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Chapter 22: Problem 123

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 isomers of C3H8O that react with KMnO4 are propanol and propan-2-ol. The products of their reactions with excess KMnO4 are propanoic acid (C2H5COOH) and propanone (CH3COCH3).

Step by step solution

01

Identify the isomers of C3H8O.

First, we need to list all the possible isomers of the formula C3H8O. There are three possible isomers: 1. Propanol (C3H7OH) - linear structure with the OH group (alcohol) attached to the first carbon atom 2. Propan-2-ol (C3H7OH) - linear structure with the OH group (alcohol) attached to the second carbon atom 3. Methyl ethyl ether (CH3OCH2CH3) - linear structure with an O atom in between a methyl (CH3) group and an ethyl (CH2CH3) group
02

Determine which isomers react with KMnO4.

The next step is to determine which of these isomers will react with potassium permanganate (KMnO4), a strong oxidizing agent. Oxidizing agents react with alcohols but not with ethers. Therefore, the two isomers that will react with KMnO4 are propanol and propan-2-ol.
03

Identify the products of the reactions with KMnO4.

When primary alcohols (like propanol) react with KMnO4, they are oxidized to carboxylic acids. So, the product of the reaction of propanol with KMnO4 will be propanoic acid (C2H5COOH). When secondary alcohols (like propan-2-ol) react with KMnO4, they are oxidized to ketones. So, the product of the reaction of propan-2-ol with KMnO4 will be propanone (CH3COCH3).
04

Conclusion

The two isomers of C3H8O that react with KMnO4 are propanol and propan-2-ol. The corresponding products of their reactions with excess KMnO4 are propanoic acid and propanone.

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Most popular questions from this chapter

The two isomers having the formula \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) boil at \(-23^{\circ} \mathrm{C}\) and \(78.5^{\circ} \mathrm{C}\). Draw the structure of the isomer that boils at \(-23^{\circ} \mathrm{C}\) and of the isomer that boils at \(78.5^{\circ} \mathrm{C}\).

If you had a group of hydrocarbons, what structural features would you look at to rank the hydrocarbons in order of increasing boiling point?

Polyesters containing double bonds are often crosslinked by reacting the polymer with styrene. a. Draw the structure of the copolymer of \(\mathrm{HO}-\mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{OH} \quad\) and \(\quad \mathrm{HO}_{2} \mathrm{C}-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}_{2} \mathrm{H}\) b. Draw the structure of the crosslinked polymer (after the polyester has been reacted with styrene).

Draw the five structural isomers of hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\).

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