The major product (P) in the following reaction is :
Benzil-Benzilic Acid Rearrangement
Benzil-Benzilic Acid Rearrangement

Solution & Explanation

### 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 1. The aldehyde carbon (-CHO) is much more electrophilic than the ketone carbon (-CO-) due to less steric hindrance and absence of phenyl group electron donation. 2. Hydroxide ion (mathrmOH^-) selectively attacks the aldehyde carbonyl carbon, forming a tetrahedral intermediate.
Benzil-Benzilic Acid Rearrangement
Benzil-Benzilic Acid Rearrangement
### Step 1: Hydride Transfer Mechanism The tetrahedral intermediate collapses, prompting an intramolecular hydride (H^-) transfer to the adjacent ketone carbonyl carbon: mathrmPh-CO-C(O^-)(OH)H rightarrow mathrmPh-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^+ 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-hydroxy carboxylate salt. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions

Q61 jee_main_2026_21_jan_morning Chemical Reactions of Aldehydes and Ketones
An organic compound “P” of molecular formula C_6H_12O_3 gives positive Iodoform test but negative Tollen’s test. When “P” is treated with dilute acid, it produces “Q”. “Q” gives positive Tollen’s test and also iodoform test. The structure of “P” is :
  • A. mathrmCH_3-CO-CH(OCH_3)-CH_2(OCH_3)
  • B. mathrmCH_3-CO-CH_2-CH(OCH_3)_2
  • C. mathrmH-CO-CH_2-CH_2-C(OCH_3)_2-CH_3
  • D. mathrmCH_3-CO-CO-CH_3 text (with acetal structure)

Solution

### Core Logic Compound P (C_6H_12O_3) gives a positive iodoform test, indicating it has a methyl ketone group (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 (an acetal of aldehyde). This compound 'P' has a 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 The product Q (mathrmCH_3-CO-CH_2-CHO) has a methyl ketone (positive iodoform test) and an aldehyde (positive Tollen's test).
Chemical Reactions of Aldehydes and Ketones diagram for Q61 - JEE Main 2026 Morning
Chemical Reactions of Aldehydes and Ketones diagram for Q61 - JEE Main 2026 Morning
### 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 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
Q jee_main_2025_02_april_evening Preparation of Carboxylic Acids
Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product? (A) mathrmR - C equiv N xrightarrow[textmild condition]mathrm(i) H^+ / H_2O (B) mathrmR - MgX xrightarrow[mathrm(ii) H_3O^+]mathrm(i) CO_2 (C) mathrmR - C equiv N xrightarrow[mathrm(ii) H_3O^+]mathrm(i) SnCl_2 / HCl (D) mathrmR cdot CH_2 cdot OH xrightarrowmathrmPCC (E)
Preparation of Carboxylic Acids
Preparation of Carboxylic Acids
Choose the correct answer from the options given below:
  • A. textA and D only
  • B. textA, B and E only
  • C. textB, C and E only
  • D. textB and E only

Solution

### Related Formula mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_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 (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 - **Reaction (C)**: Stephen reduction converts nitrile to aldehyde: mathrmR-Cequiv N xrightarrowSnCl_2/HCl R-CH=NH xrightarrowH_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 - **Reaction (E)**
Preparation of Carboxylic Acids
Preparation of Carboxylic Acids

: Rosenmund reduction reduces acid chloride to aldehyde first: 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 ### 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) 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
Q jee_main_2025_02_april_morning Reactions of Phenolic Benzaldehydes
Given below are two statements : Statement (I): Vanillin
Reactions of Phenolic Benzaldehydes
Reactions of Phenolic Benzaldehydes
will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin
Reactions of Phenolic Benzaldehydes
Reactions of Phenolic Benzaldehydes
will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. (1)\ textStatement I is incorrect but Statement II is correct
  • B. (2)\ textStatement I is correct but Statement II is incorrect
  • C. (3)\ textBoth Statement I and Statement II are incorrect
  • D. (4)\ textBoth Statement I and Statement II are correct

Solution

### Related Formula Phenolic protons react with standard strong bases: mathrmAr-OH + NaOH rightarrow Ar-ONa + H_2O Aldol condensation structural requirement: Requires presence of acidic 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 group allows acid-base reaction with mathrmNaOH directly
Vanillin structural functional group verification for Q36
Vanillin structural functional group verification for Q36
. 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-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
Q jee_main_2025_08_april_evening Aldol Condensation
When the dicarbonyl compound shown below undergoes an base-catalyzed intramolecular aldol condensation reaction, the major structural product formed is: {{Q_IMG1}}
Dicarbonyl reactant structural profile for Q43
The image features a symmetrical open-chain diketone containing branching elements, poised for intramolecular cyclization.
  • A.
  • B.
  • C.
  • D.

Solution

### 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-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), introducing an alpha,beta-unsaturated carbonyl arrangement that provides stabilization via conjugated resonance.
Detailed intramolecular cyclization mechanism mapping
The image features a symmetrical open-chain diketone containing branching elements, poised for intramolecular cyclization.
### 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
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)