An organic compound (X) with molecular formula C_3H_6O is not readily oxidised. On reduction it gives C_3H_8O (Y) which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis gives 2, 3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:

Solution & Explanation

### Core Logic Let's deduce the identities stepwise: 1. Compound (X) has the formula C_3H_6O and is resistant to mild oxidation, which identifies it as a ketone: **Acetone** (CH_3COCH_3). 2. Reduction of Acetone yields a secondary alcohol, Propan-2-ol (CH_3CH(OH)CH_3, Compound Y). 3. Treatment of Propan-2-ol with HBr substitutes the hydroxyl group to form 2-Bromopropane (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). 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
Grignard reaction pathways structural connectivity diagram for Q41
Grignard reaction pathways structural connectivity diagram for Q41

Reference Study Guides

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

More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2025_04_april_morning

Practice all Aldehydes, Ketones and Carboxylic Acids previous-year questions →

Rankbit System
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%)