You have two C_6H_{10}O ketones,I and II. Bot | vedclass

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(B) $2-$Methylcyclopentanone $(I)$ has an $\alpha-$hydrogen atom at the chiral center. In the presence of acid,it undergoes keto-enol tautomerization,

Vedclass · Accepted Answer

$I$ is $2-$Methylcyclopentanone,$II$ is $3-$Methylcyclopentanone

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(B) $2-$Methylcyclopentanone $(I)$ has an $\alpha-$hydrogen atom at the chiral center. In the presence of acid,it undergoes keto-enol tautomerization,which leads to the loss of chirality at the $\alpha-$carbon,resulting in racemization.$3-$Methylcyclopentanone $(II)$ also has a chiral center,but it is not at the $\alpha-$position relative to the carbonyl group. Therefore,it cannot form an enol that would lead to racemization at the chiral center under acidic conditions.Both $2-$methylcyclopentanone and $3-$methylcyclopentanone,upon Wolff-Kishner reduction $(N_2H_4, OH^-, \Delta)$,yield the same achiral hydrocarbon,methylcyclopentane $(C_6H_{12})$.Thus,$I$ is $2-$methylcyclopentanone and $II$ is $3-$methylcyclopentanone.

List-$I$	List-$II$
$A$. Grignard reagent	$1$. $H_2 / Pd-BaSO_4$
$B$. Clemmensen reduction	$2$. $N_2H_4 / KOH / (CH_2OH)_2$
$C$. Rosenmund reduction	$3$. $CH_3MgX$
$D$. Wolff-Kishner reduction	$4$. $Zn-Hg / \text{conc. } HCl$
	$5$. $H_2 / Ni$

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List-$I$ List-$II$
$A$. Grignard reagent $1$. $H_2 / Pd-BaSO_4$
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