(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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9 7 8 C 9 8 200 190 B 5 A -197.72 9 8 7 15 4 3 0: ང་ 200 190 180 147.52 134.98 170 160 150 140 130 120 110 100 90 90 OH 10 4 3 1 2 -143.04 140. 180 170 160 150 140 130 120 110 100 90 CI 3 5 1 2 141.89 140.07 200 190 180 170 160 150 140 130 120 110 100 ៖- 90 129. 126.25 80 70 60 -60 50 40 10 125.19 -129.21 80 70 3.0 20 20 -8 60 50 10 ppm -20 40 128.31 80 80 70 60 50 40 40 -70.27 3.0 20 10 ppm 00˚0-- 77.17 30 20 20 -45.36 10 ppm -0.00 26.48 22.32 ―30.10 ―-0.00

Assign all the carbons

C 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00

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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