See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the parent chain and functional groups: The molecule contains a carboxylic acid (-COOH) and an aldehyde (-CHO) at opposite ends. The longest carbon chain connecting these two groups must be identified. Counting carbons: COOH-CH=CH-CH(Br)-CH(Cl)-CH=CH-CH(cyclopropyl)-CH2-CHO gives 10 carbons total, so the parent chain is deca-dienoic acid with the -oic acid at C1 and the aldehyde (oxo) at C10. Step 2 - Locate the double bonds: There are two double bonds in the chain. One is between C2 and C3 (adjacent to the carboxylic acid end), and the other is between C6 and C7 (in the middle of the chain). Step 3 - Assign E/Z geometry: For the C2-C3 double bond: C1 is COOH (higher priority) vs H on C2, and C4 (Br-bearing carbon, higher priority) vs H on C3 — the higher priority groups are on opposite sides, so E configuration. For the C6-C7 double bond: C5 (Cl-bearing, higher priority) vs H on C6, and C8 (cyclopropyl-bearing, higher priority) vs H on C7 — the higher priority groups are on opposite sides, so E configuration. Thus the compound is (2E,6E). Step 4 - Locate substituents: Bromine is at C4, chlorine is at C5, cyclopropyl is at C8, and the aldehyde (oxo) is at C10. Step 5 - Assemble the name: Parent chain = deca-2,6-dienoic acid; substituents at C4 (bromo), C5 (chloro), C8 (cyclopropyl), C10 (oxo); stereochemistry (2E,6E). Full name: (2E,6E)-4-bromo-5-chloro-8-cyclopropyl-10-oxodeca-2,6-dienoic acid. Therefore, the correct answer is (2E,6E)-4-bromo-5-chloro-8-cyclopropyl-10-oxodeca-2,6-dienoic acid.