See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the functional groups. The molecule has a terminal alkyne (C≡CH) at one end and a terminal alkene (CH2=CH-) at the other end, with a 6-carbon chain total. Step 2: Count the carbons. The structure is CH2=CH-CH2-CH2-C≡CH, which has 6 carbons: C1=C2-C3-C4-C5≡C6 or C6=C5-C4-C3-C2≡C1. Step 3: Apply IUPAC numbering rules. According to IUPAC rules, when both a double bond and a triple bond are present, the chain is numbered to give the lower locants to the multiple bonds as a set. If the locant sets are equal, the double bond gets the lower number. Step 4: Number from the alkene end: C1=C2-C3-C4-C5≡C6 gives double bond at C1 and triple bond at C5. Locant set: {1, 5}. Number from the alkyne end: C1≡C2-C3-C4-C5=C6 gives triple bond at C1 and double bond at C5. Locant set: {1, 5}. Step 5: Since both numbering schemes give the same locant set {1,5}, the IUPAC rule states that the double bond is given the lower locant. Therefore, number from the alkene end: double bond at C1, triple bond at C5. The name is Hex-1-en-5-yne. Step 6: Eliminate wrong answers. (a) Hex-5-en-1-yne would imply numbering from the alkyne end giving triple bond at 1 and double bond at 5 — this violates the rule that double bond gets lower number when sets are equal. (c) Hex-6-en-1-yne and (d) Hex-1-en-6-yne are impossible for a 6-carbon chain (position 6 cannot have a double bond starting there for a terminal alkene in a 6C chain). Therefore, the correct answer is B.