In the following series of reactions identify the major products A & B respectively:

### Related Formula Orienting effects in electrophilic aromatic substitution: - Bromine (-Br$-Br$) is ortho/para-directing (para-dominated due to steric hindrance). - Sulfonic acid group (-SO_3H$-SO_3H$) is a strong deactivating, meta-directing group.

### Core Logic Analyzing the first step: - Sulfonation of bromobenzene with mathrmSO_3/mathrmH_2mathrmSO_4$\mathrm{SO}_3/\mathrm{H}_2\mathrm{SO}_4$ yields 4-bromobenzenesulfonic acid as the major product (A) due to steric hindrance at the ortho-position. ### Step 1: Determine orientation for the second step In 4-bromobenzenesulfonic acid, we have two substituents: - -Br$-Br$ (ortho/para director) - -SO_3H$-SO_3H$ (meta director) The positions meta to the deactivating -SO_3H$-SO_3H$ group correspond to the positions ortho to the -Br$-Br$ group. Both directing effects align on the same position (carbon-3/carbon-5). Since -Br$-Br$ is activating relative to -SO_3H$-SO_3H$, it controls the orientation. ### Step 2: Identify Product B Halogenation with mathrmBr_2/mathrmFe$\mathrm{Br}_2/\mathrm{Fe}$ introduces a bromine atom ortho to the existing bromine atom (meta to -SO_3H$-SO_3H$): textProduct B = text3,4-dibromobenzenesulfonic acid$$\text{Product B} = \text{3,4-dibromobenzenesulfonic acid}$$ This matches Option (2). ### Pattern Recognition When an activating group (-Br$-Br$) and a deactivating group (-SO_3H$-SO_3H$) compete on a benzene ring, the orienting influence of the activating group wins. Position ortho to the bromine atom is favored over meta positions of the sulfonic acid group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes