See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the functional groups: The structure shown is HC≡C-CH2-CH=CH2. It contains a terminal triple bond (alkyne) and a terminal double bond (alkene) within a five-carbon chain. Step 2 - Count the carbons: C1≡C2-C3-C4=C5. There are five carbons in the longest chain containing both unsaturations, so the parent chain is 'pent'. Step 3 - Number the chain to give lowest locants to the principal characteristic group and multiple bonds. IUPAC rules prioritize giving the lowest possible locants. Numbering from the alkene end: C1=C2-C3-C4≡C5 would give double bond at 1, triple bond at 4. Numbering from the alkyne end: C1≡C2-C3-C4=C5 would give triple bond at 1, double bond at 4. When an enyne is named, the double bond (ene) gets the lower locant according to IUPAC 2013 recommendations when the set of locants is the same by either direction; here {1,4} vs {1,4} are the same set, so the double bond (ene suffix) gets the lower number. Therefore numbering starts from the alkene end: double bond at C1, triple bond at C4. Step 4 - Construct the name: parent chain = pent, double bond at position 1 = pent-1-en, triple bond at position 4 = pent-1-en-4-yne. Step 5 - Verify: pent-1-en-4-yne corresponds to CH2=CH-CH2-C≡CH, which matches the drawn structure HC≡C-CH2-CH=CH2 read from either end. Therefore, the correct answer is pent-1-en-4-yne.