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Chapter 22: Problem 121

Three different organic compounds have the formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\). Only two of these isomers react with \(\mathrm{KMnO}_{4}\) (a strong oxidizing agent). What are the names of the products when these isomers react with excess \(\mathrm{KMnO}_{4} ?\)

Short Answer

Expert verified
The products formed when the two reactive isomers of \(C_3H_8O\) (propan-1-ol and propan-2-ol) react with excess \(\mathrm{KMnO}_4\) are propanoic acid and propanone, respectively.

Step by step solution

01

Oxidation of Propan-1-ol

Propan-1-ol \((CH_3-CH_2-CH_2OH)\) is a primary alcohol, and primary alcohols are oxidized by \(\mathrm{KMnO}_4\) to first aldehydes and then further to carboxylic acids. In this case, propan-1-ol will first be oxidized to propanal, and in the excess of \(\mathrm{KMnO}_4\), propanal will be further oxidized to propanoic acid. Hence, the product formed when propan-1-ol reacts with excess \(\mathrm{KMnO}_4\) is propanoic acid.
02

Oxidation of Propan-2-ol

Propan-2-ol \((CH_3-CHOH-CH_3)\) is a secondary alcohol, and secondary alcohols are oxidized by \(\mathrm{KMnO}_4\) to ketones. In this case, propan-2-ol will be oxidized to propanone. Hence, the product formed when propan-2-ol reacts with excess \(\mathrm{KMnO}_4\) is propanone. Thus, the products formed when the two reactive isomers of \(C_3H_8O\) (propan-1-ol and propan-2-ol) react with excess \(\mathrm{KMnO}_4\) are propanoic acid and propanone, respectively.

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