In the IUPAC system, aliphatic carboxylic acids are named by replacing the ending –e in the name of the corresponding alkane with –oic acid. In numbering the carbon chain, the carboxylic carbon is numbered one. For naming compounds containing more than one carboxyl group, the alkyl chain leaving carboxyl groups is numbered and the number of carboxyl groups is indicated by adding the multiplicative prefix, dicarboxylic acid, tricarboxylic acid, etc. to the name of parent alkyl chain. The positions of –COOH groups are indicated by the arabic numeral before the multiplicative prefix.
What is the IUPAC name of the given compound?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Hexa-2,4-dienoic acid
The passage states that the carboxylic carbon is numbered one and the name ends in –oic acid. The compound is \( \ce{CH3-CH=CH-CH=CH-COOH} \).
Chain with the carboxyl carbon: six carbons, so the parent is hexanoic acid.
Double bonds: numbering from COOH, they start at C-2 and C-4.
Two double bonds: “-an-” becomes “-dien-”, and “a” is added after “hex”: hexa-2,4-dienoic acid.
Why the other options are incorrect
Option (1) Penta-2,4-dienoic acid — Incorrect. It counts only five carbons; the carboxyl carbon is part of the chain, which has six.
Option (2) Pentenedioic acid — Incorrect. “Dioic acid” means two –COOH groups; the compound has one, and two C=C bonds.
Option (4) Hexenedioic acid — Incorrect. It also names a dicarboxylic acid with one C=C bond; here there is one –COOH group and two C=C bonds.