See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the carbon skeleton. Count the total carbons in the longest chain that includes both the double bond and the triple bond. Starting from the end nearest the double bond: C1-C2=C3-C4-C5≡C6-C7-C8-C9-C10. This gives a 10-carbon (dec-) chain. Step 2 – Identify functional groups and their positions. There is a carbon-carbon double bond (en-) between C2 and C3, and a carbon-carbon triple bond (-yne) between C5 and C6. Numbering from the end closest to the double bond gives the lower locants (2 for the double bond, 5 for the triple bond) compared to numbering from the other end. Step 3 – Determine stereochemistry of the double bond. The double bond at C2-C3 has a methyl group on C3 side and a hydrogen on both carbons; applying E/Z (CIP) priority rules, the higher-priority groups on each carbon of the double bond are on opposite sides, making it the (E) configuration (trans). Step 4 – Assemble the IUPAC name. Parent chain = decane (10 carbons), double bond at position 2 (en), triple bond at position 5 (yne), E geometry at the double bond → (2E)-dec-2-en-5-yne. Step 5 – Verify numbering. Numbering from the alkene end gives locants 2 (double bond) and 5 (triple bond), sum = 7. Numbering from the other end would give locants 8 (double bond) and 5 (triple bond), sum = 13. Lower sum confirms correct numbering. Therefore, the correct answer is (2E)-dec-2-en-5-yne.