Final answer:
Under mild acidic conditions, 5-Hydroxypentanoic acid cyclizes to form 4-Hydroxytetrahydrofuran, as the intramolecular attack of the hydroxy group on the carboxylic acid group forms a five-membered lactone ring.
Step-by-step explanation:
Under mild acidic conditions, 5-Hydroxypentanoic acid cyclizes to form a cyclic ester, which is more commonly referred to as a lactone. The cyclization would involve the formation of a five-membered ring as the hydroxy group (OH) from the 5th carbon atom attacks the carbonyl carbon of the carboxylic acid (COOH). This intramolecular esterification yields a compound known as 4-Hydroxytetrahydrofuran.
The correct answer to the question is option a. 4-Hydroxytetrahydrofuran. This is because when a five-carbon acid like 5-Hydroxypentanoic acid cyclizes, it forms a four-membered ring with one oxygen (resulting from the -OH group and the carbonyl group) and this structure matches that of tetrahydrofuran with a hydroxy group at the 4-position.