Chapter 24, Problem 24.67P

Answer the following questions about $2$-acetylcyclopentanone.

$2$-acetylcyclopentanone

a. What starting materials are needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (a)?

b. What starting materials are needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (b)?

c. What product is f ormed when $2$-acetylcyclopentanone is treated with ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$, followed by ${\text{CH}}_{\text{3}}\text{I}$?

d. Draw the Robinson annulation product(s) formed by reaction of $2$-acetylcyclopentanone with methyl vinyl ketone $\left({\text{CH}}_2={\text{CHCOCH}}_3\right)$.

e. Draw the structure of the most stable enol tautomer(s).

Interpretation Introduction

(a)

Interpretation: The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (a) is to be predicted.

Concept introduction: In crossed Claisen condensation reaction, a base abstracts an acidic proton from an $\text{α}$ carbon atom of carbonyl compound to form an enolate. This enolate reacts with carbonyl group of other carbonyl compounds that lead to the formation of $\text{β}$-ketoester.

Answer to Problem 24.67P

The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (a) is,

Figure 1

Explanation of Solution

The given compound is,

Figure 2

The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (a) must have one ester group and chain of seven carbon atoms as shown below.

Figure 1

The base abstracts the proton from $\text{C5}$ and forms an enolate. This enolate acts as nucleophile and attacks the carbonyl carbon of ester group resulting in the formation of five membered ring. The corresponding chemical reaction is shown below.

Figure 3

Conclusion

The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (a) is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation: The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (b) is to be predicted.

Concept introduction: In crossed Claisen condensation reaction, a base abstracts an acidic proton from an $\text{α}$ carbon atom of carbonyl compound to form an enolate. This enolate reacts with carbonyl group of other carbonyl compounds that lead to the formation of $\text{β}$-ketoester.

Answer to Problem 24.67P

The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (b) is,

Figure 4

Explanation of Solution

The given compound is,

Figure 2

The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (b) must have one ester group and ketone of five membered ring as shown below.

Figure 4

The base abstracts the proton from $\text{α}$ carbon atom of ketone and forms an enolate. This enolate acts as nucleophile and attacks the carbonyl carbon of ester group resulting in the formation of $2$-acetylcyclopentanone. The corresponding chemical reaction is shown below.

Figure 5

Conclusion

The starting materials needed to form $2$-acetylcyclopentanone by a Claisen reaction that forms bond (b) is shown in Figure 8.

Interpretation Introduction

(c)

Interpretation: The product formed on treatment of $2$-acetylcyclopentanone with ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}\text{I}$ is to be determined.

Concept introduction: Alkylation of carbonyl compounds takes place in the presence of a strong base. The use of appropriate base, solvent and temperature can yield one major product regioselectively. The first step is abstraction of proton and second step is attack of electrophile to form alkylated product.

Answer to Problem 24.67P

The product formed on treatment of $2$-acetylcyclopentanone with ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}\text{I}$ is,

Figure 6

Explanation of Solution

The product formed on treatment of $2$-acetylcyclopentanone with ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}\text{I}$ is shown below.

Figure 7

The base ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ abstracts a proton between the two carbonyl groups of $2$-acetylcyclopentanone to form an enolate. This enolate reacts with methyl iodide to form alkylated product.

Conclusion

The product formed on treatment of $2$-acetylcyclopentanone with ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}\text{I}$ is shown in Figure 8.

Interpretation Introduction

(d)

Interpretation: The products formed on Robinson annulations of $2$-acetylcyclopentanone with methyl vinyl ketone are to be determined.

Concept introduction: Robinson annulation is a combination of Michael addition and intramolecular aldol condensation reactions. It takes place between $\text{α,β-unsaturated}$ carbonyl compound and carbonyl compound. The enolate formation takes place from carbonyl compound which attacks the $\text{α,β-unsaturated}$ carbonyl compound to form dicarbonyl which undergoes intramolecular aldol condensation which leads to the ring formation.

Answer to Problem 24.67P

The products formed on Robinson annulations of $2$-acetylcyclopentanone with methyl vinyl ketone are,

Figure 8

Explanation of Solution

The first step is the Michael addition between $2$-acetylcyclopentanone with methyl vinyl ketone as shown below.

Figure 9

The base abstracts the proton between the two carbonyl groups to form enolate. The enolate formed gives the conjugate addition on methyl vinyl ketone followed by protonation resulting in $1,5$-dicarbonyl group. This compound undergoes intramolecular aldol condensation as shown below.

Figure 10

The base abstracts the acidic proton from the $\text{α}$ carbon atom $\text{C3}$ to form enolate. This enolate undergoes intramolecular cyclization with $\text{C1}$ to form compound $\text{I}$ and $\text{II}$. This enolate undergoes intramolecular cyclization with $\text{C1}$ to form compound $\text{III}$ and $\text{IV}$.

Conclusion

The products formed on Robinson annulations of $2$-acetylcyclopentanone with methyl vinyl ketone are shown in Figure 8.

Interpretation Introduction

(e)

Interpretation: The structure of the most stable enol form tautomer is to be drawn.

Concept introduction: Keto and enol form of a compound exists in the chemical equilibrium with each other. Enol form refers to the compound in which carbonyl group exist as hydroxyl group adjacent to carbon-carbon double bond.

Answer to Problem 24.67P

The structure of the most stable enol form tautomer is,

Figure 11

Explanation of Solution

The enol form tautomer of $2$-acetylcyclopentanone are shown below.

Figure 12

The stability of enols depends on the stability of alkenes. The alkene with endo double bond is more stable in case of five membered ring.

Thus, the most stable enol form tautomer is,

Figure 11

Conclusion

The structure of the most stable enol form tautomer is shown in Figure 11.