See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the functional groups: The structure shows both a carbon-carbon triple bond (alkyne) and a carbon-carbon double bond (alkene), making this an enyne compound. Step 2 - Count the carbon chain: Starting from the left terminal carbon, there are 6 carbons total: C1(=CH2)-C2=C3-C4≡C5-C6(CH3). This gives a hexane backbone, so the parent chain is hex. Step 3 - Locate and number the functional groups: The triple bond (yne) is between C4 and C5, and the double bond (ene) is between C2 and C3. Numbering from the end that gives lower locants to the double bond: double bond at C2-C3 (locant 2) and triple bond at C4-C5 (locant 4). The suffix becomes -en-4-yne, with the ene at position 2 giving hex-2-en-4-yne. Step 4 - Determine stereochemistry of the double bond: At C2, substituents are H and C1 (CH3 group, lower priority) vs at C3, substituents are H and C4≡C5-C6 chain. Applying E/Z rules: at C2, the chain (C1) has higher priority than H; at C3, the alkynyl chain has higher priority than H. The two higher-priority groups (C1-chain and C4-chain) are on opposite sides of the double bond, making it the E (trans) configuration. Step 5 - Assemble the IUPAC name: Parent chain = hex, double bond at 2 with E geometry, triple bond at 4 → (2E)-hex-2-en-4-yne. Therefore, the correct answer is (2E)-hex-2-en-4-yne.