The name of the compound $C_{4}H_{9}COOC_{3}H_{7}$ is

Analyze the question: We need the IUPAC name of an ester written as C₄H₉COOC₃H₇.

Concept: An ester has the general formula R–COO–R′.

• The alkyl group R′ attached to the oxygen (the alcohol part) is named first, as the "alkyl" word.
• The R–COO part (the acid part) is named second, with the ending "-oate". Its carbon count includes the carboxyl carbon.

Solution:

1. Acid part: C₄H₉–COO– • C₄H₉ has 4 carbons, and the COO carbon adds 1 more, so 5 carbons in total. • A 5-carbon acid is pentanoic acid, so this part becomes pentanoate.

2. Alcohol part: –C₃H₇ • 3 carbons, so it is a propyl group.

3. Combine them, alkyl group first: propyl pentanoate.

Check the other options:

• B (pentyl propanoate) would be C₂H₅COOC₅H₁₁, so the groups are swapped.
• C (butyl propanoate) would be C₂H₅COOC₄H₉.
• D (propyl butanoate) would be C₃H₇COOC₃H₇, which has only 4 carbons in the acid part.

Answer: A. propyl pentanoate

Tip: Count the carbon of the –COO– group as part of the acid chain. Name the group after the oxygen first, then the acid part. For example, CH₃COOC₂H₅ is ethyl ethanoate.