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.
Find the equivalent capacitance between A and B, where \( C = 16 \, \mu F \).
If the equation of the parabola with vertex \( \left( \frac{3}{2}, 3 \right) \) and the directrix \( x + 2y = 0 \) is \[ ax^2 + b y^2 - cxy - 30x - 60y + 225 = 0, \text{ then } \alpha + \beta + \gamma \text{ is equal to:} \]