17P - Organic Chemistry (Mcmurry)

Question

Show the steps in preparing each of the following substances using either a malonic ester synthesis or an acetoacetic ester synthesis:

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Step-by-Step Solution

Answer

You need to provide the mechanism for preparing each of the given substances using either a malonic ester synthesis or an acetoacetic ester synthesis.

Answer of subpart (a): Step 1: Explanation of malonic ester synthesis or acetoacetic ester synthesis

In both these two reactions, there is a conversion of alkyl halide. In acetoacetic ester synthesis, the alkyl halide converts into a methyl ketone having three more carbons. In malonic ester synthesis, the alkyl halide converts into carboxylic acid having to more carbons.

 

Both the two reactions have same steps such as

1. Formation of an enolate ion
2. S_N² attack of the enolate ion on alkyl halide
3. Hydrolysis and decarboxylation

Step 2: Mechanism of the reaction

 

The compound represent by the model is 6-methylhept-5-ene-2-one.

 

Step 3: Explanation of the solution

The given compound is a methyl ketone hence it can be prepared by acetoacetic ester synthesis.

The ethoxide ion abstracts a proton from the methylene group from the ester and forms an enolate ion.

The enolate ion attacks the 1-bromo-3-methyl-2-butene and removes the bromide ion.

Then after hydrolysis and decarboxylation, you get the desired product.

 

 

Step 4: Conclusion

The given compound can be prepared by acetoacetic ester synthesis like

Answer of subpart (b): Step 5: Explanation of malonic ester synthesis or acetoacetic ester synthesis

In both these two reactions, there is a conversion of alkyl halide. In acetoacetic ester synthesis, the alkyl halide converts into a methyl ketone having three more carbons. In malonic ester synthesis, the alkyl halide converts into carboxylic acid having to more carbons.

 

Both the two reactions have same steps such as

1. Formation of an enolate ion
2. S_N² attack of the enolate ion on alkyl halide
3. Hydrolysis and decarboxylation

Step 6: Mechanism of the reaction

The compound represent by the model is 2-methyl-3-phenylpropanoic acid.

Step 7: Explanation of the solution

The given compound is a carboxylic acid hence it can be prepared by malonic ester synthesis.

The ethoxide ion abstracts a proton from the methylene group from the ester and forms an enolate ion.

The enolate ion attacks the benzyl bromide and removes the bromide ion.

Then, another proton abstraction happens by the base.

The new enolate ion attacks the methyl bromide and removes the bromide ion.

Lastly, after hydrolysis and decarboxylation, you get the desired product.

 

 

 

 

Step 8: Conclusion

The given compound can be prepared by malonic ester synthesis like