ORG.CHEM W/TEXT+SOLU.MANUAL
ORG.CHEM W/TEXT+SOLU.MANUAL
15th Edition
ISBN: 9780393252125
Author: KARTY
Publisher: W.W.NORTON+CO.
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Chapter 9, Problem 9.64P
Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided. The stereochemistry of the products is to be predicted where appropriate. If the reaction yields exclusively one product or a mixture of products, then the major product is to be determined.

Concept introduction:

For the prediction of the outcome of the SN2, SN1, E2, and E1 reactions, four factors are considered. First, determine if the substrate molecule has a good leaving group. For the prediction of SN2, SN1, E2, and E1 reactions, the leaving groups must be stable. Next, the carbon atom attached to the leaving group needs to be sp3 hybridized. After that, if the leaving group is attached to a primary carbon, then SN1 and E1 reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is bonded to the tertiary carbon, then SN2 reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are likewise possible. Next, examine the influence of the strength of the nucleophile, concentrations of the attacking species, the ability of the leaving group, and the effect of the solvent. Finally, if all SN2, SN1, E2, and E1 reactions appear to be favored, then heat is the influencing factor. Elimination reactions are favored at higher temperatures, and substitutions reactions are carried at low temperatures.

Expert Solution
Check Mark

Answer to Problem 9.64P

The complete, detailed mechanism for the reaction is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  1

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  2

The absolute configuration of the substrate is S. This is solvolysis reaction because ethanol acts as a nucleophile and a solvent. The leaving group, -OTs, is a good leaving group. So it is suitable for nucleophilic substitution and elimination reaction. -OTs is attached on the secondary carbon, so we consider all four SN2, SN1, E2, and E1 mechanisms.

The attacking species CH3CH2OH is a weak nucleophile and weak base favoring SN1 and E1 reactions. The concentrations of the weak base do not matter. -OTs is an excellent leaving group, which favors SN1. Finally, the solvent CH3CH2OH is a polar protic solvent, which also favors SN1. In the given reaction, heat is not provided, so substitution is more favored than elimination. So, SN1 mechanism is the predominant mechanism for the given reaction.

The loss of the leaving group, -OTs, produces a secondary carbocation.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  3

A tertiary carbocation is highest stability than a secondary and primary carbocation. Next, 1, 2-methyl shift takes place to produce a more stable carbocation.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  4

Now, the weak nucleophile, CH3CH2OH attacks the carbocation to produce the substitution product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  5

Finally, deprotonation of the SN1 product takes place to yield an uncharged substitution product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  6

The detailed, complete mechanism is as follows:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  7

Conclusion

The outcome of the given reaction is predicted by considering factors like the nature of the leaving group, substrate, the strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided. The stereochemistry of the products is to be predicted where appropriate. If the reaction yields exclusively one product or a mixture of products, then the major product is to be determined.

Concept introduction:

For the prediction of the outcome of the SN2, SN1, E2, and E1 reactions, four factors are considered. First, determine if the substrate molecule has a good leaving group. For the prediction of SN2, SN1, E2, and E1 reactions, the leaving groups must be stable. Next, the carbon atom attached to the leaving group needs to be sp3 hybridized. After that, if the leaving group is attached to a primary carbon, then SN1 and E1 reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is bonded to the tertiary carbon, then SN2 reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are likewise possible. Next, examine the influence of the strength of the nucleophile, concentrations of the attacking species, the ability of the leaving group, and the effect of the solvent. Finally, if all SN2, SN1, E2, and E1 reactions appear to be favored, then heat is the influencing factor. Elimination reactions are favored at higher temperatures, and substitutions reactions are carried at low temperatures.

Expert Solution
Check Mark

Answer to Problem 9.64P

The complete, detailed mechanism for the reaction is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  8

The stereochemistry of the product is R configuration.

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  9

The absolute configuration of the substrate is S. The leaving group, -OTs, is a good leaving group. So it is suitable for nucleophilic substitution and elimination reaction. -OTs is attached on the secondary carbon, so we consider all four SN2, SN1, E2, and E1 mechanisms.

The attacking species CN- is a strong nucleophile and weak base, favoring SN2 and E1 reactions. The high concentration of the weak base favors SN2 mechanism. -OTs is an excellent leaving group, which favors SN1 and E1. Finally, the solvent DMF is a polar aprotic solvent, which also favors SN2 and E2. In the given reaction, heat is not provided, so substitution is more favored than elimination. So, SN2 mechanism is the predominant mechanism for the given reaction.

