1.

4-Pentenoic acid when treated with `I_(2)` and `NaHCO_(3)` gives:A. 4,5-diiodopentanoic acidB. 5-iodomethyl-dihydrofuran-2-oneC. 5-iodo-tetrahydropyran-2-oneD. 4-pentenolyiodide

Answer» Correct Answer - b
Iodo lactonization.


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