Question

An organic compound \(P\) with molecular formula \(C_6H_6O_3\) gives a ferric chloride test and does not have an 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 (\(-CH_3\)) groups in compound \(S\) is \_\_\_\_\_.

Hint:

When analyzing organic reaction sequences, determine the functional groups and predict the reactivity with each reagent to deduce the product.

Answer 1

Step 1: Compound \(P\) reacts with 3 equivalents of \(NH_2OH\), indicating the presence of 3 carbonyl groups, forming oxime \(Q\).
Step 2: Excess \(CH_3I\) with \(KOH\) reacts with \(Q\) to methylate all hydroxyl groups, resulting in compound \(R\).
Step 3: Compound \(R\) reacts with iso-butylmagnesium bromide (Grignard reagent) to add alkyl groups. Hydrolysis with \(H_3O^+\) yields compound \(S\). The reaction scheme is: \[ P \xrightarrow{3NH_2OH} Q \xrightarrow{\text{CH}_3I,\text{KOH}} R \xrightarrow{\text{iso-BuMgBr, H}_3\text{O}^+} S \] \includegraphics[width=0.4\linewidth]{ch13.png}
The structure of \(S\) contains a total of 12 methyl groups: \[ \text{Total \(-CH_3\) groups in compound \(S\): 12.} \]