Match the LIST-I with LIST-II

List-I (Reagents)	List-II (Name of Reaction involving carbonyl compound)
A. NH_2-NH_2$NH_{2}-NH_{2}$, KOH	I. Tollen's Test
B. Ag(NH_3)_2OH$Ag(NH_{3})_{2}OH$	II. Clemmensen Reduction
C. Aq. CuSO_4$CuSO_{4}$, Sodium Potassium tartarate, KOH	III. Wolff-Kishner Reduction
D. Zn - Hg, HCl	IV. Fehling's Test

Choose the correct answer from the options given below

### Core Logic Compound P (C_6H_12O_3$C_6H_{12}O_3$) gives a positive iodoform test, indicating it has a methyl ketone group (CH_3CO-$CH_3CO-$). It gives a negative Tollen's test, indicating no free aldehyde group. On acidic hydrolysis, P yields Q. Compound Q gives both positive iodoform test and Tollen's test, meaning Q contains both a methyl ketone and an aldehyde group. Looking at the options, if P is an acetal of an aldehyde, acidic hydrolysis will regenerate the aldehyde. Option 2 is mathrmCH_3-CO-CH_2-CH(OCH_3)_2$\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2}$ (an acetal of aldehyde). This compound 'P' has a CH_3CO-$CH_3CO-$ group (positive iodoform) and an acetal (protected aldehyde, negative Tollen's). On hydrolysis: mathrmCH_3-CO-CH_2-CH(OCH_3)_2 xrightarrowmathrmH_2O/H^+ mathrmCH_3-CO-CH_2-CHO + 2mathrmCH_3OH$$\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2} \xrightarrow{\mathrm{H_2O/H^+}} \mathrm{CH_3-CO-CH_2-CHO} + 2\mathrm{CH_3OH}$$ The product Q (mathrmCH_3-CO-CH_2-CHO$\mathrm{CH_3-CO-CH_2-CHO}$) has a methyl ketone (positive iodoform test) and an aldehyde (positive Tollen's test).

### Pattern Recognition Whenever an aldehyde test becomes positive *after* hydrolysis, it points to a protected aldehyde, usually an acetal or hemiacetal. A compound with molecular formula C_n H_2n O_3$C_n H_{2n} O_3$ often represents a keto-acetal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

### Core Logic - A. textH_2, textPd-textBaSO_4 rightarrow$\text{H}_2, \text{Pd}-\text{BaSO}_4 \rightarrow$ II. Rosenmund Reduction - B. textSnCl_2, textHCl rightarrow$\text{SnCl}_2, \text{HCl} \rightarrow$ IV. Stephen Reaction - C. textCrO_2textCl_2, textCS_2 rightarrow$\text{CrO}_2\text{Cl}_2, \text{CS}_2 \rightarrow$ I. Etard Reaction - D. textCO, textHCl, textAnhyd. textAlCl_3 rightarrow$\text{CO}, \text{HCl}, \text{Anhyd. } \text{AlCl}_3 \rightarrow$ III. Gatterman–Koch Reaction ### Step 1: Final Conclusion Combining the correct matching gives A-II, B-IV, C-I, D-III, which is option (4). ### Pattern Recognition Sees: standard name reactions for aldehyde preparation. Trap: Confusing Etard reagent with Gatterman-Koch or Stephen reduction. ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula textAr-CHO + textconc. NaOH xrightarrowtextCannizzaro textAr-CH_2textOH + textAr-COO^-$$\text{Ar-CHO} + \text{conc. NaOH} \xrightarrow{\text{Cannizzaro}} \text{Ar-CH}_2\text{OH} + \text{Ar-COO}^-$$ ### Core Logic Step 1: Compound A has molecular formula textC_8textH_8textO_2$\text{C}_8\text{H}_8\text{O}_2$ and undergoes Cannizzaro reaction with conc. NaOH implies$\implies$ A lacks alpha$\alpha$-hydrogen and is an aromatic aldehyde. Step 2: Compound A is 4-methoxy benzaldehyde (textCH_3textO-C_6textH_4text-CHO$\text{CH}_3\text{O-C}_6\text{H}_4\text{-CHO}$). Step 3: Cross-Aldol condensation with acetophenone yields a single aldol condensation product (B)$(B)$ structure. Step 4: Cannizzaro reaction with conc. NaOH produces 4-methoxybenzyl alcohol and 4-methoxybenzoate anion.

### Pattern Recognition Sees: textC_8textH_8textO_2$\text{C}_8\text{H}_8\text{O}_2$ undergoing Cannizzaro to form substituted benzyl alcohol. Shortcut: Presence of -textCHO$-\text{CHO}$ without alpha$\alpha$-H and -textOCH_3$-\text{OCH}_3$ ring substituent points directly to 4-methoxy benzaldehyde. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### 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