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Chapter 5: Problem 81

The reagent that can be used to distinguish between pentan-2-one and pentan-3-one is (a) \(\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}\) (b) \(\mathrm{SeO}_{2}\) (c) \(\mathrm{I}_{2} / \mathrm{NaOH}\) (d) \(\mathrm{KMnO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
Answer: I2 / NaOH

Step by step solution

01

Analyze Reaction with Zn-Hg / HCl

In this option, the reagent given is a mixture of zinc amalgam \(\mathrm{Zn}-\mathrm{Hg}\) and hydrochloric acid \(\mathrm{HCl}\). This combination is known as Clemmensen reduction, which is used for reducing carbonyl compounds to their corresponding hydrocarbons. However, it cannot distinguish between pentan-2-one and pentan-3-one since both of them are carbonyl compounds and thus will be reduced to the respective hydrocarbons. So, this is not the correct reagent.
02

Analyze Reaction with SeO2

Selenium dioxide \(\mathrm{SeO}_{2}\) is a selective oxidizing agent for aldehydes to form the corresponding carboxylic acids. In this case, pentan-2-one and pentan-3-one are both ketones, and neither of them have aldehyde functional groups. Therefore, \(\mathrm{SeO}_{2}\) would not react with either of them, so this is not the correct reagent.
03

Analyze Reaction with I2 / NaOH

The reagent combination of iodine \(\mathrm{I}_{2}\) and sodium hydroxide \(\mathrm{NaOH}\) forms a reaction known as the iodine-alkali reaction or iodoform test. This reaction is useful to distinguish methyl ketones from other ketones. Pentan-2-one is a methyl ketone ([CH_3CO] group), while pentan-3-one is not. The iodoform test would react with pentan-2-one, forming a yellow precipitate, but not with pentan-3-one. This makes this reagent suitable for distinguishing between the two compounds.
04

Analyze Reaction with KMnO4 / H2SO4

Potassium permanganate \(\mathrm{KMnO}_{4}\) in the presence of sulfuric acid \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a strong oxidizing agent. It oxidizes aldehydes and alcohols to carboxylic acids, but it does not have much affinity for ketones like pentan-2-one and pentan-3-one. Therefore, this reagent cannot distinguish between them. Since we've analyzed all the reagent options, we can conclude that the correct answer is: (c) \(\mathrm{I}_{2} / \mathrm{NaOH}\)

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