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) → 3,3-Dimethylbutanoic acid
The passage states that an aliphatic acid is named from the alkane with the ending –e replaced by –oic acid, and that the carboxylic carbon is numbered one. The compound is \( \ce{(CH3)3C-CH2-COOH} \).
Longest chain with the carboxyl carbon: COOH (C-1), CH2 (C-2), the branched carbon (C-3) and one of its methyl groups (C-4): four carbons, so butanoic acid.
Substituents: the other two methyl groups both sit on C-3: 3,3-dimethyl.
Name: 3,3-dimethylbutanoic acid.
Why the other options are incorrect
Option (1) Hexanoic acid — Incorrect. It has the same six carbons, but hexanoic acid is the unbranched chain \( \ce{CH3(CH2)4COOH} \).
Option (2) 2,2-Dimethylbutanoic acid — Incorrect. That compound has both methyl groups on C-2, next to the carboxyl carbon; here a CH2 group lies between them.
Option (4) 4-Methylpentanoic acid — Incorrect. That is \( \ce{(CH3)2CH-CH2-CH2-COOH} \), with a five-carbon chain and one methyl branch; the given compound has a four-carbon chain with two methyl groups on one carbon.