See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the functional group: The structure contains a -C(=O)-Br group, which is an acyl halide (acid bromide). IUPAC nomenclature for acyl halides uses the suffix '-oyl halide' derived from the parent carboxylic acid. Step 2 - Identify the parent chain: The carbon chain including the carbonyl carbon is: C(=O)Br - CH2 - CH(CH3) - CH3. Counting the carbonyl carbon, the longest chain containing the acyl group has 4 carbons (butanoyl). The parent acid would be butanoic acid, so the base name is 'butanoyl bromide'. Step 3 - Number the chain: In acyl halides, the carbonyl carbon is always C-1. So numbering from the carbonyl end: C1 = C(=O)Br, C2 = CH2, C3 = CH(CH3), C4 = CH3. Step 4 - Identify substituents: There is a methyl branch at C3. Step 5 - Assemble the name: 3-methylbutanoyl bromide. Step 6 - Evaluate other options: (a) 2-Methylbutanoyl bromide would place the methyl at C2, which is incorrect since CH2 at C2 has no substituent. (b) 2-Methylbutan-4-oyl bromide is not valid IUPAC nomenclature; the carbonyl carbon in acyl halides is always C1 by definition. (c) 1-Bromo-3-Methylbutanone treats the compound as a haloketone, which is incorrect; the -C(=O)Br is an acyl bromide, not a bromoketone. (d) 3-Methylbutanoyl bromide correctly names the compound with the methyl substituent at C3 of the butanoyl chain. Therefore, the correct answer is D.