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 textCompounds with CH_3-CH(OH)- text or CH_3-CO- text groups undergo the iodoform reaction to form CHI_3 downarrow text (Yellow Solid)$$\text{Compounds with } CH_3-CH(OH)- \text{ or } CH_3-CO- \text{ groups undergo the iodoform reaction to form } CHI_3 \downarrow \text{ (Yellow Solid)}$$ ### Core Logic Let's check the structural groups of each given option: - **(A)** CH_3 - CH(OH) - C_2H_5$CH_3 - CH(OH) - C_2H_5$: Contains the methylcarbinol group (CH_3-CH(OH)-$CH_3-CH(OH)-$). Gives a positive iodoform test. - **(B)** CH_3 - CH_2 - CH_2 - OH$CH_3 - CH_2 - CH_2 - OH$: Linear primary alcohol, does not contain the required group. - **(C)** CH_3 - CO - C_2H_5$CH_3 - CO - C_2H_5$: Contains the methyl ketone group (CH_3-CO-$CH_3-CO-$). Gives a positive iodoform test. - **(D)** CH_3 - OH$CH_3 - OH$: Methanol does not give the test. - **(E)** CH_3 - CH_2 - H$CH_3 - CH_2 - H$: Ethane does not give the test. Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform (CHI_3$CHI_3$). ### Step 1: Chemical Equations The balanced haloform pathways occur as follows: CH_3-CH(OH)-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+$$CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$$ CH_3-CO-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+$$CH_3-CO-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$$ ### Pattern Recognition The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal -CH_3$-CH_3$ affixed directly to a carbonyl oxygen index (C=O$C=O$) or a hydroxyl carbon (CH-OH$CH-OH$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

### Core Logic Intramolecular aldol condensation involves a dicarbonyl compound reacting within itself to yield stable cyclic alpha, beta$\alpha, \beta$-unsaturated rings (most commonly 5- or 6-membered rings due to minimal ring strain). * Products (1), (2), and (3) can all be cleanly synthesized via intramolecular cyclization path workflows from their respective dialdehyde/diketone precursors. * Product (4) features an exo-cyclic group structure arrangement formed strictly through an **intermolecular** condensation sequence step between two distinct reactant units, rather than an internal cyclization path layout. ### Pattern Recognition Look closely at the ring substitution system. Intermolecular steps are forced when intramolecular cyclization path workflows would generate highly strained small rings or structurally impossible orientations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Let's deduce the identities stepwise: 1. Compound (X) has the formula C_3H_6O$C_3H_6O$ and is resistant to mild oxidation, which identifies it as a ketone: **Acetone** (CH_3COCH_3$CH_3COCH_3$). 2. Reduction of Acetone yields a secondary alcohol, Propan-2-ol (CH_3CH(OH)CH_3$CH_3CH(OH)CH_3$, Compound Y). 3. Treatment of Propan-2-ol with HBr$HBr$ substitutes the hydroxyl group to form 2-Bromopropane (CH_3CH(Br)CH_3$CH_3CH(Br)CH_3$, Compound Z). 4. Reacting 2-Bromopropane with Magnesium in ether creates the branched Grignard reagent, Isopropylmagnesium bromide ((CH_3)_2CHMgBr$(CH_3)_2CHMgBr$). 5. Finally, nucleophilic addition of this Grignard reagent to Acetone followed by aqueous workup yields the highly branched tertiary alcohol: **2,3-dimethylbutan-2-ol**. ### Pattern Recognition Resistance to mild oxidation immediately distinguishes ketones from isomeric aldehydes. Nucleophilic addition of an isopropyl Grignard to acetone cleanly yields the 2,3-dimethylbutan-2-ol framework. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

### Related Formula textCarbonyl compound + text2,4-DNP ightarrow textHydrazone derivative (Positive test) $$\text{Carbonyl compound} + \text{2,4-DNP} ightarrow \text{Hydrazone derivative (Positive test)} $$ textAldehyde + textTollens' Reagent ightarrow textSilver Mirror (Positive test)$$\text{Aldehyde} + \text{Tollens' Reagent} ightarrow \text{Silver Mirror (Positive test)}$$ textKetone + textTollens' Reagent ightarrow textNo reaction (Negative test) $$\text{Ketone} + \text{Tollens' Reagent} ightarrow \text{No reaction (Negative test)} $$ ### Core Logic Analyzing individual functional constraints: - Positive 2,4-DNP test shows compound contains a carbonyl group (aldehyde or ketone). - Negative Tollens' test clarifies it is not an aldehyde; hence it must be a ketone. - The compound is optically active, meaning it must possess a chiral center (carbon with 4 distinct groups). Let's evaluate the options via structural configurations:

### Step 1: Structural Verification Option (2) represents 3-methylpentan-2-one: textCH3-textC(=textO)-oversetasttextCH(textCH3)(textCH2textCH3) $$\text{CH}3-\text{C}(=\text{O})-\overset{\ast}{\text{CH}}(\text{CH}3)(\text{CH}2\text{CH}3) $$ The third carbon (C3) is linked to: -textH$-\text{H}$, -textCH_3$-\text{CH}_3$, -textCH_2textCH_3$-\text{CH}_2\text{CH}_3$, and -textCOCH_3$-\text{COCH}_3$. It has 4 distinct structural fields, making it chiral and optically active. ### Pattern Recognition Tollens' negative + DNP positive = Ketone. Once categorized as a ketone, look directly for the structure holding a carbon with four unique groups to secure the optical activity constraint. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let's match each aldehyde preparation method with its official named organic reaction: * **(A) mathrmRCN ightarrow mathrmRCHO$\mathrm{RCN} ightarrow \mathrm{RCHO}$ using mathrmSnCl_2/mathrmHCl$\mathrm{SnCl}_{2}/\mathrm{HCl}$ followed by hydrolysis:** This is the classic **Stephen reaction** ightarrow$ ightarrow$ **(IV)**. * **(B) Reducing an acyl chloride (mathrmRCOCl$\mathrm{RCOCl}$) to an aldehyde using mathrmH_2$\mathrm{H}_{2}$ over mathrmPd-BaSO_4$\mathrm{Pd-BaSO}_{4}$:** This partial reduction is known as the **Rosenmund reduction** ightarrow$ ightarrow$ **(III)**. * **(C) Oxidizing toluene to benzaldehyde using chromyl chloride (mathrmCrO_2mathrmCl_2$\mathrm{CrO}_{2}\mathrm{Cl}_{2}$) in mathrmCS_2$\mathrm{CS}_{2}$:** This selective oxidation method is the **Etard reaction** ightarrow$ ightarrow$ **(I)**. * **(D) Converting benzene to benzaldehyde using mathrmCO$\mathrm{CO}$ and mathrmHCl$\mathrm{HCl}$ in the presence of anhydrous mathrmAlCl_3/mathrmCuCl$\mathrm{AlCl}_{3}/\mathrm{CuCl}$:** This formylation process is the **Gatterman-Koch reaction** ightarrow$ ightarrow$ **(II)**. Combining these assignments yields the final sequence: (A)-(IV), (B)-(III), (C)-(I), (D)-(II). ### Step-by-Step Layout The visual reaction components correspond directly to the official structural transformations:

### Pattern Recognition Quick identification keys: - Nitrile ightarrow$ ightarrow$ Aldehyde = Stephen - Acid Chloride ightarrow$ ightarrow$ Aldehyde = Rosenmund - Toluene ightarrow$ ightarrow$ Chromyl Complex = Etard - Benzene ightarrow$ ightarrow$ Carbon Monoxide = Gatterman-Koch ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids