Question

Ethene and Aniline can be distinguished by (A) \(\mathrm{NaHCO}_{3}\) test (B) \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\) test (C) Bayer's reagent test (D) Both (B) and (C)

Short answer

Both the Bromine water test (Br2/H2O) and Bayer's reagent test can effectively distinguish between Ethene and Aniline. The bromine water test will give a positive result (bromine will lose its color) for Ethene, whereas Aniline will not react, and the water will remain colored. The Bayer's reagent test, an alkaline solution of potassium permanganate (KMnO4), will cause the purple-colored permanganate to lose color when in the presence of an alkene like Ethene, but not with Aniline. Therefore, the correct answer is option (D), as both tests (B) and (C) can differentiate these substances.

Step-by-step solution

Test A: Sodium bicarbonate test (NaHCO3)

The sodium bicarbonate test is used to distinguish between carboxylic acids and other functional groups. When a carboxylic acid reacts with sodium bicarbonate, a salt, water, and carbon dioxide are produced. Because neither Ethene nor Aniline is a carboxylic acid, both substances do not react with NaHCO3. This test will not be able to distinguish between Ethene and Aniline, making option (A) incorrect.

Test B: Bromine water test (Br2/H2O)

The bromine water test is a reaction between bromine and an alkene. When an alkene is present, the bromine reacts with the double bond, breaking the double bond and creating a vicinal dibromo compound. Since Ethene is an alkene, the bromine water test will give a positive result (bromine will lose its color) in the case of Ethene. However, Aniline will not react with bromine; hence, the water will remain colored. This test is effective to distinguish between Ethene and Aniline. So, option (B) is correct.

Test C: Bay­er's reagent test

Bayer's reagent, which is an alkaline solution of potassium permanganate (KMnO4), is used as an oxidizing agent to identify the presence of alkenes. The reagent reacts with the double bond in an alkene, causing the purple-colored permanganate to lose color as the reaction proceeds. As Ethene is an alkene, it will cause the permanganate solution to lose color, making the solution clear. On the other hand, Aniline would not react with this reagent since it does not have a double carbon-carbon bond. As the solution remains purple, it can be concluded that the presence of Aniline has been identified. Therefore, option (C) is also correct. So, both the Bromine water test and Bayer's reagent test can effectively distinguish between Ethene and Aniline. Hence, the correct answer is option (D), as both tests (B) and (C) can be used to differentiate these substances.

Key concepts

Bromine Water Test

To understand the bromine water test, it's fundamental to know it is used for detecting the presence of unsaturated compounds, such as alkenes. When bromine water (Br₂/H₂O) comes into contact with an alkene, a characteristic reaction occurs. The bromine molecule (Br₂) reacts with the carbon-carbon double bond to form a colorless dibromo compound. This reaction also results in the decolorization of the originally brown bromine water, serving as a visual indicator that an alkene is present.

In the given exercise, ethene's reaction with bromine water leads to a loss of the brown color, confirming the presence of a carbon-carbon double bond, while aniline does not contain this bond and does not react, leaving the bromine water colored. Thus, the bromine water test effectively distinguishes between ethene, an alkene, and aniline, an aromatic amine.

Bayer's Reagent Test

Moving on to Bayer's reagent test, which involves an alkaline solution of potassium permanganate (KMnO₄), it is another crucial method for identifying alkenes. The vibrant purple color of potassium permanganate fades to colorless or brown manganese dioxide when it reacts with alkenes. This test relies on the oxidative ability of the permanganate ion under alkaline conditions to add across the carbon-carbon double bond.

For the compounds in question, ethene will react with Bayer's reagent by changing the purple color to colorless or a brown precipitate, indicating a double bond's presence. In contrast, aniline lacks this double bond and will not affect the potassium permanganate's color. Bayer's reagent test not only confirms the presence of double bonds but can also hint at the bond's specific location within a molecule.

Functional Group Identification

Functional group identification is the cornerstone of organic chemistry, as it allows chemists to determine the structural features responsible for the chemical behavior of a molecule. Through various chemical tests, functional groups such as alkenes, alcohols, carboxylic acids, amines, and others can be discerned.

In our context, the bromine water and Bayer's reagent tests are instrumental in identifying the alkene functional group present in ethene and absent in aniline. While these tests are specific for alkenes, other compounds necessitate additional specialized tests. True mastery in organic chemistry involves understanding each functional group's unique reactions and choosing appropriate tests to identify them.

Alkene Reactions

Alkene reactions form a significant segment of organic chemistry, known for their diversity and usefulness in synthesis. One of the primary characteristics of alkenes is their carbon-carbon double bonds, which are regions of high electron density and thus reactive. They undergo additions reactions, where elements or compounds add across the double bond.

For example, in reactions with bromine water and Bayer's reagent, an alkene like ethene undergoes an addition reaction. The double bond opens up to accommodate the new atoms, resulting in a vicinal dibromo compound with bromine and a diol with Bayer's reagent. By understanding the specifics of such reactions, we can infer the presence of alkenes and predict the outcomes of their chemical behavior with accuracy.