JEE Main 2023 · 8 April · evening · Single correct · +4 / −1
The correct IUPAC nomenclature for the following compound is:
- (a) 2 - Formyl -5-methylhexan-6-oic acid
- (b) 5-Methyl-2-oxohexan-6-oic acid
- (c) 5-Formyl -2-methylhexanoic acid
- (d) 2-Methyl-5-oxohexanoic acid
Answer and solution
Answer (d) 2-Methyl-5-oxohexanoic acid
Solution
Chemistry ma’am · 4 steps
The COOH carbon is always C1, so number from it and the ketone lands on C5 as an oxo group.
- Principal group. The molecule has a carboxylic acid and a ketone. -COOH ranks higher, so it sets the suffix "-oic acid" and its carbon becomes C1.
- Longest chain. Start at the COOH carbon and go through CH(CH₃), CH₂, CH₂, the C=O carbon and the end CH₃. That is 6 carbons, so the parent is hexanoic acid.
HOOC(1)-CH(2)(CH₃)-CH₂(3)-CH₂(4)-C(5)(=O)-CH₃(6)
- Name the substituents. A methyl sits on C2. The ketone C=O is inside the chain at C5, so it is written as the prefix oxo. Arrange them alphabetically: methyl before oxo.
- So. The name is 2-Methyl-5-oxohexanoic acid, option (d).
The idea to keep. When the principal group is COOH, its carbon is C1 and you never write a locant for it. A ketone in the same chain becomes the prefix oxo, and formyl is used only for a -CHO group.
Answer from the NTA final answer key. Solution by PrepWiser.