A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below:

The structure of (mathbfPi^prime primemathbfP^prime prime)$(\mathbf{\Pi}^{\prime \prime}\mathbf{P}^{\prime \prime})$ is

### Related Formula CH_3-CHO + H_2N-NH-CO-NH_2 rightarrow CH_3-CH=N-NH-CO-NH_2 + H_2O$$CH_3-CHO + H_2N-NH-CO-NH_2 \rightarrow CH_3-CH=N-NH-CO-NH_2 + H_2O$$ ### Core Logic Ethanal (CH_3CHO$CH_3CHO$) reacts with semicarbazide (H_2N-NH-CO-NH_2$H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone. ### Step 1: Product Analysis The product is Ethanal semicarbazone: CH_3-CH=N-NH-CO-NH_2$CH_3-CH=N-NH-CO-NH_2$. Counting the nitrogen atoms in this structure: 1. The imine nitrogen (=N-$=N-$) 2. The amine nitrogen (-NH-$-NH-$) 3. The amide nitrogen (-NH_2$-NH_2$) Total = 3 Nitrogen atoms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic The reaction of benzene with carbon monoxide (CO$CO$) and hydrogen chloride (HCl$HCl$) in the presence of anhydrous aluminium chloride (AlCl_3$AlCl_3$) and cuprous chloride (CuCl$CuCl$) to give benzaldehyde is known as the Gatterman-Koch reaction.

### Pattern Recognition CO + HCl rightarrow$CO + HCl \rightarrow$ Formyl chloride intermediate (in situ) with Lewis acid rightarrow$\rightarrow$ formylation of benzene. This is definitively the Gatterman-Koch formylation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic The given reactant is p-hydroxybenzaldehyde, which contains both a phenolic -OH$-OH$ group (acidic) and an aldehyde group (electrophilic). When one equivalent of Grignard reagent (PhMgBr$PhMgBr$) is added, it behaves primarily as a strong base due to the presence of an acidic proton. Acid-base reactions are extremely fast compared to nucleophilic additions. The acidic phenolic -OH$-OH$ reacts with PhMgBr$PhMgBr$: PhMgBr + HO-C_6H_4-CHO rightarrow Ph-H (Benzene) + BrMg-O-C_6H_4-CHO$$PhMgBr + HO-C_6H_4-CHO \rightarrow Ph-H (Benzene) + BrMg-O-C_6H_4-CHO$$ Upon workup with aq. NH_4Cl$aq. NH_4Cl$, the phenoxide ion simply regenerates the starting p-hydroxybenzaldehyde. However, the question asks for the number of hydroxyl groups present in the formed product (Benzene).

The distinct product formed in the reaction is Benzene. Benzene has 0 hydroxyl groups. ### Pattern Recognition Whenever a Grignard reagent encounters a molecule with an acidic hydrogen (alcohol, phenol, amine, alkyne), it will invariably act as a base first. If only 1 equivalent is used, nucleophilic addition to carbonyls will NOT happen. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers