An organic compound P having molecular formula C6H6O3 gives ferric chloride test and does not have intramolecular hydrogen bond. The compound P reacts with 3 equivalents of NH2OH 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 H3O+ gives compound S as the major product. The total number of methyl (-CH3) group(s) in compound S is .
Answer
12
Solution
P (C6H6O3, FeCl₃ test +, no intramolecular H-bond: OH groups meta) = phloroglucinol, which reacts through its keto form:
[figure] [figure]
[figure] 3NH2OH [figure] (Q)
The three C=O groups of the keto form give a trioxime.
[figure] CH3I (excess), KOH [figure] (R)
Each active CH₂ (between two C=O) is C-methylated twice: 6 CH₃ on the ring.
[figure] [(ii) H3O+](i) (CH3)2CHCH2MgBr (excess) [figure] (S)
R = -CH2CH(CH3)2 (isobutyl)
The Grignard reagent adds to all three C=O; each gives a tertiary alcohol carrying an isobutyl group.
CH₃ count in S = 6 (ring) + 3 × 2 (isobutyl) = 12
Answer: 12