Chapter 16, Problem 41P

Interpretation Introduction

(a)

Interpretation:

The product formed in the reaction of ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{CHO}$ with ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ needs to be determined.

Concept Introduction:

Potassium dichromate is a good oxidizing agent and it has the tendency to oxidize alcohols to aldehydes and aldehydes to carboxylic acids as well.An oxidizing agent is a chemical species which has the tendency to oxidize other to possible higher oxidation states by making them loose electron and thus itself gets reduced.

Answer to Problem 41P

  ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}\text{COOH}$

Explanation of Solution

Aldehydes when made to react with potassium dichromate are known to form the corresponding carboxylic acids. The reaction of hexanal in the presence of potassium dichromate will result in the formation of hexanoic acid as shown below in the chemical equation:

  ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}\text{CHO}\stackrel{{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\to }{\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}\text{COOH}$

Hence the reaction of hexanal with potassium dichromate results in the formation of the corresponding carboxylic acid which is hexanoic acid.

Interpretation Introduction

(b)

Interpretation:

The product formed in the reaction of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ with the following aldehyde needs to be determined.

  

Concept Introduction:

Potassium dichromate is a good oxidizing agent and it has the tendency to oxidize alcohols to aldehydes and aldehydes to carboxylic acids as well. An oxidizing agent is a chemical species which has the tendency to oxidize other to possible higher oxidation states by making them loose electron and thus itself gets reduced.

Answer to Problem 41P

  

Explanation of Solution

Aldehydes when made to react with potassium dichromate are known to form the corresponding carboxylic acids. The reaction of 2-cyclopentylacetaldehyde in the presence of potassium dichromate will result in the formation of 2-cyclopentanoic acid as shown below in the chemical equation:

  

Hence the reaction of 2-cyclopentylacetaldehyde with potassium dichromate results in the formation of the corresponding carboxylic acid which is 2-cyclopentylacetic acid.

Interpretation Introduction

(c)

Interpretation:

The product formed in the reaction of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ with the following compound needs to be determined.

  

Concept Introduction:

Potassium dichromate is a good oxidizing agent and it has the tendency to oxidize alcohols to aldehydes and aldehydes to carboxylic acids as well. An oxidizing agent is a chemical species which has the tendency to oxidize other to possible higher oxidation states by making them loose electron and thus itself gets reduced.

Answer to Problem 41P

No product formed.

Explanation of Solution

The given compound is 2-ethylcyclopentanone and it cannot react with potassium dichromate as it contains ketone as a functional group and potassium dichromate does not oxidize ketones. Ketones do not contain the hydrogen atom attached to the carbonyl group hence they resist to oxidize and only the strong oxidizing agents such as potassium permanganate are able to oxidize them. Therefore, no reaction will occur.

Interpretation Introduction

(d)

Interpretation:

The product formed in the reaction of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ with the following compound needs to be determined.

  ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{2}}\text{OH}$

Concept Introduction:

Potassium dichromate is a good oxidizing agent and it has the tendency to oxidize alcohols to aldehydes and aldehydes to carboxylic acids as well. An oxidizing agent is a chemical species which has the tendency to oxidize other to possible higher oxidation states by making them loose electron and thus itself gets reduced.

Answer to Problem 41P

  ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}\text{COOH}$

Explanation of Solution

The given structure clearly shows that this compound is hexanol. It is an aliphatic alcohol and it will react with potassium reagent to form the corresponding aldehyde which will further get oxidized and the reaction will finally result in the formation of a carboxylic acid.

  ${\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}{\text{CH}}_{\text{2}}\text{OH}\stackrel{{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\to }{\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}\text{CHO}\stackrel{{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\to }{\text{CH}}_{\text{3}}{{\text{(CH}}_{\text{2}}\text{)}}_{\text{4}}\text{COOH}$