Chapter 22.SE, Problem 17VC

Interpretation Introduction

a)

Interpretation:

The steps involved in preparing the compound represented by the model, using either malonic ester synthesis or an acetoacetic ester synthesis, are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

Explanation of Solution

The compound represented by the model is 6-methylhept-5-ene-2-one. It is a methyl ketone and hence it can be prepared using acetoacetic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group of the ester to form the enolate ion. The enolate ion then attacks 1-bromo-3-methy-2-butene and displaces the bromine as bromide ion. The product obtained upon hydrolysis with dilute acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using acetoacetic ester synthesis are given below.

Interpretation Introduction

b)

Interpretation:

The steps involved in preparing the compound represented by the model using either malonic ester synthesis or an acetoacetic ester synthesis are to be given.

Concept introduction:

Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.

Both reactions involve the same steps such as i) enolate ion formation ii) SN₂ attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.

Answer to Problem 17VC

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.

Explanation of Solution

The compound represented by the model is 2-methyl-3-phenylpropanoic acid and hence it can be prepared using malonic ester synthesis. The ethoxide ion abstracts a proton from the active methylene group to form the enolate ion. The enolate ion then attacks benzyl bromide and displaces the bromine as bromide ion. The abstraction of another acidic hydrogen in the product by the base and the nucleophilic displacement of bromine from methyl bromide by enolate ion introduces a methyl group at α- position of the diester. The alkylated diester obtained upon hydrolysis with aqueous acids and decarboxylation by heating yields the product.

Conclusion

The steps involved in preparing the compound represented by the model using malonic ester synthesis are given below.