See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the parent chain and principal characteristic group. The molecule contains a C=O–NH group (amide), so the parent is an amide. Counting carbons on the acyl side: C1(carbonyl)–C2(CH2)–C3(CH–cyclopentadienyl)–C4(CH2Br) gives a four-carbon chain → butanamide. Step 2 – Name the acyl portion substituents. • C3 carries a cyclopenta-2,4-dien-1-yl group (a cyclopentadienyl ring attached through C1 of the ring, with double bonds at positions 2,3 and 4,5 of the ring). • C4 carries a bromo substituent and is the terminal carbon (–CH2Br). So the acyl part is: 4-bromo-3-(cyclopenta-2,4-dien-1-yl)butanoyl. Step 3 – Name the amine portion (N-substituent). On nitrogen: the carbon directly attached to N bears a cyclopropyl group and a –CH2Cl group, making it a 2-cyclopropyl-3-chloropropyl group. Numbering from N: C1(NH–CH)–C2(cyclopropyl)–C3(CH2Cl) → 3-chloro-2-cyclopropylpropyl. Step 4 – Assemble the IUPAC name. Combine acyl name + amide + N-substituent: 4-bromo-N-(3-chloro-2-cyclopropylpropyl)-3-(cyclopenta-2,4-dien-1-yl)butanamide. Step 5 – Verify locants and stereodescriptors. No stereocenters are specified in the image; all locants are consistent with the structure drawn. The answer matches the given ground truth. Therefore, the correct answer is 4-bromo-N-(3-chloro-2-cyclopropylpropyl)-3-(cyclopenta-2,4-dien-yl)butanamide.