(a) Interpretation: How to carry out the given transformation which may require more than one reaction is to be shown. Concept introduction: Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene . In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

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Answer 1

Question

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

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1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)

19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

Li+ is a hard acid. With this in mind, which if the following compounds should be most soluble in water? Group of answer choices LiBr LiI LiF LiCl

Answer 2

Question

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

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Answer 3

Question

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

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Answer 4

Question

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

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Answer 5

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

Answer 6

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Answer 7

Chapter 10, Problem 10.73P

Interpretation Introduction

(a)

Interpretation:

How to carry out the given transformation which may require more than one reaction is to be shown.

Concept introduction:

Usually secondary and tertiary alcohols show dehydration reaction through E1 mechanism. In presence of any strong acid, the dehydration of alcohols occurs. In step one, the strong acid protonates the hydroxyl oxygen atom and converts it to a good leaving group. In the next step, the leaving group leaves resulting in the formation of a carbocation intermediate. The carbocation intermediate which is formed can rearrange itself via 1, 2-hydride shift or 1, 2-methyl shift to yield a more stable carbocation. Third step is a deprotonation step in which water removes the proton which is adjacent to the carbocation and forms the most substituted alkene. In this reaction, the most substituted alkene is the major product. According to the Zaitsev rule, the proton gets removed from the carbon atom with the least hydrogen atoms.

Answer 8

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

Answer 9

Interpretation Introduction

(b)

Interpretation:

How to carry out the given transformation requiring more than one reaction is to be shown.

Concept introduction:

Alcohols can be converted to least substituted alkene via Hoffman elimination. Hoffman elimination requires an amine precursor which can be generated from the starting alcohol after using PBr3 to convert the alcohol to bromide. The alkyl bromide when treated with ammonia converts to a primary amine. When primary amines are heated with excess of methyl triiodide in presence of aqueous silver oxide followed by heat, it results in the least substituted alkene as the major product.

Answer 10

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Answer 13

Ch. 10 - Prob. 10.1P Ch. 10 - Prob. 10.2P Ch. 10 - Prob. 10.3P Ch. 10 - Prob. 10.4P Ch. 10 - Prob. 10.5P Ch. 10 - Prob. 10.6P Ch. 10 - Prob. 10.7P Ch. 10 - Prob. 10.8P Ch. 10 - Prob. 10.9P Ch. 10 - Prob. 10.10P

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