The IUPAC name for a carboxylic acid is based on the name of the hydrocarbon with the same number of carbon atoms. The ending -oic is appended, as in ethanoic acid, which is the IUPAC name for acetic acid. Draw the structure of the following acids: (a) methanoic acid, \((\mathbf{b})\) pentanoic acid, \((\mathbf{c}) 2\) -chloro-3-methyldecanoic acid.

Short Answer

Expert verified
The structures of the carboxylic acids are as follows: (a) Methanoic Acid: HCOOH, (b) Pentanoic Acid: CH₃CH₂CH₂CH₂COOH, and (c) 2-Chloro-3-Methyldecanoic Acid: CH₃(CH₂)₇COOH | CH₃ | Cl

Step by step solution

01

(a) Methanoic Acid

To draw the structure of methanoic acid, we first need to find the number of carbon atoms. Since the prefix "meth" corresponds to one carbon atom, and it's a carboxylic acid, we know there must be a COOH group attached to it. The structure of methanoic acid will therefore be: HCOOH
02

(b) Pentanoic Acid

To draw the structure of pentanoic acid, we first need to find the number of carbon atoms. The prefix "pent" corresponds to five carbon atoms. As it's a carboxylic acid, we will have a COOH group attached to one end of the carbon chain. The structure of pentanoic acid will therefore be: CH₃CH₂CH₂CH₂COOH
03

(c) 2-Chloro-3-Methyldecanoic Acid

To draw the structure of 2-chloro-3-methyldecanoic acid, we need to: 1. Identify the number of carbon atoms: The prefix "decan" corresponds to 10 carbon atoms. 2. Identify the substituents: At the 2nd carbon atom, there is a chlorine atom (2-chloro). At the 3rd carbon atom, there is a methyl group (3-methyl). Once we have identified the substituents and the number of carbon atoms, we can draw the structure: CH₃(CH₂)₇COOH | CH₃ | Cl

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