Question

The IUPAC name of the following compound is :

• A. 3–Bromo–5–methylcyclopentanoic acid
• B. 4–Bromo–2–methylcyclopentane carboxylic acid
• C. 3–Bromo–5–methylcyclopentane carboxylic acid
• D. 5–Bromo–3–methylcyclopentanoic acid

Hint:

A \(\mathrm{COOH}\) group attached to a ring is named as '-carboxylic acid' and that carbon is C-1; then give the substituents the lowest locant set.

Answer 1

The Correct Option is B

Step 1: The ring carries a \(\mathrm{COOH}\) group, so the compound is a cycloalkane carboxylic acid, not a '-oic acid': parent = cyclopentanecarboxylic acid, with the ring carbon bearing \(\mathrm{COOH}\) as C-1. This rules out options (1) and (4) ('cyclopentanoic acid').
Step 2: Substituents are \(\mathrm{CH_3}\) on the carbon adjacent to C-1 and Br on the carbon two bonds further around the ring, on the far side from \(\mathrm{CH_3}\).
Step 3: Numbering towards the methyl gives locants \(\{2, 4\}\); numbering the other way gives \(\{3, 5\}\). The lowest locant set is \(\{2,4\}\).
Step 4: Alphabetical order: bromo before methyl, so the name is 4-bromo-2-methylcyclopentane carboxylic acid.
Answer: (2) 4–Bromo–2–methylcyclopentane carboxylic acid