See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the functional groups and carbon skeleton. The structure shows a terminal alkyne (C≡C) at one end and a double bond (C=C) elsewhere, with five carbons total. Reading from the structure: C1≡C2–C3=C4–C5(H3), giving pent-3-en-1-yne as the base name. Step 2 – Number the chain. IUPAC rules require giving the principal characteristic group (alkyne) the lowest possible locant. Numbering from the alkyne end: C1≡C2, then C3=C4, then C5. This gives 'pent-3-en-1-yne' (alkyne at C1, double bond at C3). Step 3 – Determine the geometry of the double bond. The double bond is between C3 and C4. C3 bears H and C2 (part of the alkyne chain); C4 bears H and C5 (methyl). Looking at the drawn structure, the two hydrogen atoms (or the two higher-priority groups) are on the same side of the double bond — the chain bends toward the same face, indicating Z (cis) geometry. Using CIP priority: on C3, –C≡CH > –H; on C4, –CH3 > –H. In the drawn structure both higher-priority groups (the alkyne side and the methyl side) are on the same side of the double bond, confirming Z configuration. Step 4 – Assemble the full name. Combining: (3Z)-pent-3-en-1-yne. Step 5 – Why not E? If the groups were on opposite sides of the double bond, it would be (3E)-pent-3-en-1-yne. The drawn wedge/line geometry shows a Z (cis) arrangement. Therefore, the correct answer is (3Z)-pent-3-en-1-yne.