The carbon atom attached to the leaving group is a chiral center. CN- is a strong nucleophile, and a backside attack on the carbon leads to an inversion of stereochemistry.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  10

The stereochemistry of the product is R configuration.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  11

Conclusion

The outcome of the given reaction is predicted by considering factors like the nature of the leaving group, substrate, the strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided. The stereochemistry of the products is to be predicted where appropriate. If the reaction yields exclusively one product or a mixture of products, then the major product is to be determined.

Concept introduction:

For the prediction of the outcome of the SN2, SN1, E2, and E1 reactions, four factors are considered. First, determine if the substrate molecule has a good leaving group. For the prediction of SN2, SN1, E2, and E1 reactions, the leaving groups must be stable. Next, the carbon atom attached to the leaving group needs to be sp3 hybridized. After that, if the leaving group is attached to a primary carbon, then SN1 and E1 reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is bonded to the tertiary carbon, then SN2 reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are likewise possible. Next, examine the influence of the strength of the nucleophile, concentrations of the attacking species, the ability of the leaving group, and the effect of the solvent. Finally, if all SN2, SN1, E2, and E1 reactions appear to be favored, then heat is the influencing factor. Elimination reactions are favored at higher temperatures, and substitutions reactions are carried at low temperatures.

Expert Solution
Check Mark

Answer to Problem 9.64P

The complete, detailed mechanism for the reaction is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  12

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  13

The absolute configuration of the substrate is S. The leaving group, -OTs, is a good leaving group. So it is suitable for nucleophilic substitution and elimination reaction. -OTs is attached on the secondary carbon, so we consider all four SN2, SN1, E2, and E1 mechanisms.

The attacking species (CH3)3CO- is a strong and bulky nucleophile, favoring E2 mechanism. The high concentration of the attacking species favors E2 over the E1 mechanism. Normally alkoxide favor SN2 and E2, but because of the excessive bulkiness of (CH3)3CO- species, it favors only the E2 reaction. -OTs is an excellent leaving group, which favors SN1 and E1. Finally, the solvent DMSO is a polar aprotic solvent, which also favors SN2 and E2. In the given reaction, heat is not provided, so substitution is more favored than elimination. So, the E2 mechanism is the predominant mechanism for the given reaction.

In the substrate, the alpha C atom has one H atom attached to it. Since (CH3)3CO- is a strong and bulky nucleophile, it abstracts the H atom from the terminal C atom and yields alkene as the major product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  14

Conclusion

The outcome of the given reaction is predicted by considering factors like the nature of the leaving group, substrate, the strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided. The stereochemistry of the products is to be predicted where appropriate. If the reaction yields exclusively one product or a mixture of products, then the major product is to be determined.

Concept introduction:

For the prediction of the outcome of the SN2, SN1, E2, and E1 reactions, four factors are considered. First, determine if the substrate molecule has a good leaving group. For the prediction of SN2, SN1, E2, and E1 reactions, the leaving groups must be stable. Next, the carbon atom attached to the leaving group needs to be sp3 hybridized. After that, if the leaving group is attached to a primary carbon, then SN1 and E1 reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is bonded to the tertiary carbon, then SN2 reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are likewise possible. Next, examine the influence of the strength of the nucleophile, concentrations of the attacking species, the ability of the leaving group, and the effect of the solvent. Finally, if all SN2, SN1, E2, and E1 reactions appear to be favored, then heat is the influencing factor. Elimination reactions are favored at higher temperatures, and substitutions reactions are carried at low temperatures.

Expert Solution
Check Mark

Answer to Problem 9.64P

The complete, detailed mechanism for the reaction is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  15

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  16

The absolute configuration of the substrate is S. The leaving group, -OTs, is a good leaving group. So it is suitable for nucleophilic substitution and elimination reaction. -OTs is attached on the secondary carbon, so we consider all four SN2, SN1, E2, and E1 mechanisms.

The attacking species CO32- is a weak nucleophile and weak base, favoring SN1 and E1 reactions. The concentration of the weak base does not matter. -OTs is an excellent leaving group, which favors SN1 and E1. Finally, the solvent CH3CH2OH is a polar protic solvent, which also favors SN1 and E1. In the given reaction, heat is provided, so elimination is more favored than substitution. So, the E1 mechanism is the predominant mechanism for the given reaction.

