4- pentenoic acid when treated with $I_{2}$ and $N a H C O_{3}$ gives :

4, 5 -diiodopentanoic acid

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5 - idomethyl - dihydrofuran -2-one

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5-iodo tetrahydropyran -2-one

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4-pentenolyiodide

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Solution

The correct option is B 5 - idomethyl - dihydrofuran -2-one
$- ∙ ∙ O ∙ ∙ H$ acts as a good nucleophile which attaches the double bond and results in a furan ring.

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I_{2} / N a O H

4-Pentenoic acid S O C l_{2} - --- \to (X) Piperidine - ----- \to (Y)

(Y) i . L A H - ------ \to i i . H^{+} / H_{2} O

H_{2} / N i

N a H {C O}_{3}

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