An organic compound P having molecular formul | vedclass

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(A) The compound $P$ is cyclohexane-$1,3,5$-trione,which exists in tautomeric equilibrium with phloroglucinol. It gives a positive $FeCl_3$ test due t

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(A) The compound $P$ is cyclohexane-$1,3,5$-trione,which exists in tautomeric equilibrium with phloroglucinol. It gives a positive $FeCl_3$ test due to the enol form and lacks intramolecular hydrogen bonding in the tri-keto form.$P$ reacts with $3$ equivalents of $NH_2OH$ to form the tri-oxime $Q$.Treatment of $P$ with excess $CH_3I$ and $KOH$ results in the exhaustive methylation of the $\alpha$-carbons,yielding $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-trione $(R)$.Reaction of $R$ with excess iso-butylmagnesium bromide $(iso-BuMgBr)$ acts as a base due to the steric hindrance of the six methyl groups on the $\alpha$-carbons,leading to the reduction of the carbonyl groups to hydroxyl groups (enolization/reduction) rather than nucleophilic addition. The final product $S$ is $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-triol.In $2,2,4,4,6,6$-hexamethylcyclohexane-$1,3,5$-triol,there are $6$ methyl groups attached to the ring carbons. Each methyl group is a $-CH_3$ unit. Thus,the total number of methyl groups is $6$.

Column-$1$ (Reaction of carbonyl compound with)	Column-$2$ (Product)
$A$. Hydroxylamine	$I$. Hydrazone
$B$. Alcohol	$II$. Schiff's base (Substituted imine)
$C$. Hydrazine	$III$. Oxime
$D$. Amine	$IV$. Ketal

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Match the following:
Column-$1$ (Reaction of carbonyl compound with) Column-$2$ (Product)
$A$. Hydroxylamine $I$. Hydrazone
$B$. Alcohol $II$. Schiff's base (Substituted imine)
$C$. Hydrazine $III$. Oxime
$D$. Amine $IV$. Ketal
The correct match is:

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