An organic compound P having molecular formula $C_6H_6O_3$ gives ferric chloride test and does not have intramolecular hydrogen bond. The compound P reacts with 3 equivalents of $NH_2OH$ to produce oxime Q. Treatment of P with excess methyl iodide in the presence of KOH produces compound R as the ma...

The correct numerical answer is 12. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Chemistry Question 13 Solution

$C_6H_6O_3$ (P) with positive $FeCl_3$ test and no intramolecular H-bond is Phloroglucinol (1,3,5-trihydroxybenzene).
Phloroglucinol tautomerizes to cyclohexane-1,3,5-trione. Reaction with 3 equivalents of $NH_2OH$ confirms three $C=O$ groups.
Treatment of Phloroglucinol with excess $MeI/KOH$ leads to exhaustive C-methylation at the alpha positions (2,2,4,4,6,6-hexamethylcyclohexane-1,3,5-trione). This is compound R, which has 6 methyl groups.
Compound R reacts with excess iso-butylmagnesium bromide at the 3 carbonyl groups. Each $C=O$ becomes $C(OH)(i-Bu)$ .
An iso-butyl group is $-CH_2-CH(CH_3)_2$ , containing 2 methyl groups.
Total methyl groups in S = 6 (from R) + 3 (iso-butyl groups) $\times$ 2 (methyls per group) = $6 + 6 = 12$ .