Question

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 1

Correct Answer: 12

Reaction Scheme and Compound Identification 

Step 1: Reaction with Hydroxylamine (NH2OH)

Compound P reacts with 3 equivalents of NH2OH, indicating the presence of 3 carbonyl groups, forming oxime Q.

Step 2: Methylation Reaction

Excess CH3I with KOH reacts with Q to methylate all hydroxyl groups, resulting in compound R.

Step 3: Reaction with Iso-butylmagnesium Bromide

Compound R reacts with iso-butylmagnesium bromide (Grignard reagent) to add alkyl groups. Hydrolysis with \( \text{H}_3\text{O}^+ \) yields compound S.

Reaction Scheme:

\[ P \xrightarrow{\text{3 NH}_2\text{OH}} Q \xrightarrow{\text{CH}_3\text{I}, \text{KOH}} R \xrightarrow{\text{iso-BuMgBr}, \text{H}_3\text{O}^+} S \]

Final Answer:

The structure of compound S contains a total of 12 methyl groups:

Total \( -\text{CH}_3 \) groups in compound S: 12