Question

Write the chemical equation representing the reaction of ethanol with an acidified solution of potassium dichromate.

Answer 1

Hint:If we look at ethanol, we can say that it is derived from the parent molecule ethane, a hydrocarbon having two carbons. And by looking at the suffix (-ol) of Ethanol we can say that it is alcohol. Potassium dichromate (${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$) is an oxidizing agent that has been acidified by an acid.

Complete step-by-step answer:The for us to write the chemical reaction, let’s look at the reactant first,
1. Ethanol (${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{OH}}$)
2. ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ and ${{\text{H}}^{\text{ + }}}$ from any acid.
Now, we know that potassium dichromate is an oxidizing agent, therefore it will oxidize the ethanol,and we also know that primary alcohol will oxidize to give the corresponding aldehyde, and secondary alcohol will oxidize to the corresponding ketone.
Let’s get to our chemical equation now, since Ethanol is the primary alcohol it will oxidize to ethanal. Hence, the required reaction can be written as:
$C{H_3} - \mathop C\limits^{\mathop |\limits^{OH} } - C{H_3} \to C{H_3} - \mathop C\limits^{\mathop {||}\limits^O } - C{H_3}$

Additional information:A primary alcohol has the $\alpha $ carbon and is bonded with only one-carbon group. Secondary alcohol has $\alpha $ a carbon bonded with 2 carbons. And tertiary alcohols have the $\alpha $ carbon bonded with 3-carbon groups.
Let’s see a reaction of secondary alcohol undergoing oxidation with the same reagents, and the reaction can be written as:
${\text{C}}{{\text{H}}_{\text{3}}} - \mathop {\text{C}}\limits^{\mathop {\text{|}}\limits^{{\text{OH}}} } - {\text{C}}{{\text{H}}_{\text{3}}} \to {\text{C}}{{\text{H}}_{\text{3}}} - \mathop {\text{C}}\limits^{\mathop {{\text{||}}}\limits^{\text{O}} } - {\text{C}}{{\text{H}}_{\text{3}}}$

Note: Tertiary alcohols do not undergo oxidation. And if we use a strong oxidizing agent like potassium permanganate (${\text{KMn}}{{\text{O}}_4}$), both the primary alcohols and secondary alcohols will directly get oxidized to the corresponding carboxylic acid.