Chapter 24, Problem 24.46P

Draw the product of each Robinson annulation from the given starting materials using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution.

a. c.

b. d.

Interpretation Introduction

(a)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ 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 $\text{α}-\text{carbon}$ 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 that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.46P

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is $8\text{a}-\text{methyl}-3,4,8,8\text{a}-\text{tetrahydronaphthalene}-1,6\left(2H,7H\right)-\text{dione}$.

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown as,

Figure 1

In this reaction, the given dicarbonyl compound is treated with a base, ${}^{-}\text{OH}$ 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 $\text{β}-\text{carbon}$ of the given $\text{α},\text{β}-\text{unsaturated}$ 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, $8\text{a}-\text{methyl}-3,4,8,8\text{a}-\text{tetrahydronaphthalene}-1,6\left(2H,7H\right)-\text{dione}$.

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ 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 $\text{α}-\text{carbon}$ 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 that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.46P

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is $4-\text{ethyl}-3-\text{phenylcyclohex}-2-\text{enone}$.

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown as,

Figure 2

In this reaction, the given carbonyl compound is treated with a base, ${}^{-}\text{OH}$ 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 $\text{β}-\text{carbon}$ of the given $\text{α},\text{β}-\text{unsaturated}$ 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, $4-\text{ethyl}-3-\text{phenylcyclohex}-2-\text{enone}$.

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ 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 $\text{α}-\text{carbon}$ 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 that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.46P

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is,

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown as,

Figure 3

In this reaction, the given carbonyl compound is treated with a base, ${}^{-}\text{OH}$ 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 $\text{β}-\text{carbon}$ of the given $\text{α},\text{β}-\text{unsaturated}$ 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, $4-\text{methyl}-4,4\text{a},5,6,7,8-\text{hexahydronaphthalene}-2\left(3H\right)-\text{one}$.

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ 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 $\text{α}-\text{carbon}$ 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 that results in the formation of the new carbon-carbon bond and the desired Robinson annulation product.

Answer to Problem 24.46P

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is,

Explanation of Solution

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown as,

Figure 4

In this reaction, the given carbonyl compound is treated with a base, ${}^{-}\text{OH}$ 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 $\text{β}-\text{carbon}$ of the given $\text{α},\text{β}-\text{unsaturated}$ 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, $10-\text{methyl}-1,2,10,10\text{a}-\text{tetrahydrophenanthren}-3\left(9H\right)-\text{one}$.

Conclusion

The product that is formed by the Robinson annulation reaction from the given starting materials by using ${}^{-}\text{OH}$ in ${\text{H}}_{\text{2}}\text{O}$ solution is shown in Figure 4.