Chapter 24, Problem 24.24P

Draw the products when each pair of compounds is treated with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ in a Robinson annulation reaction.

a. c.

b. d.

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ 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.

Answer to Problem 24.24P

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is-

Explanation of Solution

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown as,

Figure 1

In this reaction, the carbonyl compound, $2-\text{methylcyclopentane}-1,3-\text{dione}$ is treated with a base, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ which leads to the elimination of a proton from the $\text{α}-\text{carbon}$ 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 $\text{β}-\text{carbon}$ of the $\text{α},\text{β}-\text{unsaturated}$ carbonyl compound, $\text{but}-3-\text{en}-\text{one}$ that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Conclusion

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ 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.

Answer to Problem 24.24P

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is-

Explanation of Solution

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown as,

Figure 2

In this reaction, the carbonyl compound, $\text{ethyl}-3-\text{oxobutanoate}$ is treated with a base, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ which leads to the elimination of a proton from the $\text{α}-\text{carbon}$ 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 $\text{β}-\text{carbon}$ of the $\text{α},\text{β}-\text{unsaturated}$ carbonyl compound, $\text{but}-3-\text{en}-\text{one}$ that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Conclusion

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ 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.

Answer to Problem 24.24P

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is-

Explanation of Solution

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown as,

Figure 3

In this reaction, the carbonyl compound, cyclopentanone is treated with a base, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ which leads to the elimination of a proton from the $\text{α}-\text{carbon}$ 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 $\text{β}-\text{carbon}$ of the $\text{α},\text{β}-\text{unsaturated}$ carbonyl compound, $3-\text{methylbut}-3-\text{ene}-2-\text{one}$ that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Conclusion

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ 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.

Answer to Problem 24.24P

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is-

Explanation of Solution

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown as,

Figure 4

In this reaction, the carbonyl compound is treated with a base, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$ which leads to the elimination of a proton from the $\text{α}-\text{carbon}$ 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 $\text{β}-\text{carbon}$ of the $\text{α},\text{β}-\text{unsaturated}$ carbonyl compound, $\text{but}-3-\text{en}-\text{one}$ that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Conclusion

The product that is formed by the Robinson annulation reaction of the given pair of compounds with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{O}}^{-}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is shown in Figure 4.