Home » Uncategorized » Why must the oxidizing agent be in excess during the oxidation of alcohol to carboxylic acid?

Why must the oxidizing agent be in excess during the oxidation of alcohol to carboxylic acid?

Okay.. Minor things first.. As you guys know, for simplicity’s sake, the oxidation of alcohols to carboxylic acids using an oxidizing agent such as acidified potassium dichromate (K2Cr2O7) is represented by an equation that looks like this:

post6d1

However, I detected dissent from people in class.. Like.. Why don’t we write the full equation? What so difficult about it? Tell me why?! That is because the full equation looks something like:

post6d2

I’m sure u guys agree that it is better that we stick to the simplified form.. no?

Anyways.. Moving on.. Why must the oxidizing agent be in excess?

That is because the oxidation of alcohols to carboxylic acids takes place via a 2 stage process.

In stage 1:

Alcohols are oxidized to form aldehydes. In the example below, ethanol is oxidized to an aldehyde called ethanal.

post6d3

In stage 2

The aldehyde is then oxidized to form the carboxylic acid. In the example below, ethanal is oxidized to form ethanoic acid.

post6d4

If there is insufficient oxidizing agent, the reaction stops at the aldehyde instead of forming the desired carboxylic acid! Similarly, reflux condition is needed to prevent the ethanal from escaping before it is completely oxidized to form the carboxylic acid!

Note: For JC students, this mechanism applies to primary alcohols. Often, secondary alcohols do not fully oxidize to form carboxylic acids but stop at ketones.

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