Question

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

Answer: I2 / NaOH

Step-by-step solution

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.

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.

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.

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) \(\mathrm{I}_{2} / \mathrm{NaOH}\)

Key concepts

Iodoform Test

The Iodoform Test is a fascinating and practical method used in organic chemistry to identify the presence of methyl ketones in a compound. A methyl ketone is a molecule with the [CH_3CO] group. The test involves the use of iodine (\(\mathrm{I}_{2}\)) in the presence of a base like sodium hydroxide (\(\mathrm{NaOH}\)).

• Reaction Process: When methyl ketones are treated with iodine and sodium hydroxide, the methyl group's hydrogen is replaced by iodine atoms.
• Significance: A positive result forms a yellow precipitate known as iodoform (\(\mathrm{CHI}_3\)), which is distinct in its yellow color and antiseptic smell.
• Application: This test can be used to distinguish between pentan-2-one, which is a methyl ketone, and pentan-3-one, which is not, as only the former will produce the iodoform.

It's a simple, yet powerful test thanks to the definite visual clue—a yellow iodoform—that confirms the presence of a specific functional group in a molecule.

Clemmensen Reduction

The Clemmensen Reduction is a widely utilized reaction in organic chemistry, specifically for transforming carbonyl compounds into hydrocarbons.
This method employs zinc amalgam (Zn-Hg) and hydrochloric acid (HCl) as reagents.

• Main Function: It reduces ketones and aldehydes into their corresponding alkanes.
• Process Detail: The carbonyl group (C=O) is converted to a methylene group (CH2), ultimately simplifying the molecule by removing the oxygen.
• Limitations: This method cannot distinguish between similar ketones, such as pentan-2-one and pentan-3-one. Both would simply transform into pentane, leaving no differences to observe.

Despite its limitations in distinguishing similar compounds, the Clemmensen Reduction is crucial for specific organic synthesis applications, offering a straightforward path to remove oxygen from carbonyl groups.

Selective Oxidizing Agents

Selective oxidizing agents play a vital role in transforming specific functional groups within a molecule without affecting others. These reagents ensure that chemists can achieve precise modifications when synthesizing or analyzing organic compounds.

• Example - Selenium Dioxide (\(\mathrm{SeO}_{2}\)): Known for its ability to oxidize aldehydes to carboxylic acids without disturbing other functional groups like ketones. This selectivity is valuable when working with complex molecules.
• Limitation in Ketones: With pentan-2-one and pentan-3-one, \(\mathrm{SeO}_{2}\) won’t react, making it ineffective for distinguishing these compounds.

Selective oxidizing agents like \(\mathrm{SeO}_{2}\) are critical in organic synthesis and analysis, where precise changes in a molecular structure are needed while retaining the rest of the compound untouched.

Methyl Ketones Identification

Identifying methyl ketones is a pivotal task in organic chemistry. Methyl ketones are easily recognizable due to the [CH_3CO] group they contain, making them susceptible to certain chemoselective reactions.

• Common Methods - Iodoform Test: The use of the iodoform test allows for the detection of methyl ketones due to the formation of a yellow precipitate (\(\mathrm{CHI}_3\)), offering an efficient identification process.
• Importance: Methyl ketone identification is significant in applications ranging from organic synthesis to industrial processes, where specific structural features are necessary for desired reactions.
• Identifiable Features: These compounds exhibit certain reactivity patterns that are well-utilized in synthesis and analysis techniques.

Understanding methyl ketone identification enhances one's ability to determine structural details and predict reactivity, which is essential in advancing both research and practical applications in chemistry.