The loss of the leaving group, -OTs, produces a secondary carbocation.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  17

A tertiary carbocation is greatest stability than a secondary and primary carbocation. Next, 1, 2-methyl shift takes place to produce a more stable carbocation.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  18

Now, the weak nucleophile CO32- abstracts the highly substituted proton and produces the most highly substituted alkene.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  19

The detailed, complete mechanism is as follows:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  20

Conclusion

The outcome of the given reaction is predicted by considering factors like the nature of the leaving group, substrate, the strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(e)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided. The stereochemistry of the products is to be predicted where appropriate. If the reaction yields exclusively one product or a mixture of products, then the major product is to be determined.

Concept introduction:

For the prediction of the outcome of the SN2, SN1, E2, and E1 reactions, four factors are considered. First, determine if the substrate molecule has a good leaving group. For the prediction of SN2, SN1, E2, and E1 reactions, the leaving groups must be stable. Next, the carbon atom attached to the leaving group needs to be sp3 hybridized. After that, if the leaving group is attached to a primary carbon, then SN1 and E1 reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is bonded to the tertiary carbon, then SN2 reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are likewise possible. Next, examine the influence of the strength of the nucleophile, concentrations of the attacking species, the ability of the leaving group, and the effect of the solvent. Finally, if all SN2, SN1, E2, and E1 reactions appear to be favored, then heat is the influencing factor. Elimination reactions are favored at higher temperatures, and substitutions reactions are carried at low temperatures.

Expert Solution
Check Mark

Answer to Problem 9.64P

The complete, detailed mechanism for the reaction is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  21

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  22

This is a solvolysis reaction because ethanol acts as a nucleophile and the solvent. The leaving group, Cl-, is a good leaving group. So it is suitable for nucleophilic substitution and elimination reaction. Cl- is attached on the tertiary carbon, so we consider three mechanisms -- SN1, E2, and E1 -- and rule out SN2.

The attacking species CH3CH2OH is a weak nucleophile and weak base favoring SN1 and E1 reactions. The concentration of the attacking species does not matter. Cl- is moderate leaving group, which favors all four reactions. Finally, the solvent CH3CH2OH is a polar protic solvent, which also favors SN1 and E1. In the given reaction, heat is provided, so elimination is more favored than substitution. So, the E1 mechanism is the predominant mechanism for the given reaction.

The loss of the leaving group, Cl-, produces a tertiary carbocation.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  23

Now, the weak nucleophile CH3CH2OH abstracts the proton from the terminal CH3 group and produces the most highly substituted alkene.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  24

The detailed, complete mechanism is as follows:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  25

Conclusion

The outcome of the given reaction is predicted by considering factors like the nature of the leaving group, substrate, the strength of the reagent used, solvent, and temperature.

Interpretation Introduction

(f)

Interpretation:

The complete, detailed mechanism for the reaction is to be provided. The stereochemistry of the products is to be predicted where appropriate. If the reaction yields exclusively one product or a mixture of products, then the major product is to be determined.

Concept introduction:

For the prediction of the outcome of the SN2, SN1, E2, and E1 reactions, four factors are considered. First, determine if the substrate molecule has a good leaving group. For the prediction of SN2, SN1, E2, and E1 reactions, the leaving groups must be stable. Next, the carbon atom attached to the leaving group needs to be sp3 hybridized. After that, if the leaving group is attached to a primary carbon, then SN1 and E1 reactions are not feasible unless the resulting carbocation is resonance stabilized. If the leaving group is bonded to the tertiary carbon, then SN2 reactions are not feasible. If it is attached to a secondary carbon atom, then all four reactions are likewise possible. Next, examine the influence of the strength of the nucleophile, concentrations of the attacking species, the ability of the leaving group, and the effect of the solvent. Finally, if all SN2, SN1, E2, and E1 reactions appear to be favored, then heat is the influencing factor. Elimination reactions are favored at higher temperatures, and substitutions reactions are carried at low temperatures.

Expert Solution
Check Mark

Answer to Problem 9.64P

The complete, detailed mechanism for the reaction is shown below:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  26

Explanation of Solution

The given reaction is

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  27

The leaving group, Cl-, is a suitable leaving group. So it is suitable for nucleophilic substitution and elimination reaction. Cl- is attached on the secondary carbon, so we consider all four SN2, SN1, E2, and E1 mechanisms.

