Question

Draw the organic product (if any) expected from the following reaction: (include all hydrogen atoms) 

$$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q) \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\\ &\text { Note: } \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \text { is present in excess. } \end{aligned} $$

Answer 1

Concepts and reason

The problem is based on the concept of oxidation of alcohol. Oxidation of alcohol with strong oxidizing reagent results in the formation of carboxylic acid.

Fundamentals

An oxidizing reagent has the ability to gain electrons. Therefore, it reduces itself and oxidize the other compound.

The reagent, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is a strong oxidizing reagent. It oxidizes primary alcohol to carboxylic acid.

 

Consider the reaction for the oxidation of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) as:

$$ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{H}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \quad \mathrm{CrO}_{3}+\mathrm{H}_{2} \mathrm{O} $$

The reagent \(\mathrm{CrO}_{3}\) is the strong oxidizing reagent.

In the reaction, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) acts as a catalyst. The reagent, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is oxidized to \(\mathrm{CrO}_{3},\) which is a strong oxidizing agent.

 

Consider the reaction for the oxidation of alcohol as follows:

$$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{CrO}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} $$

Consider the mechanism for the oxidation of aldehyde to carboxylic acid as follows:

Consider the reaction for the formation of carboxylic acid as follows:

$$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q) \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} $$

The organic product from the reaction is as follows:

$$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q) \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} $$

In the reaction, \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts in situ and forms \(\mathrm{CrO}_{3}\) which is a strong oxidizing reagent. The oxidation of alcohol results in the formation of aldehyde as an intermediate. The aldehyde is further oxidized to a carboxylic acid.