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 ______.

Step 1: Identification of Compound P

• Molecular Formula: $$\text{C}_6\text{H}_6\text{O}_3$$
• Ferric Chloride ($$\text{FeCl}_3$$) Test: Indicates the presence of phenolic or enolic hydroxyl ($$-\text{OH}$$) groups.
• No Intramolecular Hydrogen Bonding: This rules out adjacent arrangements like pyrogallol ($$1,2,3$$-trihydroxybenzene).

Consequently, compound $$P$$ is phloroglucinol ($$1,3,5$$-trihydroxybenzene). Phloroglucinol readily exhibits keto-enol tautomerism to exist in its keto form, cyclohexane-$$1,3,5$$-trione. Because it has $$3$$ carbonyl groups in this form, it reacts perfectly with $$3$$ equivalents of hydroxylamine ($$\text{NH}_2\text{OH}$$) to yield the trioxime Q.

Step 2: Formation of Compound R

When phloroglucinol ($$P$$) is treated with excess methyl iodide ($$\text{CH}_3\text{I}$$) in the presence of a strong base like $$\text{KOH}$$, it undergoes exhaustive C-alkylation at its active methylene positions rather than O-alkylation.

Each of the three $$-\text{CH}_2-$$ positions in the cyclohexane-$$1,3,5$$-trione tautomer is methylated twice, leading to $$2,2,4,4,6,6$$-hexamethylcyclohexane-$$1,3,5$$-trione as compound $$R$$.

• Methyl groups in $$R$$: $$6$$ methyl groups attached directly to the ring carbons.

Step 3: Formation of Compound S

Compound $$R$$ contains $$3$$ ketone carbonyl groups. When treated with excess iso-butylmagnesium bromide ($$\text{i-BuMgBr}$$, a Grignard reagent) followed by acid hydrolysis ($$\text{H}_3\text{O}^+$$), each carbonyl carbon undergoes a nucleophilic addition reaction to form a tertiary alcohol.

• Each ketone group adds one iso-butyl group ($$-\text{CH}_2-\text{CH}(\text{CH}_3)_2$$).
• Since there are $$3$$ ketone groups, a total of $$3$$ iso-butyl groups are added.

Step 4: Total Methyl Group Count in S

Let's add up all the methyl ($$-\text{CH}_3$$) groups present in the final structure $$S$$:

• From the core structure of $$R$$: $$6 \text{ methyl groups}$$
• From the $$3$$ attached iso-butyl groups: $$3 \times 2 = 6 \text{ methyl groups}$$

$$\text{Total number of methyl groups} = 6 + 6 = 12$$