See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substituents and their directing effects: The ring bears two substituents: an acetamido group (NHC(=O)CH3) and a methyl group (CH3) on adjacent carbons (positions 1 and 2 respectively). Step 2 - Directing effects: The acetamido group (NHCOMe) is an ortho/para director and activator (lone pair on nitrogen donates into the ring despite the carbonyl, net effect is o/p director). The methyl group (CH3) is also an ortho/para director and activator. Step 3 - Assign positions: Let C1 = NHCOMe, C2 = CH3. The remaining positions are C3 (labeled A), C4 (labeled B), C5 (labeled C), C6 (unlabeled, adjacent to C1). Step 4 - Determine positions activated by each group: - NHCOMe at C1 directs to ortho (C2, C6) and para (C4 = B). C2 is already substituted. - CH3 at C2 directs to ortho (C1, C3 = A) and para (C5 = C). C1 is already substituted. Step 5 - Find positions activated by BOTH groups (cooperative/synergistic activation): - Position A (C3): ortho to CH3 (activated by CH3) but meta to NHCOMe (not activated by NHCOMe). Only one group activates. - Position B (C4): para to CH3? No, C4 is para to C1 (NHCOMe), and meta to C2 (CH3). Wait - re-examining: para to NHCOMe at C1 is C4 = B. Meta to CH3 at C2 is also C5, not C4. C4 is ortho to C3, meta to C2(CH3) going one way... Actually C4 is para to C1(NHCOMe) and also it is 2 positions from C2(CH3) making it meta to CH3. So B is para to NHCOMe (strongly activated) but meta to CH3 (not deactivated, just not extra activated). - Position C (C5): para to CH3 at C2 (activated by CH3), and meta to NHCOMe at C1 (not activated by NHCOMe). - Position C6: ortho to NHCOMe (activated), meta to CH3 (not extra activated). Step 6 - The acetamido group is a stronger activating/directing group than methyl due to nitrogen lone pair donation. The para position to the acetamido group (position B, C4) receives the strongest activation from the dominant director (NHCOMe). Additionally, being para to the stronger group provides maximal electron density at B. While B is meta to CH3, methyl does not deactivate meta positions strongly - it simply doesn't activate them as much. The dominant activation from NHCOMe to its para position (B) makes B the most electron-rich and most favorable for EAS. Step 7 - Why not A or C: Position A is ortho to CH3 but meta to NHCOMe - the stronger group doesn't favor it. Position C is para to CH3 but meta to NHCOMe - again the stronger group doesn't favor it. Position B benefits from the para direction of the stronger NHCOMe group. Therefore, the correct answer is B.