Chapter 23, Problem 56A

Interpretation Introduction

Interpretation: To draw the structures of the products that are expected for each reaction given.

Concept Introduction: The functional group that is present in the given reactant is the ester. When an ester is hydrolyzed it gives acid and alcohol. Similarly, when the ester is treated with a base like $\text{KOH}$ or $\text{NaOH}$ the product obtained will also be acid and alcohol.

Answer to Problem 56A

  ${\text{CH}}_{\text{3}}{\text{COOCH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\stackrel{\text{HCl}}{\to }{\text{CH}}_{\text{3}}{\text{COOH+CH}}_{\text{3}}\text{OH}$

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}\stackrel{{\text{H}}^{\text{+}}}{\to }{\text{CH}}_{\text{3}}\text{COOH}+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$

  ${\text{HCOOCH}}_{\text{2}}{\text{CH}}_{\text{3}}+\text{KOH}\to {\text{HCOOH+CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$

Explanation of Solution

Methyl ethanoate on hydrolysis will produce Ethanoic acid and methanol. Similarly, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ which is an ester will also form acid and alcohol as a product. The acid is ethanoic acid and the alcohol is propanol. ${\text{HCOOCH}}_{\text{2}}{\text{CH}}_{\text{3}}$ on reaction with a strong base like $\text{KOH}$ will produce formic acid and ethanol. This clearly shows that ester on reduction will produce primary alcohols as the alcohols produced in all three reactions are primary.

Conclusion

This reduction of ester will take place in two steps. In the first step, the ester will be converted aldehyde and in the second step, this aldehyde will be converted to alcohol by using reducing agents like ${\text{LiAlH}}_4$ . If esters are oxidized then they are converted into carboxylic acids.