mechanism is to be drawn.

Concept introduction:

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

Expert Solution
Check Mark

Answer to Problem 17.60P

The nucleophile adds irreversibly and undergoes a 1, 4-addition reaction.

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  8

Explanation of Solution

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon.

In this reaction, both carbonyl carbon as well as beta carbon are electrophilic. In this reaction, the nucleophile adds reversibly; therefore this reaction undergoes 1, 4-addition. In this reaction, O atom forms in the nucleophilic addition step. It is then protonated in a subsequent proton transfer reaction to form uncharged alcohol. The reaction mechanism is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  9

Here the enol tautomerizes back to the ketone via back to back proton transfer steps. It is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  10

Conclusion

From the location of the electrophilic carbon, the major product for the given reaction is predicted, and the reaction mechanism is drawn.

Interpretation Introduction

(e)

Interpretation:

For the given reaction, the major product and the reaction mechanism is to be drawn.

Concept introduction:

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon. Attack at the carbonyl carbon yields the direct addition (or 1, 2—addition) product whereas attack at the beta carbon yields the conjugate addition (or 1, 4-addition). Direct addition to an alpha, beta-unsaturated polar π bond dominates if the nucleophile adds irreversibly to the polar π bond. Otherwise, conjugate addition is favored.

Expert Solution
Check Mark

Answer to Problem 17.60P

The nucleophile adds irreversibly and undergoes a 1, 4-addition reaction.

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  11

Explanation of Solution

The weak resonance contributor with separated charges generates a small partial positive charge on the beta carbon in the resonance hybrid. Thus, a nucleophile can attack at either the carbonyl carbon or the beta carbon. A polar π bond that is alpha, beta-unsaturated is susceptible to a nucleophilic attack at both the electron-poor atom of the polar π bond and the beta carbon.

In this reaction, both carbonyl carbon as well as beta carbon are electrophilic. In this reaction, the nucleophile adds reversibly; therefore this reaction undergoes 1, 4-addition. In this reaction, O atom forms in the nucleophilic addition step. It is then protonated in a subsequent proton transfer reaction to form uncharged alcohol. The reaction mechanism is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  12

Here the enol tautomerizes back to the ketone via back to back proton transfer steps. It is shown below:

Get Ready for Organic Chemistry, Chapter 17, Problem 17.60P , additional homework tip  13

Conclusion

From the location of the electrophilic carbon, the major product for the given reaction is predicted, and the reaction mechanism is drawn.

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Chapter 17 Solutions

Get Ready for Organic Chemistry

Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22PCh. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - Prob. 17.25PCh. 17 - Prob. 17.26PCh. 17 - Prob. 17.27PCh. 17 - Prob. 17.28PCh. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Prob. 17.31PCh. 17 - Prob. 17.32PCh. 17 - Prob. 17.33PCh. 17 - Prob. 17.34PCh. 17 - Prob. 17.35PCh. 17 - Prob. 17.36PCh. 17 - Prob. 17.37PCh. 17 - Prob. 17.38PCh. 17 - Prob. 17.39PCh. 17 - Prob. 17.40PCh. 17 - Prob. 17.41PCh. 17 - Prob. 17.42PCh. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Prob. 17.45PCh. 17 - Prob. 17.46PCh. 17 - Prob. 17.47PCh. 17 - Prob. 17.48PCh. 17 - Prob. 17.49PCh. 17 - Prob. 17.50PCh. 17 - Prob. 17.51PCh. 17 - Prob. 17.52PCh. 17 - Prob. 17.53PCh. 17 - Prob. 17.54PCh. 17 - Prob. 17.55PCh. 17 - Prob. 17.56PCh. 17 - Prob. 17.57PCh. 17 - Prob. 17.58PCh. 17 - Prob. 17.59PCh. 17 - Prob. 17.60PCh. 17 - Prob. 17.61PCh. 17 - Prob. 17.62PCh. 17 - Prob. 17.63PCh. 17 - Prob. 17.64PCh. 17 - Prob. 17.65PCh. 17 - Prob. 17.66PCh. 17 - Prob. 17.67PCh. 17 - Prob. 17.68PCh. 17 - Prob. 17.69PCh. 17 - Prob. 17.70PCh. 17 - Prob. 17.71PCh. 17 - Prob. 17.72PCh. 17 - Prob. 17.73PCh. 17 - Prob. 17.74PCh. 17 - Prob. 17.75PCh. 17 - Prob. 17.76PCh. 17 - Prob. 17.77PCh. 17 - Prob. 17.78PCh. 17 - Prob. 17.79PCh. 17 - Prob. 17.80PCh. 17 - Prob. 17.81PCh. 17 - Prob. 17.82PCh. 17 - Prob. 17.83PCh. 17 - Prob. 17.84PCh. 17 - Prob. 17.1YTCh. 17 - Prob. 17.2YTCh. 17 - Prob. 17.3YTCh. 17 - Prob. 17.4YTCh. 17 - Prob. 17.5YTCh. 17 - Prob. 17.6YTCh. 17 - Prob. 17.7YTCh. 17 - Prob. 17.8YTCh. 17 - Prob. 17.9YTCh. 17 - Prob. 17.10YTCh. 17 - Prob. 17.11YTCh. 17 - Prob. 17.12YTCh. 17 - Prob. 17.13YTCh. 17 - Prob. 17.14YT
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