See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The molecule shown has four carbons total. Reading the structure: there is a CH2Cl group attached to a carbon bearing an OH group, which is in turn attached to a carbon that bears both a methyl group and a terminal double bond (CH2=C). Step 2: Determine the parent chain. The longest carbon chain containing the principal characteristic group (OH) and the double bond is four carbons: ClCH2-CH(OH)-C(CH3)=CH2. This gives a but-3-ene backbone. Step 3: Number the chain to give the principal characteristic group (OH) the lowest possible locant. Numbering from the CH2Cl end: C1=CH2Cl, C2=CH(OH), C3=C(CH3)=, C4=CH2. The OH is at C2, the double bond is between C3 and C4 (but-3-en-2-ol), and the methyl branch is at C3. Step 4: Identify all substituents and their positions. - C1: chloromethyl → chloro at position 1 - C2: hydroxyl → ol at position 2 - C3: methyl branch → 3-methyl - C3-C4: double bond → but-3-en Step 5: Assemble the IUPAC name following the order: locant for halo substituent + halo + locant for methyl + methyl + parent chain + ene locant + ene + ol locant + ol → 1-chloro-3-methylbut-3-en-2-ol. There are no other structural interpretations since the chain length, the positions of OH, Cl, methyl, and double bond are unambiguous from the drawn structure. Therefore, the correct answer is 1-chloro-3-methylbut-3-en-2-ol.