Question

In the given reactions identify A and B.

• A. A: 2–Pentyne    B: trans–2–butene
• B. A: n–Pentane    B: trans–2–butene
• C. A: 2–Pentyne    B: Cis–2–butene
• D. A: n–Pentane    B: Cis–2–butene

Answer 1

The Correct Option is A

To identify compounds A and B from the given reactions, we need to analyze the transformations shown:

1. The first reaction involves hydrogenation using Pd/C as a catalyst. Pd/C typically catalyzes the addition of hydrogen to alkynes, converting them into alkenes. 
2. Here, compound A undergoes partial hydrogenation. Given that the product from this reaction is an alkene, A must be an alkyne.
3. The structural formula shown for the product suggests a symmetrical molecule with a trans configuration, which indicates trans–2–butene. This partial hydrogenation is specific to alkynes turning into trans alkenes when done via Lindlar's catalyst or related methods.
4. Now looking at the structure of A, the triple bond location confirms it to be 2–Pentyne, as any other location would not produce this product.

Therefore, for the given reactions, A is identified as 2–Pentyne and B as trans–2–butene, correlating with the correct option:

• A: 2–Pentyne    B: trans–2–butene

Answer 2

Step-by-step Analysis:
1. Reaction of A with Hydrogen (H$_2$) in the Presence of Pd/C:
This reaction involves the partial hydrogenation of an alkyne (2-pentyne) to an alkene (pent-2-ene) using a palladium catalyst supported on carbon (Pd/C). The product of this reaction is a cis-alkene if a Lindlar catalyst is used, but in this case, the hydrogenation results in a trans-alkene (trans-2-butene) under typical catalytic conditions.
2. Reduction of CH$_3$--C $\equiv$ C--CH$_3$ (2-Butyne) with Sodium in Liquid Ammonia:
This reaction is known as the Birch reduction, which selectively converts alkynes to trans-alkenes. Therefore, the product B formed is trans-2-butene.
Conclusion:
Compound A is 2-pentyne.
Compound B is trans-2-butene.