Chapter 20, Problem 20.1P

Interpretation Introduction

(a)

Interpretation:

The detailed mechanism and the major product of the given reaction is to be drawn.

Concept introduction:

The carbonyl carbon atom in compounds such as aldehydes, ketones, and esters is relatively electron-poor. It can add a nucleophile, forming a tetrahedral transition state. This is followed by the elimination of the leaving group, itself a nucleophile, to form the product. When an ester is treated with an alkoxide, it replaces the original alkoxide group in from the ester substrate. Since the reaction results in formation of another ester, it is known as transesterification.

Answer to Problem 20.1P

The mechanism of the given reaction can be drawn as

The major product of the reaction is

Explanation of Solution

The given reaction is

The carbonyl carbon in the ester substrate is partially positively charged, i.e., electron-poor. The reagent ${\text{NaOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ is ionic and essentially behaves as the negatively charged nucleophile ${}^{-}{\text{OCH}}_{\text{2}}{\text{CH}}_{\text{3}}$. It will add to the carbonyl carbon, converting it to a tetrahedral intermediate with the negative charge shifted to the carbonyl oxygen.

In the next step, the leaving group, the phenoxy ion, from the original ester is eliminated. This step forms the major product, another ester.

Since the substrate and the product are both esters, of comparable stability, the two steps are reversible.

Thus, the complete detailed mechanism for the reaction can be drawn as

And the major product of the reaction is

Conclusion

The mechanism of the reaction and the major product were drawn based on nucleophilic addition-elimination step.

Interpretation Introduction

(b)

Interpretation:

The detailed mechanism of the reaction and its major product are to be drawn.

Concept introduction:

The carbonyl carbon atom in compounds such as aldehydes, ketones, and esters is relatively electron-poor. It can add a nucleophile, forming a tetrahedral transition state. This is followed by the elimination of the leaving group, itself a nucleophile, to form the product. When an ester is treated with an alkoxide, it replaces the original alkoxide group in from the ester substrate. Since the reaction results in formation of another ester, it is known as transesterification.

Answer to Problem 20.1P

The complete mechanism for the reaction can be drawn as

And the major product of the reaction as

Explanation of Solution

The given reaction is

The substrate is an ester, with an electron-poor carbonyl carbon. The reagent is ionic and essentially acts as the negativey charged nucleophile ${\text{CH}}_{\text{3}}{\text{CH(O}}^{-}{\text{)CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$. The nucleophile will attack and add to the electrophilic (electron-poor) carbonyl carbon in the first step. This forms a tetrahedral intermediate with the negative charge moved onto the carbonyl oxygen.

Next, the leaving group (ethoxide) from the original ester is eliminated to formt eh major product.

The major product is another ester, of comparable stability. Therefore, both steps will be reversible steps.

Thus, the complete mechanism for the reaction can be drawn as

And the major product of the reaction as

Conclusion

The complete mechanism and the major product of the reaction were drawn based on nucleophilic addition-elimination steps.