JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
The IUPAC name of the following compound is :
- (a) 3-Bromo-5-methylcyclopentane carboxylic acid
- (b) 3-Bromo-5-methylcyclopentanoic acid
- (c) 4-Bromo-2-methylcyclopentane carboxylic acid
- (d) 5-Bromo-3-methylcyclopentanoic acid
Answer and solution
Answer (c) 4-Bromo-2-methylcyclopentane carboxylic acid
Solution
Chemistry ma’am · 3 steps
The COOH sits outside the ring, so name it 'carboxylic acid' and number from its carbon towards the methyl.
- Choose the suffix. The -COOH carbon is attached to the ring but is not one of the five ring carbons. When a ring is the parent, this group gets the suffix carbonoic form 'carboxylic acid', giving cyclopentane carboxylic acid. The ending '-oic acid' is only for a COOH carbon that belongs to the parent chain itself.
- Fix C1. The ring carbon carrying -COOH is C1, because the principal group must get the lowest locant. Number round the ring in the direction that gives the substituents the lowest set of locants.
Direction from C1 Methyl at Bromo at Locant set Towards the CH₃ carbon 2 4 {2, 4} Towards the CH₂ carbon 5 3 {3, 5} - So. The locants are {2, 4} and the prefixes go in alphabetical order, bromo before methyl. The name is 4-Bromo-2-methylcyclopentane carboxylic acid, option (c).
The idea to keep. When COOH hangs off a ring, its carbon is not counted in the ring: use 'carboxylic acid', call the attachment carbon C1, then go the way that gives the lowest locant set.
Answer from the NTA final answer key. Solution by PrepWiser.