Draw the product of each Robinson annulation from the given starting materials using − OH in H 2 O solution. a. c. b. d.

Question

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Answer 1

Textbook Question

Chapter 24, Problem 24.45P

Draw the product of each Robinson annulation from the given starting materials using $− OH$ in $H 2 O$ solution.

a. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 1 c.

b. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 3 d.

Expert Solution

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Expert Solution

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Expert Solution

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Expert Solution

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.

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A. B. b. Now consider the two bicyclic molecules A. and B. Note that A. is a dianion and B. is a neutral molecule. One of these molecules is a highly reactive compound first characterized in frozen noble gas matrices, that self-reacts rapidly at temperatures above liquid nitrogen temperature. The other compound was isolated at room temperature in the early 1960s, and is a stable ligand used in organometallic chemistry. Which molecule is the more stable molecule, and why?

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Answer 2

Textbook Question

Chapter 24, Problem 24.45P

Draw the product of each Robinson annulation from the given starting materials using $− OH$ in $H 2 O$ solution.

a. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 1 c.

b. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 3 d.

Expert Solution

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Expert Solution

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Expert Solution

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Expert Solution

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.

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

Answer 3

Textbook Question

Chapter 24, Problem 24.45P

Draw the product of each Robinson annulation from the given starting materials using $− OH$ in $H 2 O$ solution.

a. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 1 c.

b. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 3 d.

Expert Solution

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Expert Solution

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Expert Solution

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Expert Solution

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.

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

Answer 4

Textbook Question

Chapter 24, Problem 24.45P

Draw the product of each Robinson annulation from the given starting materials using $− OH$ in $H 2 O$ solution.

a. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 1 c.

b. Chapter 24, Problem 24.45P, Draw the product of each Robinson annulation from the given starting materials using , example 3 d.

Expert Solution

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Expert Solution

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Expert Solution

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Expert Solution

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Expert Solution

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Expert Solution

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Expert Solution

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.

Answer 8

Expert Solution

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer 13

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 1

Answer 14

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 2

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $8a−methyl−3,4,8,8a−tetrahydronaphthalene−1,6(2H,7H)−dione$ .

Answer 15

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 1.

Answer 16

Expert Solution

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer 21

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is $4−ethyl−3−phenylcyclohex−2−enone$ .

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 3

Answer 22

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 4

Figure 2

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−ethyl−3−phenylcyclohex−2−enone$ .

Answer 23

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 2.

Answer 24

Expert Solution

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer 29

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 5

Answer 30

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 6

Figure 3

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $4−methyl−4,4a,5,6,7,8−hexahydronaphthalene−2(3H)−one$ .

Answer 31

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 3.

Answer 32

Expert Solution

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is to be drawn.

Concept introduction: The combination of the Michael reaction and intramolecular aldol reaction in which the formation of a ring takes place is known as Robinson annulation reaction.

In this reaction, the carbonyl compound is treated with a base which leads to the elimination of a proton from the $α−carbon$ atom. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with the $β−carbon$ of the $α,β−unsaturated$ carbonyl compound that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer 37

Answer to Problem 24.45P

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 7

Answer 38

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown as,

Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card, Chapter 24, Problem 24.45P , additional homework tip 8

Figure 4

In this reaction, the given carbonyl compound is treated with a base, $−OH$ that leads to the elimination of an acidic proton present between the two carbonyl groups. This elimination leads to the formation of an enolate ion. Then, this enolate ion acts as the nucleophile and undergoes conjugate addition reaction with $β−carbon$ of the given $α,β−unsaturated$ carbonyl compound by joining the indicated carbon atoms of both the compounds. The joining of two carbon atoms results in the formation of the new carbon-carbon bond and the desired Robinson annulation product, $10−methyl−1,2,10,10a−tetrahydrophenanthren−3(9H)−one$ .

Answer 39

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using $−OH$ in $H2O$ solution is shown in Figure 4.