See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Count the total carbons in the structure. The molecule shown has a terminal alkyne on the right side (vinyl/allyl type), a chain, a cis double bond, and a butyl chain on the left. Counting all carbons: HC≡C-CH2-CH2-CH=CH-CH2-CH2-CH2-CH2-CH2-CH3 — that gives 12 carbons total, so the parent chain is dodecane (dodec-). Step 2: Identify the functional groups. There is a terminal triple bond (alkyne) and an internal double bond (alkene). The compound is therefore an enyne. Step 3: Number the chain to give the principal characteristic group (or the first point of difference) the lowest locant. The terminal alkyne is at C1-C2 (since alkynes get lower numbers when in conflict with alkenes per IUPAC rules, and here numbering from the alkyne end gives triple bond at C2 and double bond at C6). Numbering from the alkyne end: C1≡C2-C3-C4-C5=C6-C7-C8-C9-C10-C11-C12. Step 4: Assign the double bond geometry. The double bond is between C6 and C7. Looking at the drawn structure, the two higher-priority groups on each carbon of the double bond are on the same side (cis), which corresponds to Z configuration. Step 5: Construct the IUPAC name. Parent chain: dodecane → dodec. Triple bond at position 2: -2-yne. Double bond at position 6 with Z geometry: (6Z)-6-en. Combining: (6Z)-dodec-6-en-2-yne. Step 6: Verify. The structure is HC≡C-CH2-CH2-CH=CH-(CH2)4-CH3 where the double bond at C6 is Z (cis), and the triple bond is at C2 (terminal). Total carbons = 12. Name = (6Z)-dodec-6-en-2-yne. ✓ Therefore, the correct answer is (6Z)-dodec-6-en-2-yne.