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.
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: