See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The image shows a molecule with a triple bond (≡) on the left and a double bond (//) on the right, connected together, forming a 4-carbon chain. This gives the connectivity: CH≡C–CH=CH2. Step 2: Count the carbons. There are 4 carbons total: C1≡C2–C3=C4. Step 3: Identify the functional groups. There is a triple bond (alkyne) between C1 and C2, and a double bond (alkene) between C3 and C4. Step 4: Apply IUPAC nomenclature rules. The parent chain is butane (4 carbons). When both a double bond and a triple bond are present, the chain is named as an 'en-yne'. Numbering starts from the end that gives the lowest locants to the multiple bonds. Numbering from the left (triple bond end): triple bond at C1, double bond at C3. Numbering from the right (double bond end): double bond at C1, triple bond at C3. The set {1,3} is the same either way, but when there is a tie, the double bond gets the lower number. However, in this case numbering from the triple bond end gives triple bond at 1 and double bond at 3, while numbering from the alkene end gives double bond at 1 and triple bond at 3. IUPAC 2013 recommendations state that when there is a choice, lower locants are given to double bonds over triple bonds only as a tiebreaker. Here both numbering schemes give locant set {1,3}, so we prefer the lower locant for the double bond: but-1-en-3-yne would have double bond at 1 and triple bond at 3 (CH2=CH–C≡CH). But the structure shown is CH≡C–CH=CH2, which with numbering from the triple bond end gives but-1-en-3-yne where the ene is at 3 and yne at 1... Re-examining: the name but-1-en-3-yne means double bond at C1 (CH2=CH–) and triple bond at C3 (–C≡CH). The structure CH2=CH–C≡CH matches but-1-en-3-yne. The image shows ≡ on the left and = on the right, which is the same molecule (CH2=CH–C≡CH), just drawn from the alkyne terminus. Both representations describe the same compound. Step 5: Confirm the name. The 4-carbon compound with a double bond starting at C1 and a triple bond starting at C3 is named but-1-en-3-yne. Therefore, the correct answer is but-1-en-3-yne.