In the given reactions identify A and B.
- A: 2–Pentyne B: trans–2–butene
- A: n–Pentane B: trans–2–butene
- A: 2–Pentyne B: Cis–2–butene
- A: n–Pentane B: Cis–2–butene
The Correct Option is A
Solution and Explanation
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.
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