See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substituents and their positions on the benzene ring. From the structure, four substituents are present: F (top), Cl (right), Br (bottom), I (left). Going around the ring, the relative positions are: F and Cl are adjacent (ortho), Cl and Br are adjacent (ortho), Br and I are adjacent (ortho), and I and F are separated by one carbon (para relationship with Cl). Step 2 - Apply IUPAC numbering rules for polysubstituted benzenes. The ring must be numbered to give the substituents the lowest possible set of locants, and when there is a tie, alphabetical order of substituents determines the principal group to receive the lowest number. Step 3 - Assign numbers based on alphabetical priority and lowest locant set. Alphabetically: Bromo < Chloro < Fluoro < Iodo. Assign C1 to Chloro (Cl). Going in the direction that gives lower numbers to the other substituents: C1 = Cl, C2 = Br (adjacent, going one way), C3 = I (adjacent to Br), and C5 = F (two carbons from I, across the ring from Cl). This gives locant set {1,2,3,5} for Cl, Br, I, F respectively. Step 4 - Verify this matches option B: 2-Bromo-1-chloro-5-fluoro-3-iodo benzene. With C1=Cl, C2=Br, C3=I, C5=F, listed alphabetically: 2-Bromo, 1-Chloro, 5-Fluoro, 3-Iodo benzene. This matches option B exactly. Step 5 - Why other options fail: - Option (a) gives 1-Bromo-2-chloro-3-fluoro-6-iodo: locant set {1,2,3,6} but F and Cl are not adjacent (they are para-like), so this arrangement is geometrically incorrect. - Option (c) gives 4-Bromo-2-chloro-5-iodo-1-fluoro: locant set {1,2,4,5} which is higher than {1,2,3,5}. - Option (d) gives 2-Bromo-3-chloro-1-iodo-5-fluoro: placing I at C1 violates alphabetical priority rule (Cl should get lower number than I). Therefore, the correct answer is B.