Number of Bromine atoms present in product

The correct answer is 5.

Let us analyze the chemical reaction sequence step-by-step starting from Nitrobenzene, as shown in the given image:

Step (i): Bromination of Nitrobenzene with Br₂ / Fe
The nitro group (-NO₂) on the benzene ring is a meta-directing and strongly deactivating group. Electrophilic aromatic substitution with Br₂ in the presence of Fe catalyst introduces one bromine atom at the meta-position relative to the -NO₂ group.
Product formed: m-Bromonitrobenzene (contains 1 Br atom).

Steps (ii) & (iii): Reduction with Sn / HCl followed by pH neutralisation
The nitro group (-NO₂) is reduced to an amino group (-NH₂).
Product formed: m-Bromoaniline (contains 1 Br atom).

Step (iv): Reaction with Bromine water (Br₂ / H₂O)
The amino group (-NH₂) is a very strong ortho/para-directing and activating group. In m-bromoaniline, the positions ortho and para to the -NH₂ group are positions 2, 4, and 6 (assuming -NH₂ is at position 1 and -Br is at position 3). Bromine water polybrominates all available ortho and para positions relative to the -NH₂ group.
Since position 3 is already occupied by bromine, electrophilic substitution occurs at positions 2, 4, and 6, introducing 3 additional bromine atoms.
Product formed: 2,3,4,6-Tetrabromoaniline (contains 1 + 3 = 4 Br atoms).

Step (v): Diazotisation with NaNO₂ + HCl
The primary aromatic amine group (-NH₂) reacts with nitrous acid (NaNO₂ + HCl) at low temperature to form a diazonium salt.
Product formed: 2,3,4,6-Tetrabromobenzenediazonium chloride (contains 4 Br atoms).

Step (vi): Sandmeyer Reaction with CuBr
Treatment of the diazonium salt with cuprous bromide (CuBr) replaces the diazonium group (-N₂⁺Cl^-) with a bromine atom (-Br).
Product formed: 1,2,3,4,5-Pentabromobenzene.

Conclusion:
The final product obtained is pentabromobenzene, which contains a total of 5 bromine atoms.