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 _____.

The correct answer is 12.

Step 1: Identifying Compound P
The molecular formula of compound P is given as C₆H₆O₃. It gives a positive ferric chloride (FeCl₃) test, which indicates the presence of phenolic or enolic groups. Additionally, P has no intramolecular hydrogen bonds, which means the hydroxyl groups are not adjacent to each other (i.e., they are in meta-positions).
P reacts with 3 equivalents of hydroxylamine (NH₂OH) to form a trioxime Q. This indicates that the keto tautomer of P contains 3 carbonyl groups. These properties identify compound P as phloroglucinol (benzene-1,3,5-triol), which exists in tautomeric equilibrium with its triketo form, cyclohexane-1,3,5-trione.

Step 2: Identifying Compound R
When P is treated with excess methyl iodide (CH₃I) in the presence of KOH, C-alkylation occurs. Cyclohexane-1,3,5-trione has three active methylene (-CH₂-) groups located between the carbonyl groups. The hydrogens on these carbons are highly acidic.
Under basic conditions, all 6 acidic hydrogens of these three active methylene groups are replaced by methyl groups. This complete C-alkylation yields compound R, which is 1,1,3,3,5,5-hexamethylcyclohexane-1,3,5-trione.
Compound R contains 6 methyl (-CH₃) groups on the ring carbons.

Step 3: Identifying Compound S and Counting Methyl Groups
Compound R is treated with excess iso-butylmagnesium bromide ((CH₃)₂CHCH₂MgBr), a Grignard reagent, followed by acidic workup (H₃O⁺).
The Grignard reagent attacks each of the 3 carbonyl groups in R. Upon protonation, each carbonyl group (C=O) is converted into a tertiary alcohol group containing one iso-butyl group:
-C(OH)(CH₂CH(CH₃)₂)-
Each iso-butyl group contains 2 methyl (-CH₃) groups. Since there are 3 carbonyl groups in R, 3 iso-butyl groups are added to the compound to form S.

The total number of methyl groups in compound S is calculated as follows:
$\text{Total methyl groups}=\text{Methyl groups originally in R}+(\text{Number of carbonyl groups}\times \text{Methyl groups per iso-butyl group})$
$\text{Total methyl groups}=6+(3\times 2)=6+6=12$
Thus, the total number of methyl groups in compound S is 12.