Answer:
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.