It is to be explained why there is a difference between the λ max values of an organic product’s UV-vis spectrum when the reaction involves two different bases. Concept introduction: An alkyl halide , when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction. If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction. In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λ max for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

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Answer 1

Question

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

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Answer 2

Question

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

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Answer 3

Question

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

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Answer 4

Question

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

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Answer 5

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

Answer 6

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

Answer 7

Chapter 15, Problem 15.58P

Interpretation Introduction

Interpretation:

It is to be explained why there is a difference between the λmax values of an organic product’s UV-vis spectrum when the reaction involves two different bases.

Concept introduction:

An alkyl halide, when heated in the presence of a strong base such as sodium hydroxide, undergoes an elimination reaction to yield an alkene via an E2 elimination reaction. Sodium hydroxide is a strong and less bulky base, which abstracts the proton attached to the most substituted alpha carbon atom of the alkyl halide. The maximum substituted alkene is the major product of this reaction.

If Lithiumdiisopropylamide (LDA), a strong but bulky base, is used instead of NaOH, it will selectively abstract the proton attached to the least substituted alpha carbon atom of the alkyl halide. The minimum substituted alkene is the major product of this reaction.

In UV-vis spectroscopy, absorption of a photon promotes an electron from a lower-energy MO to a higher-energy MO. The λmax for the longest-wavelength absorption in the spectrum corresponds to the HOMO-LUMO. Conjugation decreases the energy difference between the HOMO-LUMO and thus increases the wavelength of the photon absorbed.

Answer 8

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Answer 11

Ch. 15 - Prob. 15.1P Ch. 15 - Prob. 15.2P Ch. 15 - Prob. 15.3P Ch. 15 - Prob. 15.4P Ch. 15 - Prob. 15.5P Ch. 15 - Prob. 15.6P Ch. 15 - Prob. 15.7P Ch. 15 - Prob. 15.8P Ch. 15 - Prob. 15.9P Ch. 15 - Prob. 15.10P

Solution Summary: The author explains why the UV-vis spectrum of an organic product differs when the reaction involves two different bases.

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Chapter 15 Solutions