Question

2-pentanone and 3 -pentanone can be distinguished by (A) \(\mathrm{NaHSO}_{3}\) (B) \(\mathrm{NaOH} / \mathrm{I}_{2}\) (C) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{OH}\) (D) Both (A) and (B)

Short answer

The reagent that can distinguish between 2-pentanone and 3-pentanone is \(\mathrm{NaOH} / \mathrm{I}_{2}\) (Sodium hydroxide and iodine), as it gives a positive iodoform test for 2-pentanone (forming yellow iodoform crystals) and no visible reaction for 3-pentanone.

Step-by-step solution

Reagent A - Sodium bisulfite (\(\mathrm{NaHSO}_{3}\))

Sodium bisulfite (\(\mathrm{NaHSO}_{3}\)) reacts with aldehydes to form a white precipitate (crystalline bisulfite addition compounds), but not with ketones. Therefore, this reagent cannot distinguish between 2-pentanone and 3-pentanone, since both of them are ketones.

Reagent B - Sodium hydroxide and iodine (\(\mathrm{NaOH}/\mathrm{I}_{2}\))

Sodium hydroxide and iodine react with methyl ketones (ketones with one methyl group and any other alkyl group) to form yellow crystals of iodoform. This reaction is commonly known as the iodoform test. In this case, 2-pentanone is a methyl ketone, while 3-pentanone is not a methyl ketone. Therefore, 2-pentanone will have a positive iodoform test (forming yellow iodoform crystals), while 3-pentanone will not give any visible reaction. So, this reagent can distinguish between the two given ketones.

Reagent C - Silver ammonia complex and hydroxide (\(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{OH}\))

This reagent is typically used for the Tollens' test, which is employed to detect aldehydes (especially reducing aldehydes). The test generally involves the formation of a silver mirror on the inner surface of the test tube, due to the reduction of silver ions to metallic silver. Neither 2-pentanone nor 3-pentanone will react with this reagent, as they are both ketones. After analyzing all the given reagents, we can conclude that: Option (A), Sodium bisulfite (\(\mathrm{NaHSO}_{3}\)), cannot distinguish between 2-pentanone and 3-pentanone. Option (B), Sodium hydroxide and iodine (\(\mathrm{NaOH}/\mathrm{I}_{2}\)), can indeed distinguish between the two ketones. Option (C), Silver ammonia complex and hydroxide (\(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{OH}\)), cannot distinguish between 2-pentanone and 3-pentanone. Therefore, the correct answer is: (B) \(\mathrm{NaOH} / \mathrm{I}_{2}\)

Key concepts

Ketone Identification

Identifying ketones is an essential part of organic chemistry, especially in distinguishing between similar molecules. Ketones are organic compounds characterized by a carbonyl group (C=O) bonded to two alkyl or aryl groups. These groups can vary, resulting in different types of ketones. Two such examples are 2-pentanone and 3-pentanone.
To distinguish between ketones, chemists often rely on specific chemical reactions that provide visual cues, like precipitate formation or color change. While both 2-pentanone and 3-pentanone are classified as ketones, they differ in their molecular structure, particularly in the position of the carbonyl group. This structural difference can sometimes influence how they react with certain reagents.
To correctly identify or differentiate them, understanding their reactivity with certain tests, like the iodoform test, becomes crucial. Such tests exploit the unique presence or absence of specific subgroups within the molecule, making them invaluable tools in organic analysis.

Iodoform Test

The iodoform test is a widely used chemical test in organic chemistry to identify the presence of methyl ketones. These are ketones that have at least one methyl group (CH₃) attached to the carbonyl group.

• In the iodoform test, a solution of sodium hydroxide (NaOH) and iodine (I₂) is added to the substance.
• When a methyl ketone is present, the reaction results in the formation of a yellow crystalline precipitate called iodoform (CHI₃).

This test is particularly useful for distinguishing 2-pentanone from 3-pentanone. 2-pentanone is a methyl ketone because its carbonyl group is adjacent to a methyl group. Hence, it will give a positive result with the iodoform test, resulting in yellow crystals. In contrast, 3-pentanone, which lacks a methyl group directly attached to the carbonyl carbon, will not produce iodoform, and thus, gives a negative result.
In this way, the iodoform test effectively differentiates between 2-pentanone and 3-pentanone by exploiting the presence of the methyl group in one and not the other.

Sodium Bisulfite Reaction

The sodium bisulfite (\(\text{NaHSO}_3\)) reaction is a common test employed to detect the presence of aldehydes in compounds. When sodium bisulfite is added, it reacts with aldehydes to form a white precipitate of bisulfite addition compounds.
Interestingly, this reaction is not effective in distinguishing between ketones such as 2-pentanone and 3-pentanone, because ketones generally do not react with sodium bisulfite. Ketones lack the reactivity of aldehydes due to their more stable carbonyl group environment, which is bounded by two alkyl groups, unlike the aldehyde’s hydrogen and alkyl group setup.
Thus, though sodium bisulfite is a powerful reagent for identifying aldehydes, it is not suitable for differentiating the ketones 2-pentanone and 3-pentanone. This limitation reinforces the need for other tests, like the iodoform test, to distinguish between different ketone compounds.