Chapter 24, Problem 24.48P

Fill in the lettered reagents needed for each reaction.

Interpretation Introduction

Interpretation: The lettered reagents that are needed for the given reactions are to be shown.

Concept Introduction: In crossed claisen condensation reaction, the base abstracts the acidic proton from $\text{α}-$ carbon atom of an ester to form enolate. This enolate reacts with carbonyl compound of other ester that leads to the formation of a $\text{β-keto}\text{ester}$. The crossed Claisen condensation always takes place between the two different carbonyl compounds.

Answer to Problem 24.48P

The lettered reagents that are needed for the given reactions are shown below.

Explanation of Solution

The lettered reagent A is shown as,

Figure 1

The given compound, cyclopentanone is treated with the base, sodium ethoxide that results in the formation of an enolate ion. Then, the enolate ion reacts with the compound, diethyl carbonate to form the desired compound, $\text{ethyl}2-\text{oxocyclopentanecarboxylate}$. Thus, the reagent A is $\text{NaOEt}/\text{diethyl}\text{carbonate}$.

The lettered reagent B is shown as,

Figure 2

The compound, $\text{ethyl}2-\text{oxocyclopentanecarboxylate}$ is again treated with the base, sodium ethoxide that results in the formation of an enolate ion. Then, the enolate ion reacts with the compound, ethyl bromide to form the desired compound, $\text{ethyl}1-\text{ethyl}-2-\text{oxocyclopentanecarboxylate}$. Thus, the reagent B is $\text{NaOEt}/{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{Br}$.

The lettered reagent C is shown as,

Figure 3

The compound, $\text{ethyl}1-\text{ethyl}-2-\text{oxocyclopentanecarboxylate}$ undergoes hydrolysis in the presence of hydronium ions followed by the heating of the reaction mixture. The desired product that is obtained by this reaction is $2-\text{ethylcyclopentanone}$. Thus, the reagent C is ${\text{H}}_{\text{3}}{\text{O}}^{+}/\text{heat}$.

The lettered reagent D is shown as,

Figure 4

The compound, $2-\text{ethylcyclopentanone}$ is treated with methyl magnesium bromide. Then, the hydrolysis of the reaction mixture in the presence of the hydronium ion takes place. The desired product that is obtained by this reaction is $1,2-\text{diethylcyclopentanol}$. Thus, the reagent D is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}/{\text{H}}_{\text{3}}{\text{O}}^{+}$.

The lettered reagent E is shown as,

Figure 5

The compound, $2-\text{ethylcyclopentanone}$ is again treated with the base, LDA that abstracts a proton from the compound that leads to the formation of an enolate ion. Then, the enolate ion reacts with methyl bromide to form the desired compound, $2-\text{ethyl}-5-\text{methylcyclopentanone}$. Thus, the reagent E is $\text{LDA}/{\text{CH}}_{\text{3}}\text{Br}$.

The lettered reagent F is shown as,

Figure 6

The compound, $1,2-\text{diethylcyclopentanol}$ is treated with the acid, sulfuric acid that results in the dehydration of the compound to form the desired compound, $1,2-\text{diethylcyclopent}-1-\text{ene}$. Thus, the reagent F is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The lettered reagent G is shown as,

Figure 7

The given compound, cyclopentanone is treated with the base, sodium ethoxide that results in the formation of an enolate ion. Then, the enolate ion reacts with the compound, propanal to form the compound, $2-\left(1-\text{hydroxypropyl}\right)\text{cyclopentanone}$ followed by the dehydration of this obtained compound. The hydrolysis leads to the formation of the desired compound, $\left(E\right)-2-\text{propylidenecyclopentanone}$. Thus, the reagent G is $\text{NaOEt}/\text{propanol}/{\text{H}}_3{\text{O}}^{+}$.

The lettered reagent H is shown as,

Figure 8

The compound, $\left(E\right)-2-\text{propylidenecyclopentanone}$ is treated with ${\text{H}}_2/\text{Pd}$ for the reduction process to form the desired compound, $2-\text{propylcyclopentanone}$. Thus, the reagent H is ${\text{H}}_2/\text{Pd}$.

The lettered reagent I is shown as,

Figure 9

The compound, $2-\text{propylcyclopentanone}$ is treated with the base, LDA that abstracts a proton from the compound that leads to the formation of an enolate ion. Then, the enolate ion reacts with ethyl formate to form the desired compound, $2-\text{oxo}-3-\text{propylcyclopentanecarbaldehyde}$. Thus, the reagent I is $\text{LDA}/\text{ethyl formate}$.

The lettered reagent J is shown as,

Figure 10

The compound, $1,2-\text{diethylcyclopent}-1-\text{ene}$ is treated with ozone and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$ that abstracts a proton from the compound that leads to the formation of an enolate ion. Then, the enolate ion reacts with ethyl formate to form the desired compound, $\text{nonane}-3,7-\text{dione}$. Thus, the reagent J is ${\text{O}}_3/{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$.

The lettered reagent K is shown as,

Figure 11

The intramolecular aldol reaction takes place in the compound, $\text{nonane}-3,7-\text{dione}$ in the presence of sodium ethoxide. Then, the dehydration of the intermediate takes place in the presence of hydronium ion to form the desired product, $3-\text{ethyl}-2-\text{methylcyclohex}-2-\text{enone}$ Thus, the reagent K is $\text{NaOEt}/{\text{H}}_{\text{3}}{\text{O}}^{+}$.

The complete filled reagents that are needed for the given reactions are shown as,

Figure 12

Conclusion

The lettered reagents that are needed for the given reactions are shown in Figure 12.