For 4-pentanoic acid when treated with I₂ and NaHCO₃ gives:

• A. 4,5-diiodo pentanoic acid
• B. 5-iodomethyl-dihydrofuran-2-one
• C. 5-iodo tetrahydropyran-2-one
• D. 4-pentenoyl iodide

When 4-pentanoic acid is treated with iodine (I₂) and sodium bicarbonate (NaHCO₃), the expected product is 5-iodomethyl-dihydrofuran-2-one. This reaction typically involves the formation of a cyclic compound through a series of steps, including the introduction of iodine at the 5-position of the carbon chain. Therefore, the correct answer is:

Correct Answer

B. 5-iodomethyl-dihydrofuran-2-one

Explanation of the Reaction

The reaction mechanism involves the following key steps:

• The acid undergoes a transformation that allows for the formation of a cyclic structure.
• Iodine acts as an electrophile, adding to the carbon chain.
• The presence of sodium bicarbonate helps to facilitate the reaction conditions.

This results in the formation of the dihydrofuran derivative, which is characteristic of such reactions involving 4-pentanoic acid. Understanding these mechanisms can help in predicting the outcomes of similar organic reactions.