### Related Formula mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_3O^+ R-COOH$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$ ### Core Logic Let's analyze each reaction path to determine the major organic product: - **Reaction (A)**: Acidic hydrolysis of a nitrile under *mild conditions* yields an amide: mathrmR-Cequiv N rightarrow R-CONH_2$\mathrm{R-C\equiv N \rightarrow R-CONH_2}$ (Full conversion to carboxylic acid requires strong conditions and extended heating). - **Reaction (B)**: Carbonation of Grignard reagent using solid carbon dioxide (dry ice) followed by acid hydrolysis yields a carboxylic acid: mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_3O^+ R-COOH$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$ - **Reaction (C)**: Stephen reduction converts nitrile to aldehyde: mathrmR-Cequiv N xrightarrowSnCl_2/HCl R-CH=NH xrightarrowH_3O^+ R-CHO$\mathrm{R-C\equiv N \xrightarrow{SnCl_2/HCl} R-CH=NH \xrightarrow{H_3O^+} R-CHO}$ - **Reaction (D)**: Pyridinium chlorochromate (PCC) is a mild oxidising agent that converts primary alcohols selectively to aldehydes: mathrmR-CH_2-OH xrightarrowPCC R-CHO$\mathrm{R-CH_2-OH \xrightarrow{PCC} R-CHO}$ - **Reaction (E)**

: Rosenmund reduction reduces acid chloride to aldehyde first: rightarrow$\rightarrow$ R-CHO Subsequent oxidation with bromine water (which is a mild oxidising agent that selective oxidizes aldehydes but does not affect ketones) converts the aldehyde to carboxylic acid: mathrmR-CHO xrightarrowBr_2/water R-COOH$\mathrm{R-CHO \xrightarrow{Br_2/water} R-COOH}$ ### Step 1: Final Tally Thus, reactions (B) and (E) successfully yield carboxylic acid as the major organic product. ### Pattern Recognition Remember: Bromine water (mathrmBr_2/H_2O$\mathrm{Br_2/H_2O}$) is a mild, selective oxidising agent commonly used to oxidise aldoses and other aldehydes to monocarboxylic acids without degrading carbon-carbon chains. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula Phenolic protons react with standard strong bases: mathrmAr-OH + NaOH rightarrow Ar-ONa + H_2O$\mathrm{Ar-OH + NaOH \rightarrow Ar-ONa + H_2O}$ Aldol condensation structural requirement: Requires presence of acidic alpha$\alpha$-hydrogen atoms connected to carbonyl centers. ### Core Logic Let's analyze functional groups within the Vanillin molecular framework: * Vanillin contains a phenolic hydroxyl group, an aromatic ether, and a formyl functional group (benzaldehyde derivative). * **Statement I**: The presence of the phenolic -mathrmOH$-\mathrm{OH}$ group allows acid-base reaction with mathrmNaOH$\mathrm{NaOH}$ directly

. The aldehyde center readily reduces Tollen's reagent to produce a silver mirror. (Statement I is accurate). * **Statement II**: Vanillin lacks any alpha-hydrogens adjacent to its carbonyl carbon, preventing it from undergoing self-aldol condensation. (Statement II is false). ### Pattern Recognition Benzaldehyde and its substituted derivatives (like vanillin or benzaldehyde itself) never undergo self-aldol condensation because they lack alpha$\alpha$-carbons with abstractable protons. Instead, they typically perform Cannizzaro transformations when exposed to highly concentrated alkaline media. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula Intramolecular Cannizzaro-type reaction or Benzil-Benzilic acid rearrangement involves nucleophilic attack of hydroxide at a carbonyl group, followed by hydride transfer to the adjacent carbonyl carbon. ### Core Logic Let's analyze the starting compound, phenylglyoxal: mathrmPh-CO-CHO$\mathrm{Ph-CO-CHO}$ 1. The aldehyde carbon (-CHO$-CHO$) is much more electrophilic than the ketone carbon (-CO-$-CO-$) due to less steric hindrance and absence of phenyl group electron donation. 2. Hydroxide ion (mathrmOH^-$\mathrm{OH}^-$) selectively attacks the aldehyde carbonyl carbon, forming a tetrahedral intermediate.

### Step 1: Hydride Transfer Mechanism The tetrahedral intermediate collapses, prompting an intramolecular hydride (H^-$H^-$) transfer to the adjacent ketone carbonyl carbon: mathrmPh-CO-C(O^-)(OH)H rightarrow mathrmPh-C(O^-)H-COOH$\mathrm{Ph-CO-C(O^-)(OH)H} \rightarrow \mathrm{Ph-C(O^-)H-COOH}$ This is the rate-determining step. ### Step 2: Proton Transfer to form Product Rapid proton transfer occurs from the carboxylic acid group to the alkoxide oxygen, yielding the stable carboxylate salt: mathrmPh-CH(OH)-COO^- K^+$\mathrm{Ph-CH(OH)-COO^- K^+}$ This corresponds to Option (2). ### Pattern Recognition In asymmetrical 1,2-dicarbonyl systems with an aldehyde and a ketone, nucleophilic addition occurs preferentially at the more reactive aldehyde carbon. The hydrogen is then transferred as a hydride to the ketone carbon, yielding an alpha$\alpha$-hydroxy carboxylate salt. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Let us systematically follow the transformation steps: 1. **First Elimination**: 1,2-dibromocyclooctane reacts with alcoholic textKOH$\text{KOH}$ to remove one molecule of textHBr$\text{HBr}$, resulting in a bromocyclooctene intermediate. 2. **Second Elimination**: Treatment with the stronger base textNaNH_2$\text{NaNH}_2$ removes the second molecule of textHBr$\text{HBr}$, forming an alkyne inside the 8-membered ring: **cyclooctyne**. 3. **Kucherov Reaction**: Hydration of cyclooctyne using textHg^2+/H^+$\text{Hg}^{2+}/H^+$ creates an enol intermediate that undergoes tautomerization to form a stable ketone: **cyclooctanone**. 4. **Clemmensen Reduction**: Subjecting cyclooctanone to zinc amalgam and hydrochloric acid (textZn-Hg/HCl$\text{Zn-Hg/HCl}$) completely reduces the carbonyl group (>C=O$>C=O$) to a methylene group (-textCH_2-$-\text{CH}_2-$), finishing with **cyclooctane**.

### Pattern Recognition A vicinal dihalide treated with sequential strong bases creates an alkyne path. Alkyne hydration creates a ketone body. Finally, Clemmensen reduction takes the ketone down to a simple hydrocarbon skeleton. Recognizing this terminal reduction loop establishes cyclooctane as the undisputed answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons

### Core Logic Intramolecular aldol condensations are governed heavily by thermodynamic stability, favoring the formation of **5-membered** or **6-membered** rings over strained 3-, 4-, or large 7-membered options. 1. **Enolate Generation**: Base abstracts an alpha$\alpha$-proton to form a nucleophilic carbanion enolate. 2. **Attack Vector Selection**: Deprotonation at the outer methyl site enables a ring-closing attack on the distant carbonyl carbon, perfectly designing a highly stable cyclopentene ring skeleton. 3. **Dehydration**: Heating drives the loss of a water molecule (-textH_2textO$-\text{H}_2\text{O}$), introducing an alpha,beta$\alpha,\beta$-unsaturated carbonyl arrangement that provides stabilization via conjugated resonance.

### Pattern Recognition Count the intervening carbon chain carefully. Intramolecular aldol pathways will always selectively build 5- or 6-membered rings due to favorable ring strain kinetics. Eliminating choices based on incorrect ring sizes isolates Option (1) instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids