Consider the above sequence of reactions. (1) textBr_2 / textFeBr_3 / Delta$\text{Br}_2 / \text{FeBr}_3 / \Delta$ (2) textSn / textHCl / Delta$\text{Sn} / \text{HCl} / \Delta$ (3) textpH neutralisation rightarrow textMajor Product (P)$\text{pH neutralisation} \rightarrow \text{Major Product (P)}$ (4) textBr_2 / textH_2textO$\text{Br}_2 / \text{H}_2\text{O}$ (5) textNaNO_2 / textHBr, 0-5^circtextC$\text{NaNO}_2 / \text{HBr}, 0-5^{\circ}\text{C}$ (6) textCuBr / textNaBr$\text{CuBr} / \text{NaBr}$ The number of bromine atom(s) in the final product (P) will be:

### Core Logic Tracing the steps through nitration/bromination, reduction to amine via textSn/HCl$\text{Sn/HCl}$, subsequent extensive bromination with textBr_2/textH_2textO$\text{Br}_2/\text{H}_2\text{O}$, diazotization, and Sandmeyer bromination (textCuBr/NaBr$\text{CuBr/NaBr}$), we get substitution at multiple positions leading to 5 bromine atoms in the final product structure. ### Step 1: Final Calculation Number of Br atoms in major product (P) = 5. ### Pattern Recognition Sees: Multi-step aromatic conversion involving halogenation and diazotization. Trap: Counting substituent groups incorrectly after Sandmeyer reaction. ### Chapter Mix Class 12 Chemistry: Amines