MCQ
$4$ -Pentenoic acid when treated with $I_2$ and $NaHCO_3$ gives
- A$4, 5$ -diiodopentanoic acid
- ✓$5$ -iodomethyl-dihydrofuran- $2$ -one
- C$5$ -iodo-tetrahydropyran- $2$ -one
- D$4$ -pentenolyiodide
✓
Answer
Correct option: B.
$5$ -iodomethyl-dihydrofuran- $2$ -one
b
$(b)$ Iodo lactonization.
$(b)$ Iodo lactonization.
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