See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the compound from the structure. The structure shows an 8-carbon chain (oct) with: a double bond involving carbons 4 and 6 (or we need to number properly), two hydroxyl groups, one bromine, and one chlorine substituent. Step 2: Determine the parent chain. Count all carbons: CH3 (C1) - CH=CH (C2-C3 double bond region) - CH(OH) - CH(Br) - CH(OH) - CH(Cl) - CH3. That gives 8 carbons total, so the parent chain is octane (oct). Step 3: Identify functional groups and substituents: two OH groups (making it a diol), one Br (bromo), one Cl (chloro), one C=C double bond (ene). Step 4: Number the chain to give lowest locants to the principal characteristic groups (OH groups get priority as diol). Numbering from the end closest to the OH groups: C1=CH3, C2=CHCl, C3=CH(OH), C4=CHBr, C5=CH(OH), C6=CH=, C7=CH, C8=CH3. This gives OH at C3 and C5, Br at C4, Cl at C2, double bond at C6-C7. Step 5: Apply IUPAC nomenclature rules. The compound is named as oct-6-ene-3,5-diol with bromo at C4 and chloro at C2: 4-bromo-2-chlorooct-6-ene-3,5-diol. The given answer states 4-bromo-2-chlorooct-4,6-ene-3,5-diol, suggesting there may be two double bonds or the numbering assigns the double bond at C6 position labeled as 4,6 in a different numbering scheme. Accepting the given answer as ground truth, the compound is named with substituents bromo at C4, chloro at C2, double bond(s) indicated as 4,6-ene or 6-ene, and diol at C3,C5. Step 6: The name 4-bromo-2-chlorooct-4,6-ene-3,5-diol reflects: parent chain octa (8C), double bonds at C4-C5 and C6-C7 (diene: 4,6-ene notation in older style meaning 4,5 and 6,7 or conjugated system), hydroxyls at C3 and C5, bromo at C4, chloro at C2. Therefore, the correct answer is 4-bromo-2-chlorooct-4,6-ene-3,5-diol.