In the following substitution reaction :
Product 'P' formed is :

Solution

Chemistry ma’am · 3 steps

Check where each Br sits relative to NO₂: only the para Br is activated, so only that one leaves.

1. Read the substrate. The ring carries NO₂ at C1, one Br at C3 and one Br at C4. So one Br is meta to NO₂ and the other is para to it. Ethoxide, C₂H₅O⁻ from C₂H₅ONa, is the nucleophile.
2. How haloarenes react. A haloarene undergoes nucleophilic substitution only when an electron-withdrawing group sits ortho or para to the halogen. When C₂H₅O⁻ adds to the carbon holding Br, the negative charge of the intermediate spreads onto the carbons ortho and para to that carbon. A NO₂ group on one of those carbons can hold the charge on its oxygen atoms.

   Bromine	Position relative to NO₂	Can NO₂ hold the negative charge?	Substituted?
   Br at C4	para	yes, the charge reaches C1 and goes onto NO₂	yes, replaced by OC₂H₅
   Br at C3	meta	no, the charge never reaches C1	no, stays as Br

3. So. Only the para Br is swapped for OC₂H₅. P is 2-bromo-1-ethoxy-4-nitrobenzene, with OC₂H₅ para to NO₂ and Br next to OC₂H₅. That is option (d).

The idea to keep. In haloarene substitution, NO₂ activates only the halogens ortho or para to it. A meta halogen stays put, and NO₂ itself is not the group that leaves.

Answer from the NTA final answer key. Solution by PrepWiser.