See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The compound is CH3-C(COOC2H5)=CH-CH2-COOH. The main chain contains a carboxylic acid group (-COOH) at one end, a CH2 group, a CH= group, and a C(CH3)= group bearing an ester substituent (COOC2H5), with a terminal CH3. Step 2: Identify the principal characteristic group. The -COOH (carboxylic acid) is the principal characteristic group, so it gets suffix '-oic acid' and is numbered C-1. Step 3: Number the chain. The longest chain including the -COOH carbon: C1 = COOH carbon C2 = CH2 C3 = CH= C4 = C(CH3)(COOC2H5)= C5 = CH3 This gives a 5-carbon chain (pent) with -oic acid suffix → pentanoic acid base. Step 4: Identify the double bond. The double bond is between C3 and C4, so the compound is pent-3-enoic acid. Step 5: Identify substituents. At C4 there is a -COOC2H5 (ethoxycarbonyl) group and a methyl group (which is part of the main chain as C5). The -COOC2H5 group is named 'ethoxycarbonyl' (ethyl ester of carboxylic acid). It is located at C4. Step 6: Assemble the name. The substituent -COOC2H5 at C4 is 'ethoxycarbonyl', and the base chain is pent-3-enoic acid, giving: 4-ethoxycarbonylpent-3-enoic acid. Step 7: Evaluate other options. - (b) '4-ethanoyloxypent-3-enoic acid' is incorrect because ethanoyloxy refers to -O-CO-CH3 (acetoxy), not -COOC2H5. - (c) '3-ethoxycarbonylbut-2-enecarboxylic acid' implies a 4-carbon chain with carboxylic acid as a substituent prefix, which is not the correct approach when -COOH is the principal group in the main chain. - (d) '3-ethoxycarbonylpent-3-enoic acid' places the ethoxycarbonyl at C3 instead of C4, which is incorrect. Therefore, the correct answer is A.