JEE Advanced 2024 Paper 2 · Q13 · Named Reactions & Reagents

Question

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 major product. Reaction of R with excess iso-butylmagnesium bromide followed by treatment with $H_3O^+$ gives compound S as the major product.

The total number of methyl ($-CH_3$) group(s) in compound S is _____.

SolveKar AI · Accepted Answer

P with formula $C_6H_6O_3$ that gives a ferric-chloride test (phenolic OH) and reacts with 3 equiv NH$_2$OH to form a tris-oxime must have 3 ketonic carbonyls AND a phenolic hydroxyl tautomer; the only $C_6H_6O_3$ that fits is phloroglucinol (1,3,5-trihydroxybenzene), whose triketo tautomer (cyclohexane-1,3,5-trione) reacts with NH$_2$OH at all three C=O. Excess CH$_3$I/KOH alkylates all three OH groups AND the three CH$_2$ positions between the carbonyls in the triketo tautomer, giving 1,3,5-trimethoxy-2,2,4,4,6,6-hexamethylcyclohexane-1,3,5-trione — actually R is the C-methylated triketone with three OMe groups and the three CH$_2$ positions are dimethylated, giving R = 2,2,4,4,6,6-hexamethyl-1,3,5-trimethoxy-cyclohexane (effectively a triketone where each carbon α to two C=O is gem-dimethylated and the three OH are O-methylated). Reaction of R with excess (CH$_3$)$_2$CHCH$_2$MgBr followed by H$_3$O$^+$ adds three iso-butyl groups across the three carbonyls (each adds one $-$CH$_2$CH(CH$_3$)$_2$ — contributing 2 methyls per addition) and protonates to alcohols. Final compound S contains: 6 methyls from the gem-dimethyl carbons + 3 methyls from the OMe groups (lost on …[truncated]

JEE Advanced 2024 Paper 2 · Q13 · Named Reactions & Reagents

Solution