$\text{An organic compound}P\text{having molecular formula}{\mathrm{ C}}_6{H}_6{O}_3\text{ gives ferric chloride test and does not}$

$\text{have intramolecular hydrogen bond.}\text{The compound}P\text{reacts with}\mathrm{ 3}\text{ equivalents of}{\mathrm{ NH}}_2\mathrm{OH}\text{ to produce}$

$\text{oxime}\mathrm{ Q}.\text{Treatment of }P\text{ with excess methyl iodide in the presence of}\mathrm{ KOH}\text{ produces compound }R\text{ as}$

$\text{the major product.}\text{Reaction of}R\text{with excess}\text{iso-butylmagnesium bromide}\text{followed by treatment with}{ }_{}$

$H_3{O}^{+}\text{gives compound}\mathrm{ S}\text{ as the major product.}$
$\text{The total number of methyl}(-{\mathrm{CH}}_3)\text{group(s) in compound}\mathrm{ S}\text{ is}\text{\_\_\_\_\_}$

Correct Answer: 12

Step 1: Determine the structure of compound P

The molecular formula of compound P is C₆H₆O₃.

Let's calculate the degree of unsaturation (double bond equivalents, DBE) for C₆H₆O₃:

$\text{DBE}=C+1-\frac{H}{2}=6+1-\frac{6}{2}=4$

Key properties of compound P given in the problem statement:

1. Gives FeCl₃ test: Indicates the presence of phenolic hydroxyl group(s) (-OH attached to an aromatic ring) or an enol group.
2. Does not have intramolecular hydrogen bonding: For benzenediols/benzenetriols, ortho-hydroxyl groups can form intramolecular hydrogen bonds. Since P does not show intramolecular hydrogen bonding, the hydroxyl/oxygen groups must be symmetrically placed at meta-positions to each other.
3. Reacts with 3 equivalents of NH₂OH to form oxime Q: Hydroxylamine (NH₂OH) reacts with carbonyl groups (ketones/aldehydes) to form oximes. Since compound P reacts with 3 equivalents of NH₂OH, P must exist in a tautomeric tri-keto form with 3 carbonyl groups.

Benzene-1,3,5-triol (Phloroglucinol) tautomerizes into its tri-keto form (cyclohexane-1,3,5-trione):

$\text{Compound P: Cyclohexane-1,3,5-trione (Tautomer of Phloroglucinol)}$

Phloroglucinol has no intramolecular hydrogen bonds due to the meta arrangement of all three oxygen functionalities, and its tri-keto form possesses 3 carbonyl (-C=O) groups, reacting with 3 equivalents of hydroxylamine.

Step 2: Formation of compound R

Treatment of P (phloroglucinol) with excess methyl iodide (CH₃I) in the presence of a strong base (KOH) leads to complete methylation. Because phloroglucinol shows keto-enol tautomerism and nucleophilic carbon centers, methylation with excess CH₃I in strong base yields hexamethylcyclohexane-1,3,5-trione (or methylation at the reactive α-carbons of the tri-keto form):

Each of the three CH₂ positions in cyclohexane-1,3,5-trione has two acidic hydrogens and gets fully methylated by excess CH₃I/KOH, substituting all 6 α-hydrogens with methyl groups.

Thus, compound R is 2,2,4,4,6,6-hexamethylcyclohexane-1,3,5-trione.

Structure of R has:

- 3 carbonyl groups (-C=O) at C1, C3, and C5 positions.
- 6 methyl groups (-CH₃) attached as gem-dimethyl pairs at C2, C4, and C6 positions.

Step 3: Reaction of R with excess iso-butylmagnesium bromide followed by H₃O⁺ to give S

Compound R contains 3 ketone carbonyl groups. Reacting R with excess iso-butylmagnesium bromide (i-BuMgBr, where i-Bu = -CH₂-CH(CH₃)₂) leads to nucleophilic addition of the Grignard reagent to all 3 carbonyl groups:

After acidic workup with H₃O⁺, each of the three carbonyl carbon atoms is converted into a tertiary alcohol group bearing an iso-butyl group (-CH₂-CH(CH₃)₂).

Therefore, compound S is 2,2,4,4,6,6-hexamethyl-1,3,5-tri-isobutylcyclohexane-1,3,5-triol.

Step 4: Count the total number of methyl (-CH₃) groups in compound S

Let's count the methyl groups present in one molecule of S:

1. From the core structure of compound R, there are 6 methyl groups (two gem-methyl groups at each of the C2, C4, and C6 positions of the ring).
2. From the 3 iso-butyl groups attached to C1, C3, and C5: each iso-butyl group, -CH₂-CH(CH₃)₂, contains 2 methyl groups.
Since there are 3 iso-butyl groups, they contribute:

$3\times 2=6\text{ methyl groups}$

Adding all the methyl groups together:

$\text{Total number of methyl groups}=6\text{ (from ring carbons)}+6\text{ (from 3 iso-butyl chains)}=12$

Thus, the total number of methyl (-CH₃) groups in compound S is 12.