See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the parent ring. The structure is a six-membered carbocyclic ring (cyclohexane). Step 2: Identify substituents and special features. - There is an exocyclic double bond (=CH2, methylidene group) on one carbon of the ring. - Two hydroxyl (OH) groups are present on the ring. - One bromine (Br) atom is present on the ring. - One chlorine (Cl) atom is present on the ring. Step 3: Number the ring to give the principal characteristic groups (OH groups, as alcohols) the lowest possible locants, while also respecting the positions of the other substituents. - Assign C1 and C4 to the two OH-bearing carbons. - The adjacent carbon to C1 bears Br → C2 (Br at C2, but the image shows Br adjacent to the OH at C1 and Cl adjacent to Br). - Cl is on C3 (adjacent to Br at C2). - The methylidene (=CH2) exocyclic group is at C5. Step 4: Construct the IUPAC name. - Parent chain: cyclohexane with an exocyclic double bond → the suffix for =CH2 substituent uses 'methylidene' as a prefix, or the ring is named as cyclohex-5-ylidene... Actually, the exocyclic =CH2 is named as 'methylidene' substituent or the ring carbon bearing =CH2 is position 5 with the suffix '-ylidene'. Per IUPAC 2013 recommendations, the compound is named as a cyclohexane with a methylidene substituent: 5-methylidenecyclohexane-1,4-diol with Br at C3 and Cl at C2 (adjusting numbering): 3-bromo-2-chloro-5-methylidenecyclohexane-1,4-diol. - The two OH groups make it a diol: cyclohexane-1,4-diol. - Substituents listed alphabetically: bromo (C3), chloro (C2), methylidene (C5). - Full name: 3-bromo-2-chloro-5-methylidenecyclohexane-1,4-diol. Step 5: Verify locant set. The locants {1,2,3,4,5} are the lowest possible set for OH (1,4), Br (3), Cl (2), and =CH2 (5), consistent with IUPAC numbering rules. Therefore, the correct answer is 3-bromo-2-chloro-5-methylidenecyclohexane-1,4-diol.