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:

Solution & Explanation

### 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

More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 7

Q87 jee_main_2024_30_jan_morning Nucleophilic Addition Reactions
The compound formed by the reaction of ethanal with semicarbazide contains ________ number of nitrogen atoms.
Numerical Answer. Answer: 3 to 3

Solution

### Related Formula 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) reacts with semicarbazide (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. Counting the nitrogen atoms in this structure: 1. The imine nitrogen (=N-) 2. The amine nitrogen (-NH-) 3. The amide nitrogen (-NH_2) Total = 3 Nitrogen atoms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q73 jee_main_2024_31_jan_evening Preparation of Aldehydes and Ketones
Identify the name reaction.
Preparation of Aldehydes and Ketones diagram for Q73 - JEE Main 2024 Evening
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
  • A. text(1) Stephen reaction
  • B. text(2) Etard reaction
  • C. text(3) Gatterman-koch reaction
  • D. text(4) Rosenmund reduction

Solution

### Core Logic The reaction of benzene with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of anhydrous aluminium chloride (AlCl_3) and cuprous chloride (CuCl) to give benzaldehyde is known as the Gatterman-Koch reaction.
Preparation of Aldehydes and Ketones diagram for Q73 - JEE Main 2024 Evening
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
### Pattern Recognition CO + HCl rightarrow Formyl chloride intermediate (in situ) with Lewis acid 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
Q84 jee_main_2024_31_jan_morning Reactions with Grignard Reagent
The product of the following reaction is P.
Reactions with Grignard Reagent diagram for Q84 - JEE Main 2024 Morning
The image shows p-hydroxybenzaldehyde reacting with one equivalent of PhMgBr followed by aqueous ammonium chloride.
The number of hydroxyl groups present in the product P is
Numerical Answer. Answer: 0 to 0

Solution

### Core Logic The given reactant is p-hydroxybenzaldehyde, which contains both a phenolic -OH group (acidic) and an aldehyde group (electrophilic). When one equivalent of Grignard reagent (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 reacts with PhMgBr: PhMgBr + HO-C_6H_4-CHO rightarrow Ph-H (Benzene) + BrMg-O-C_6H_4-CHO Upon workup with 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).
Reactions with Grignard Reagent diagram for Q84 - JEE Main 2024 Morning
The image shows p-hydroxybenzaldehyde reacting with one equivalent of PhMgBr followed by aqueous ammonium chloride.
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
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