The attacking species CH3CH2O- is a strong nucleophile and strong base, favoring SN2 and E2 reactions. The high concentration of the attacking species favors SN2 and E2 mechanisms. Cl- is a moderated leaving group, which favors all four reactions. Finally, the solvent DMSO is a polar aprotic solvent, which also favors SN2 and E2. In the given reaction, heat is provided, so elimination is more favored than substitution. So, the E2 mechanism is the predominant mechanism for the given reaction.

In the substrate, only one H atom can be eliminated. The leaving group and H atom are on same side. E2 mechanism favors the anticoplanar conformation of the H and the leaving group. So, the bond rotation is carried between the C attached to the leaving group and the C attached to the H atom.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  28

The base deprotonates a hydrogen atom from the substrate and to yield the most substituted alkene as the product.

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  29

The detailed, complete mechanism is as follows:

ORG.CHEM W/TEXT+SOLU.MANUAL, Chapter 9, Problem 9.64P , additional homework tip  30

Conclusion

The outcome of the given reaction is predicted by considering factors like the nature of the leaving group, substrate, the strength of the reagent used, solvent, and temperature.

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Chapter 9 Solutions

ORG.CHEM W/TEXT+SOLU.MANUAL

Ch. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - Prob. 9.14PCh. 9 - Prob. 9.15PCh. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - Prob. 9.26PCh. 9 - Prob. 9.27PCh. 9 - Prob. 9.28PCh. 9 - Prob. 9.29PCh. 9 - Prob. 9.30PCh. 9 - Prob. 9.31PCh. 9 - Prob. 9.32PCh. 9 - Prob. 9.33PCh. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.36PCh. 9 - Prob. 9.37PCh. 9 - Prob. 9.38PCh. 9 - Prob. 9.39PCh. 9 - Prob. 9.40PCh. 9 - Prob. 9.41PCh. 9 - Prob. 9.42PCh. 9 - Prob. 9.43PCh. 9 - Prob. 9.44PCh. 9 - Prob. 9.45PCh. 9 - Prob. 9.46PCh. 9 - Prob. 9.47PCh. 9 - Prob. 9.48PCh. 9 - Prob. 9.49PCh. 9 - Prob. 9.50PCh. 9 - Prob. 9.51PCh. 9 - Prob. 9.52PCh. 9 - Prob. 9.53PCh. 9 - Prob. 9.54PCh. 9 - Prob. 9.55PCh. 9 - Prob. 9.56PCh. 9 - Prob. 9.57PCh. 9 - Prob. 9.58PCh. 9 - Prob. 9.59PCh. 9 - Prob. 9.60PCh. 9 - Prob. 9.61PCh. 9 - Prob. 9.62PCh. 9 - Prob. 9.63PCh. 9 - Prob. 9.64PCh. 9 - Prob. 9.65PCh. 9 - Prob. 9.66PCh. 9 - Prob. 9.67PCh. 9 - Prob. 9.68PCh. 9 - Prob. 9.69PCh. 9 - Prob. 9.70PCh. 9 - Prob. 9.71PCh. 9 - Prob. 9.72PCh. 9 - Prob. 9.73PCh. 9 - Prob. 9.74PCh. 9 - Prob. 9.75PCh. 9 - Prob. 9.76PCh. 9 - Prob. 9.77PCh. 9 - Prob. 9.78PCh. 9 - Prob. 9.79PCh. 9 - Prob. 9.80PCh. 9 - Prob. 9.81PCh. 9 - Prob. 9.82PCh. 9 - Prob. 9.83PCh. 9 - Prob. 9.84PCh. 9 - Prob. 9.1YTCh. 9 - Prob. 9.2YTCh. 9 - Prob. 9.3YTCh. 9 - Prob. 9.4YTCh. 9 - Prob. 9.5YTCh. 9 - Prob. 9.6YTCh. 9 - Prob. 9.7YTCh. 9 - Prob. 9.8YTCh. 9 - Prob. 9.9YTCh. 9 - Prob. 9.10YTCh. 9 - Prob. 9.11YTCh. 9 - Prob. 9.12YTCh. 9 - Prob. 9.13YTCh. 9 - Prob. 9.14YTCh. 9 - Prob. 9.15YT
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