The expected product for the given reaction has to be predicted. Concept Introduction: Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl. Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the S N 2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as, The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

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Answer 1

Question

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6

Conclusion

The product that will be formed in the given reaction was predicted.

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Answer 2

Question

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6

Conclusion

The product that will be formed in the given reaction was predicted.

Want to see more full solutions like this?

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Answer 3

Question

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6

Conclusion

The product that will be formed in the given reaction was predicted.

Want to see more full solutions like this?

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Answer 4

Question

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6

Conclusion

The product that will be formed in the given reaction was predicted.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6

Conclusion

The product that will be formed in the given reaction was predicted.

Answer 6

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6

Conclusion

The product that will be formed in the given reaction was predicted.

Answer 7

Chapter 14.5, Problem 14.27P

Interpretation Introduction

Interpretation:

The expected product for the given reaction has to be predicted.

Concept Introduction:

Ether are compounds that contain an oxygen atom that is sandwiched between two R groups. The R groups may be alkyl, vinyl or aryl.

Epoxides are very much susceptible to the reactions in which the ring is opened. This ring opening alleviates the ring strain that is present in the three-membered ring. When a strong nucleophile reacts with an epoxide, the SN2 reaction mechanism is followed resulting in ring opening. The general reaction scheme can be given as,

The first step is the attack of nucleophile on the epoxide to form an alkoxide ion by ring opening. The second step is the ion reacts with water to form the product. The second step is important to protonate the formed alkoxide.

Answer 8

Expert Solution & Answer

Answer to Problem 14.27P

The product that will be obtained from the given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 2

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given reaction is,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 3

The given starting material is an epoxide. NaSH forms a strong nucleophile namely SH−. The first step is that the nucleophile attacks the epoxide to form an alkoxide by ring opening reaction. This can be depicted as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 4

The second step is the protonation of the formed alkoxide intermediate in presence of water. Water acts as a proton donor. This can be depicted as shown below,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 5

The complete scheme can be given as,

Organic Chemistry As a Second Language: First Semester Topics, Chapter 14.5, Problem 14.27P , additional homework tip 6