See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The compound has a central quaternary carbon bearing two methyl groups (one above, one below), connected to a CH3 on the left and a CH=CH2 on the right. This gives the full carbon skeleton: CH3-C(CH3)2-CH=CH2, which has 4 carbons in the longest chain containing the double bond plus 2 methyl substituents. Step 2: Identify the longest carbon chain containing the double bond. The chain is: CH2=CH-C(CH3)2-CH3. Counting: C1=CH2, C2=CH, C3=C(CH3)2, C4=CH3. This is a 4-carbon chain (butene) with a double bond starting at C1. Step 3: Number the chain to give the double bond the lowest locant. Numbering from the CH2= end: double bond is at C1 (but-1-ene). The two methyl substituents are both on C3. Step 4: Name the compound. Parent chain: but-1-ene. Substituents: two methyl groups at C3 → 3,3-dimethyl. Full name: 3,3-Dimethylbut-1-ene. Step 5: Eliminate incorrect options. - (a) 3,3,3-Trimethylprop-1-ene: prop-1-ene has only 3 carbons; three methyls at C3 of a 3-carbon chain would give 6 carbons total — incorrect structure. - (b) 1,1,1-Trimethylprop-2-ene: same issue with carbon count and numbering. - (d) 2,2-Dimethylbut-3-ene: numbering from the wrong end; this would place the double bond at C3, but lowest locant rule requires numbering from the end nearest the double bond, giving but-1-ene not but-3-ene. Therefore, the correct answer is C.