Question

Butyne-1 and butyne-2 can be distinguished by (A) \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{NaOH}\) (C) \(\mathrm{Cu}_{2} \mathrm{Cl}_{2} / \mathrm{NH}_{4} \mathrm{OH}\) (D) \(\mathrm{H}_{2}-\mathrm{Pt}\)

Short answer

The reagent that can distinguish between butyne-1 and butyne-2 is (C) \(\mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{NH}_4 \mathrm{OH}\), as it selectively reacts with the terminal alkyne butyne-1, forming a dark red precipitate, while butyne-2 does not show any similar reaction.

Step-by-step solution

Understand the structure of butyne-1 and butyne-2

Butyne-1 is an alkyne with a triple bond between the first and second carbon atoms, and its structure can be represented as: $$ \mathrm{CH \equiv C-CH_2-CH_3} $$ On the other hand, butyne-2 is an alkyne with a triple bond between the second and third carbon atoms. Its structure can be represented as: $$ \mathrm{CH_3-C \equiv C-CH_3} $$

Analyze the reagents

Now we need to analyze the given reagents and determine which one can differentiate between butyne-1 and butyne-2. (A) \(\mathrm{Br}_2 / \mathrm{H}_2 \mathrm{O}\): Bromine water is a test for unsaturation (double or triple bonds). It will react with both butyne-1 and butyne-2 to form dibromoalkene. This reaction will occur for both the alkynes and will not be able to distinguish between them. (B) \(\mathrm{NaOH}\): Sodium hydroxide is a strong base, but it does not react with alkynes or alkene. Thus, it cannot be used to distinguish between butyne-1 and butyne-2. (C) \(\mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{NH}_4 \mathrm{OH}\): This reagent is ammoniacal cuprous chloride and selectively reacts with terminal alkynes (those with a triple bond at the end of the chain) turning dark red. Since butyne-1 is a terminal alkyne, it will react and form a dark red precipitate, while butyne-2 does not show any similar reaction. (D) \(\mathrm{H}_{2}-\mathrm{Pt}\): Hydrogenation with platinum catalysis will turn both butyne-1 and butyne-2 into alkanes. Since both isomers will undergo the same reaction, this reagent cannot be used to distinguish between them.

Conclusion

The reagent that can distinguish between butyne-1 and butyne-2 is (C) \(\mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{NH}_4 \mathrm{OH}\) because it selectively reacts with the terminal alkyne butyne-1, forming a dark red precipitate, while butyne-2 does not show any similar reaction.

Key concepts

Alkyne Reactivity

Alkynes are hydrocarbons with at least one carbon-carbon triple bond and are known for their distinctive reactivity patterns influenced by the acidity of terminal alkynes and the versatility of their triple bonds. In simple terms, the triple bond makes alkynes more reactive than alkanes and somewhat reactive compared to alkenes.

The triple bond in alkynes consists of one sigma bond and two pi bonds. The pi bonds are located above and below the plane of the atoms connected by the sigma bond. These pi bonds, due to their electron density, are the reason why alkynes readily undergo addition reactions.

For example, alkyne reactivity can be illustrated by the addition of halogens, where an electrophilic halogen molecule breaks the triple bond, resulting in a dihaloalkane. Moreover, because of their acidity, terminal alkynes can react with strong bases to form alkynides, which can be used in further synthetic applications. Understanding the basic reactivity of alkynes is crucial for students preparing for competitive exams like JEE in organic chemistry, as it forms the foundation for more complex reactions.

Terminal Alkyne Test

The terminal alkyne test is a specific chemical reaction used to identify the presence of an alkyne that has its triple bond located at the end of the carbon chain, which is known as a terminal alkyne. The test is quite simple: when treated with ammoniacal cuprous chloride \(\mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{NH}_4 \mathrm{OH}\), a terminal alkyne will form a precipitate that is dark red in color.

Here's why this happens: Terminal alkynes have a slightly negative carbon at the end of the chain due to the sp-hybridized orbital. When this carbon interacts with the copper (I) ion in the reagent, a copper acetylide complex is formed, which is insoluble and displays the characteristic color. This doesn't occur with internal alkynes, as they lack the slightly negative terminal carbon.

The test is selective and provides a clear visual indication, making it an efficient way to differentiate terminal alkynes from other types of alkynes, including internal alkynes. For students and practitioners in organic chemistry, this qualitative test is a simple yet powerful tool to analyze molecular structure without complex equipment.

Organic Chemistry JEE

Organic chemistry is a significant part of the Joint Entrance Examination (JEE), an engineering entrance assessment conducted in India. This subject tests students' understanding of carbon-containing compounds and their diverse reactions and properties. The syllabus covers a broad range of topics, including hydrocarbon classification, like alkynes, and specific chemical tests to identify different functional groups.

Success in organic chemistry requires a strong grasp of fundamental concepts, problem-solving skills, and practical applications. Students are often presented with molecules like butyne-1 and butyne-2 and asked to apply conceptual knowledge to distinguish between them using chemical reagents. Questions related to alkyne reactivity and specific tests such as the terminal alkyne test are common.

For students preparing for the JEE, practicing these concepts with textbook exercises is key. Being familiar with different reagents and their interactions with organic molecules is essential. JEE aspirants should also focus on reaction mechanisms, as they provide deeper insight into why certain reagents work or do not work in distinguishing chemical compounds. Overall, a combination of theoretical knowledge and practical application forms the core of organic chemistry preparation for JEE.