The major product Z obtained in the following reaction scheme is:

• Step 1: Diazotization (Formation of X)

  When 3-bromoaniline is treated with $$\text{NaNO}_2/\text{HCl}$$ at a low temperature ($$273 - 278\text{ K}$$), the primary aromatic amino group ($$-\text{NH}_2$$) is converted into a diazonium salt:

  $$\text{Product X} = \text{3-bromobenzenediazonium chloride}$$



• Step 2: Sandmeyer Reaction (Formation of Y)

  Treating the diazonium intermediate with cuprous bromide ($$\text{Cu}_2\text{Br}_2$$) replaces the diazonium group ($$-\text{N}_2^+\text{Cl}^-$$) with a bromine atom ($$-\text{Br}$$):

  $$\text{Product Y} = \text{1,3-dibromobenzene}$$



• Step 3: Electrophilic Aromatic Nitration (Formation of Z)

  Reacting 1,3-dibromobenzene with a nitrating mixture ($$\text{HNO}_3/\text{H}_2\text{SO}_4$$) introduces a nitro group ($$-\text{NO}_2$$) onto the benzene ring. The orientation is guided by the existing bromine substituents:

  • Bromine atoms are ortho/para-directing.
  • In 1,3-dibromobenzene, position 4 is ortho to one bromine atom and para to the other, making it highly activated by their cooperative electronic effects.
  • Position 2 (between both bromine atoms) is highly sterically hindered, minimizing any substitution there.
  $$\text{Product Z} = \text{1,3-dibromo-4-nitrobenzene (2,4-dibromonitrobenzene)}$$

Conclusion:

The sequence converts the amino group into a halogen atom and directs the subsequent nitration to the most sterically accessible and electronically activated position, yielding 1,3-dibromo-4-nitrobenzene.

Answer: Option B