Write the IUPAC names of the given carboxylic acids. A molecule has the condensed formula C H 3 C H 2 C H (O C H 3) C H 2 C O O H. IUPAC name: A six membered hydrocarbon ring with three double bonds. The group C O O H is attached to ring carbon one, and a chlorine atom is attached to ring carbon four. IUPAC name:

The correct answer is -

3-methoxypentanoic acid, and

4-chlorobenzoic acid

Explanation:

1. condensed formula of the molecule is -

CH3CH2CH(OCH3)CH2COOH and the number of the carbons would be count from the end of carboxylic acid. So, the counting as given in the image, it shows there are 5 carbon so it is Penta and has methyl group on fourth carbon so 4-methyl, therefore, its IUPAC name will be - 4-methyl pentanoic acid

2. similarly for the molecule with COOH group attached with ring carbon and chlorine is represented as in the image and it shows a benzene ring and COOH group is present at C1 and chlorine is in C4 so the name would be 4-chlorobenzoic acid.