(a) Interpretation: Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined. Concept introduction: E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C − H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

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Answer 1

Question

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

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(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³

fcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transfer

34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10

Answer 2

Question

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

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Answer 3

Question

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

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Answer 4

Question

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

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Answer 5

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

Answer 6

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Answer 7

Chapter 8, Problem 8.50P

Interpretation Introduction

(a)

Interpretation:

Whether an E2 elimination from the given alkyl halide precursor will exclusively produce (E)-anethol or a mixture of stereoisomers is to be determined.

Concept introduction:

E2 elimination is a single step mechanism involving both the substrate molecule and the reagent molecule. It involves the elimination of a leaving group along with hydrogen from a carbon adjacent to the one with the leaving group. In E2 mechanism, a strong base extracts the proton with the C−H bond pair moving to form a second bond between the two carbon atoms. Simultaneously, the leaving group departs with its pair of bond electrons. This requires that the leaving group and the extracted proton be anti to each other. This generally leads to the formation of only one isomer. However, if there are two hydrogen atoms on the carbon from which the proton is extracted, a mixture of stereoisomers is possible. The product distribution, in this case, depends on the stabilities of the conformers that lead to the two isomers.

Answer 8

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Answer 9

Interpretation Introduction

(b)

Interpretation:

Whether an E1 elimination reaction, from the given precursor, will produce a pure stereoisomer or a mixture is to be determined.

Concept introduction:

E1 elimination is a two-step reaction. The leaving group breaks off in the first step taking its bond pair with it. A trigonal planar carbocation results from this step. Since the carbocation is planar, any one of the hydrogen atoms, on an adjacent carbon, can be removed as a proton. The associated C−H bond pair can then move toward the carbocation to form the second (pi) bond between the two carbon atoms. Since there is no restriction on which hydrogen can be lost, a mixture of stereoisomers is formed.

Answer 10

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

Answer 11

Interpretation Introduction

(c)

Interpretation:

In each of the two reactions that produce a mixture, which isomer is produced in greater abundance is to be determined.

Concept introduction:

When a reaction produces a mixture of stereoisomers, the product distribution depends on two factors, the stability of the product and the stability of the conformers of the precursor in case of an E2 reaction. An E2 reaction requires that the proton and the leaving group be anti to each other. If there are two protons on the beta carbon, the relative stabilities of the two conformers will influence the product distribution.

Answer 12

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Answer 13

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Answer 15

Ch. 8 - Prob. 8.1P Ch. 8 - Prob. 8.2P Ch. 8 - Prob. 8.3P Ch. 8 - Prob. 8.4P Ch. 8 - Prob. 8.5P Ch. 8 - Prob. 8.6P Ch. 8 - Prob. 8.7P Ch. 8 - Prob. 8.8P Ch. 8 - Prob. 8.9P Ch. 8 - Prob. 8.10P

Solution Summary: The author explains that E2 elimination is a single step mechanism involving both reactants.

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