The reaction of cyclopentanone with NaOBr giv | vedclass

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(D) The reagent $NaOBr$ (sodium hypobromite) is a source of $Br_2$ in basic medium $(NaOH)$.Cyclopentanone is a ketone with $\alpha$-hydrogens. In the

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None of these

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(D) The reagent $NaOBr$ (sodium hypobromite) is a source of $Br_2$ in basic medium $(NaOH)$.Cyclopentanone is a ketone with $\alpha$-hydrogens. In the presence of $NaOBr$,it undergoes the haloform reaction or $\alpha$-halogenation depending on the conditions.However,for a simple ketone like cyclopentanone,$NaOBr$ typically leads to the formation of $\alpha$-bromo derivatives. Specifically,the reaction of cyclopentanone with $NaOBr$ results in the formation of $2$-bromocyclopentanone.Looking at the provided options,none of the structures correctly represent the expected product of the reaction between cyclopentanone and $NaOBr$. Therefore,the correct answer is $D$.

Test	Inference
$(a)$ $2, 4-DNP$ test	Coloured precipitate
$(b)$ Iodoform test	Yellow precipitate
$(c)$ Azo-dye test	No dye formation

The reaction of cyclopentanone with $NaOBr$ gives which product?
$\text{Cyclopentanone} \xrightarrow{NaOBr} \text{Product}$

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Test Inference

$(a)$ $2, 4-DNP$ test Coloured precipitate

$(b)$ Iodoform test Yellow precipitate

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The reaction of cyclopentanone with $NaOBr$ gives which product?$\text{Cyclopentanone} \xrightarrow{NaOBr} \text{Product}$