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, (b) pentanoic acid, (c) 2-chloro-3-methyldecanoic acid.

Short Answer

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The structures of the given carboxylic acids are as follows: (a) Methanoic acid: HCOOH H | C=O | O-H (b) Pentanoic acid: CH3(CH2)3COOH CH3-CH2-CH2-CH2-C=O | O-H (c) 2-chloro-3-methyldecanoic acid: CH3(CH2)7CHClCH(CH3)COOH CH3-CH2-CH2-CH2-CH2-CH2-CH2-CHCl-CH(CH3)-C=O | O-H

Step by step solution

01

(a) Drawing the structure of methanoic acid)

Methanoic acid has only one carbon atom, as the prefix "meth" indicates. In a carboxylic acid, the carbon atom is bonded to an oxygen atom in a double bond, and an OH group. So, the structure of methanoic acid (HCOOH) is as follows: H | C=O | O-H
02

(b) Drawing the structure of pentanoic acid)

Pentanoic acid has five carbon atoms, as indicated by the prefix "pent." First, draw the linear carbon chain with five carbon atoms, and add the carboxylic acid functional group (C=O and an OH group) at one end. The structure of pentanoic acid (CH3(CH2)3COOH) is as follows: CH3-CH2-CH2-CH2-C=O | O-H
03

(c) Drawing the structure of 2-chloro-3-methyldecanoic acid)

2-chloro-3-methyldecanoic acid has a decyl (10 carbon atom) chain as its base structure, with a carboxylic acid functional group at one end, a chlorine atom at position 2 and a methyl group (CH3) at position 3. Draw the base chain, add the carboxylic acid functional group at one end, and add the chloro and methyl substituents at positions 2 and 3. The structure of 2-chloro-3-methyldecanoic acid (CH3(CH2)7CHClCH(CH3)COOH) is as follows: CH3-CH2-CH2-CH2-CH2-CH2-CH2-CHCl-CH(CH3)-C=O | O-H

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