JEE Main 2025 · 22 January · morning · Single correct · +4 / −1
The IUPAC name of the following compound is :
COOH, COOCH₃; |, |; CH₃, -, CH, -, CH₂, -, CH₂, -, CH, -, CH₃
- (a) 5-Methoxycarbonly-2-methylhexanoic acid
- (b) Methyl-5-carboxy-2-methylhexanoate.
- (c) 2-Carboxy-5-methoxycarbonylhexane.
- (d) Methyl-6-carboxy-2,5-dimethylhexanoate.
Answer and solution
Answer (a) 5-Methoxycarbonly-2-methylhexanoic acid
Solution
Chemistry ma’am · 5 steps
Two carbonyl groups, but -COOH outranks the ester, so the acid takes the suffix.
- Pick the boss. The molecule has a carboxylic acid (-COOH) and an ester (-COOCH₃). In IUPAC priority, carboxylic acid ranks above ester. The name ends in -oic acid, and the ester becomes a prefix, methoxycarbonyl.
- Number the chain. The main chain must contain the -COOH carbon, and that carbon is C1. The chain runs C1 (COOH), C2 (CH), C3 (CH₂), C4 (CH₂), C5 (CH), C6 (CH₃). That makes six carbons, so the parent is hexanoic acid.
- Place substituents. C2 carries a CH₃ group, which gives 2-methyl. C5 carries -COOCH₃, which gives 5-methoxycarbonyl.
- Alphabetical order. Compare the prefixes letter by letter: "metho" comes before "methy" because o comes before y. So methoxycarbonyl is written first.
- So. The name is 5-methoxycarbonyl-2-methylhexanoic acid, which is option (a).
The idea to keep. When two functional groups compete, find the higher-priority one first (acid > ester > amide > nitrile > aldehyde > ketone). It takes the suffix and C1. Everything else becomes a prefix.
Answer from the NTA final answer key. Solution by PrepWiser.