The products A and B formed in the following reaction scheme are respectively:

1.Formation of Product A

• Reaction: Benzene undergoes nitration using $$\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4$$ to form nitrobenzene.
• Reduction: Nitrobenzene is then reduced by $$\text{Sn/HCl}$$ to convert the nitro group ($$-\text{NO}_2$$) into an amino group ($$-\text{NH}_2$$).
• Product A: Aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$).

2. Formation of Product B

• Diazotization: Aniline reacts with $$\text{NaNO}_2 + \text{HCl}$$ at a low temperature ($$0\text{-}5^\circ\text{C}$$ / $$273\text{-}278\text{K}$$) to form the stable benzene diazonium chloride salt ($$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$$).
• Azo Coupling: The diazonium ion acts as an electrophile and couples with phenol at its highly reactive, less sterically hindered para-position.
• Product B: $$p$$-Hydroxyazobenzene (an orange azo dye).