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

Step 1: Sulfonation of bromobenzene

Bromobenzene + $\ce{SO3/H2SO4}$ → sulfonation

Bromine is an ortho/para director (weakly deactivating).

In sulfonation, the para position is preferred over ortho due to:

• Less steric hindrance
• Para product is more stable

$\ce{C6H5Br + SO3/H2SO4 -> p-BrC6H4SO3H}$

Product A = para-bromobenzenesulfonic acid (4-bromobenzenesulfonic acid)

Structure: Br at position 1, $\ce{SO3H}$ at position 4

Step 2: Bromination of Product A

Para-bromobenzenesulfonic acid + $\ce{Br2/Fe}$

Now we have two substituents:

• Br at position 1: ortho/para director
• $\ce{SO3H}$ at position 4: meta director

When directors conflict:

• The more activating group dominates
• Br (ortho/para) is more activating than $\ce{SO3H}$ (meta, deactivating)

Br directs the incoming $\ce{Br}$ to positions ortho to itself:

• Position 2 (ortho to Br at C-1)
• Position 6 (ortho to Br at C-1, but also ortho to $\ce{SO3H}$)

The major product will have Br at positions 1, 2, and 4:

• C-1: original Br
• C-2: new Br (ortho to C-1 Br)
• C-4: $\ce{SO3H}$ group

This is 2,4-dibromobenzenesulfonic acid.

Product B = 2,4-dibromobenzenesulfonic acid

Answer: (b)

Key Directing Effects:

| Group | Effect | Directs to | |-------|--------|------------| | $\ce{-Br, -Cl}$ | Weakly deactivating | ortho/para | | $\ce{-SO3H}$ | Deactivating | meta | | $\ce{-OH, -OR}$ | Activating | ortho/para | | $\ce{-NO2}$ | Strongly deactivating | meta |

When directors conflict: More activating group wins.