See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the parent chain. Count the longest continuous carbon chain that includes both double bonds and the carboxylic acid terminus. Starting from the –COOH carbon: C1(COOH)–C2=C3–C4–C5–C6=C7–C8–C9–C10–C11. This gives an 11-carbon chain → undecanoic acid backbone with two double bonds → undeca-2,6-dienoic acid. Step 2 – Locate and name the double bonds. One double bond is between C2 and C3 (conjugated with the carboxylic acid), and the other is between C6 and C7 (conjugated with the aldehyde). Both must be assigned E/Z geometry. • C2=C3: C1 carries –COOH (higher priority) vs H; C3 carries C4-chain (higher priority) vs H → the two higher-priority groups are on opposite sides → E. • C6=C7: C6 carries –CHO (higher priority) vs C5 chain; C7 carries C8 chain (higher priority) vs H → higher-priority groups on opposite sides → E. Hence the prefix (2E,6E). Step 3 – Identify and number substituents. • C4: cyclopropyl group → 4-cyclopropyl. • C5: bromine atom → 5-bromo. • C6: aldehyde (formyl, –CHO) substituent on the double bond carbon → 6-formyl. • C9: bromine atom → 9-bromo. Step 4 – Verify the terminal group. C9 bears Br, and C10–C11 is an ethyl group (–CH2CH3), consistent with an undecyl chain (11 carbons total). Step 5 – Assemble the IUPAC name. Parent: undeca-2,6-dienoic acid. Geometry: (2E,6E). Substituents in alphabetical order: 5,9-dibromo (two Br atoms at C5 and C9), 4-cyclopropyl, 6-formyl. Full name: (2E,6E)-5,9-dibromo-4-cyclopropyl-6-formylundeca-2,6-dienoic acid. Therefore, the correct answer is (2E,6E)-5,9-dibromo-4-cyclopropyl-6-formylundeca-2,6-dienoic acid.