In the given reactions, we need to identify the compounds A and B.
The first reaction involves the reduction of a nitro group (\(-\text{NO}_2\)) to an amine group (\(-\text{NH}_2\)) using hydrogen and palladium (\(\text{H}_2/\text{Pd}\)) in ethanol. This suggests A is \textit{nitrobenzene}. The second reaction involves the bromination of A with bromine in the presence of sodium hydroxide (\(\text{NaOH}\)), leading to the formation of B. B is likely to be \(\text{4-bromobenzenamine}\), as the bromine typically substitutes in the para position in the presence of \(\text{NaOH}\).
Thus,
A = \textit{Nitrobenzene} (\(\text{C}_6\text{H}_5\text{NO}_2\))
B = \(\text{4-Bromobenzenamine}\) (\(\text{C}_6\text{H}_4\text{BrNH}_2\))