JEE Main 2020 · 2 September · morning · Single correct · +4 / −1
The IUPAC name for the following compound is
- (a) 2, 5-dimethyl-6-carboxy-hex-3-enal
- (b) 6-formyl-2-methyl-hex-3-enoic acid
- (c) 2, 5-dimethyl-5-carboxy-hex-3-enal
- (d) 2, 5-dimethyl-6-oxo-hex-3-enoic acid
Answer and solution
Answer (d) 2, 5-dimethyl-6-oxo-hex-3-enoic acid
Solution
Chemistry ma’am · 5 steps
-COOH outranks -CHO, so the acid gets C1 and the aldehyde carbon becomes an oxo at C6.
- Principal group. The molecule has a carboxylic acid and an aldehyde. In the IUPAC priority order -COOH is above -CHO, so the suffix is -oic acid and the acid carbon is C1.
- Pick the chain. The longest chain that holds both functional carbons and the double bond is: HOOC-CH(CH₃)-CH=CH-CH(CH₃)-CHO. That is 6 carbons, with the CHO carbon as C6.
- Name the aldehyde. The aldehyde carbon is part of the main chain, so it is named oxo (the =O on C6), not formyl. Formyl is used only when the CHO carbon lies outside the chain.
- Locants. Methyl groups sit on C2 and C5, the double bond starts at C3, and the oxo is on C6.
- So. The name is 2,5-dimethyl-6-oxohex-3-enoic acid, which is option (d).
The idea to keep. Find the senior group first: it decides the suffix and C1. A lower-priority group inside the chain becomes a prefix, and a CHO carbon in the chain is called oxo, not formyl.
Answer from the NTA final answer key. Solution by PrepWiser.