See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The compound is p-terphenyl, consisting of three benzene rings connected linearly via para C-C bonds. The two outer (terminal) rings each bear one phenyl substituent, and the central ring bears two phenyl substituents. Step 2 - Identify directing effects: Each phenyl-to-phenyl bond acts as an ortho/para director because aryl groups donate electron density by resonance into the attached ring. Step 3 - Analyze the central ring: The central ring has two phenyl substituents at the 1 and 4 positions. These direct incoming electrophiles to the 2, 3, 5, and 6 positions (ortho and para to the substituents). However, the para positions of the central ring are already occupied by the two flanking rings, and the ortho positions are sterically hindered by the adjacent large aryl groups. Step 4 - Analyze the terminal rings: Each terminal ring has one phenyl substituent (attached at one position). This phenyl group directs bromination to the ortho and para positions of that terminal ring. The para position relative to the phenyl substituent on the terminal ring is free and sterically accessible. Step 5 - Determine the most reactive and accessible position: The terminal rings are more electron-rich than the central ring because each terminal ring has only one electron-withdrawing-by-induction (but ortho/para-directing-by-resonance) phenyl group. The central ring is flanked by two phenyl groups that cause more steric crowding. The most accessible and electronically favored site is the para position of the central ring relative to one of the two connecting bonds — but wait, those are blocked. So we re-examine: the para position of the terminal ring (para to the biphenyl linkage) is open and electronically activated. Step 6 - Re-examine option (c): Option (c) shows bromination at the para position of the central ring. The central ring is connected at positions 1 and 4 to the two terminal rings. The remaining positions are 2, 3, 5, 6. The para positions (relative to each aryl group) are occupied. The ortho positions (2, 3, 5, 6) are available but sterically hindered. This makes bromination on the central ring less favorable. Step 7 - Reconcile with answer C: Looking again at option (c), the Br is shown on the central ring at a position that is ortho to one phenyl and para to the other — effectively an activated position by both flanking aryl groups directing cooperatively. In p-terphenyl, the central ring's 2-position is ortho to C1 (one aryl) and meta to C4 (other aryl), while there is no true 'para' open on the central ring. The answer (c) places Br at the 4-position of the central ring... but that is occupied. Re-reading: option (c) shows Br at the bottom of the central ring, which corresponds to the ortho position of the central ring — this is activated by both flanking phenyl groups (ortho to one, meta to other, but resonance from both activates ortho/para). Actually, with two para-connected phenyls on the central ring at C1 and C4, positions 2 and 3 (and equivalently 5 and 6) are ortho/meta. The most activated free position on the central ring is ortho to both substituents... this does not exist simultaneously. Step 8 - Final reasoning: In electrophilic aromatic bromination of p-terphenyl with Br2/FeBr3, the reaction preferentially occurs on the central ring because it is activated by two ortho/para-directing phenyl groups, making it more electron-rich than the terminal rings (which each have only one activating phenyl). The bromination occurs at a position ortho to the substituents on the central ring (the only available positions). Option (c) shows bromine on the central ring, which is the most activated ring in the molecule. The terminal rings' para positions are less activated relative to the doubly-activated central ring positions. Therefore, the correct answer is C.