Question

N₂Cl⁻

SnCl₂ / HCl NaOH (b) A P₂ 4[H]

Answer 1

The final product P₂ is aniline (C₆H₅NH₂).

Answer 2

We begin with an arenediazonium chloride (structure (b)) – i.e. a benzene ring with  –N₂Cl attached. Under strongly reducing conditions with SnCl₂/HCl in the presence of “4[H]” the diazonium group is not simply lost as N₂ but is step‐wise reduced. In fact, the –N₂Cl group is converted by a 4‑electron (4H–atom) reduction into a phenylhydrazine intermediate

$$\text{Ar–N}_2\text{Cl} \xrightarrow{\text{SnCl}_2/\text{HCl, }4[H]} \text{Ar–NH–NH}_2 \quad\text{(Product A)}$$

When this intermediate is subsequently treated with NaOH, a base–induced fragmentation occurs (the weak N–N bond is cleaved) so that ultimately an –NH₂ group replaces the hydrazino part. In other words:

$$\text{Ar–NH–NH}_2 \xrightarrow{\text{NaOH}} \text{Ar–NH}_2 \quad\text{(Product }P_2\text{)}$$

Thus, the overall transformation is the conversion of the diazonium salt into an aniline derivative.

1. Reduction: Arenediazonium chloride (b) is reduced with SnCl₂/HCl (using 4H–atoms) to yield phenylhydrazine (Product A).
2. Base treatment: Phenylhydrazine reacts with NaOH causing N–N bond cleavage to form aniline (Product P₂).