Given below are two statements : Statement (I): Vanillin will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below:

### Related Formula Phenolic protons react with standard strong bases: 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 -OH group allows acid-base reaction with 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-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

Given below are two statements : Statement (I): Vanillin will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin will undergo self aldol condensation very easily. In the light o

### Related Formula Phenolic protons react with standard strong bases: 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 -OH group allows acid-base reaction with NaOH d

Given below are two statements : Statement (I): Vanillin will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below:

### Related Formula Phenolic protons react with standard strong bases: 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 -OH group allows acid-base reaction with 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-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

Given below are two statements : Statement (I): Vanillin will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin will undergo self aldol condensation very easily. In the light o

### Related Formula Phenolic protons react with standard strong bases: 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 -OH group allows acid-base reaction with NaOH d

Given below are two statements : Statement (I): Vanillin will react with NaOH and also with Tollen's reagent. Statement…

Q28 jee_main_2025_04_april_evening Iodoform Test

Which among the following compounds give yellow solid when reacted with NaOI/NaOH? (A)

CH_3 - CH(OH) - C_2H_5

(B)

CH_3 - CH_2 - CH_2 - OH

(C)

CH_3 - CO - C_2H_5

(D)

CH_3 - OH

(E)

CH_3 - CH_2 - H

Choose the correct answer from the options given below:

A. (B), (C) and (E) Only
B. (A) and (C) Only
C. (C) and (D) Only
D. (A), (C) and (D) Only

Solution

### Related Formula

textCompounds with CH_3-CH(OH)- text or CH_3-CO- text groups undergo the iodoform reaction to form CHI_3 downarrow text (Yellow Solid)

### Core Logic Let's check the structural groups of each given option: - **(A)**

CH_3 - CH(OH) - C_2H_5

: Contains the methylcarbinol group (

CH_3-CH(OH)-

). Gives a positive iodoform test. - **(B)**

CH_3 - CH_2 - CH_2 - OH

: Linear primary alcohol, does not contain the required group. - **(C)**

CH_3 - CO - C_2H_5

: Contains the methyl ketone group (

CH_3-CO-

). Gives a positive iodoform test. - **(D)**

CH_3 - OH

: Methanol does not give the test. - **(E)**

CH_3 - CH_2 - H

: Ethane does not give the test. Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform (

CHI_3

). ### Step 1: Chemical Equations The balanced haloform pathways occur as follows:

CH_3-CH(OH)-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+

CH_3-CO-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+

### Pattern Recognition The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal

-CH_3

affixed directly to a carbonyl oxygen index (

C=O

) or a hydroxyl carbon (

CH-OH

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Q29 jee_main_2025_04_april_morning Aldol Condensation

Aldol condensation is a popular and classical method to prepare

alpha, beta

-unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?

A. $textProduct 1$ $\text{Product 1}$
B. $textProduct 2$ $\text{Product 2}$
C. $textProduct 3$ $\text{Product 3}$
D. $textProduct 4$ $\text{Product 4}$

Solution

### Core Logic Intramolecular aldol condensation involves a dicarbonyl compound reacting within itself to yield stable cyclic

alpha, beta

-unsaturated rings (most commonly 5- or 6-membered rings due to minimal ring strain). * Products (1), (2), and (3) can all be cleanly synthesized via intramolecular cyclization path workflows from their respective dialdehyde/diketone precursors. * Product (4) features an exo-cyclic group structure arrangement formed strictly through an **intermolecular** condensation sequence step between two distinct reactant units, rather than an internal cyclization path layout. ### Pattern Recognition Look closely at the ring substitution system. Intermolecular steps are forced when intramolecular cyclization path workflows would generate highly strained small rings or structurally impossible orientations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q41 jee_main_2025_04_april_morning Chemical Properties of Ketones

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:

A. $mathrmCH_3mathrmCOCH_3, mathrmCH_3mathrmCH_2mathrmCH_2mathrmOH, mathrmCH_3mathrmCH(Br)CH_3$
B. $mathrmCH_3mathrmCOCH_3, mathrmCH_3mathrmCH(OH)CH_3, mathrmCH_3mathrmCH(Br)CH_3$
C. $mathrmCH_3mathrmCH_2mathrmCHO, mathrmCH_3mathrmCH_2mathrmCH_2mathrmOH, mathrmCH_3mathrmCH_2mathrmCH_2mathrmBr$
D. $mathrmCH_3mathrmCH_2mathrmCHO, mathrmCH_3mathrmCH=CH_2, mathrmCH_3mathrmCH(Br)CH_3$

Solution

### 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

Q30 jee_main_2025_07_april_evening Identification of Carbonyl Compounds

"P" is an optically active compound with molecular formula

textC_6textH_12textO

. When "P" is treated with 2,4-dinitrophenylhydrazine, it gives a positive test. However, in presence of Tollens reagent, "P" gives a negative test. Predict the structure of "P".

A. $textCH_3text-C(=textO)text-CH_2text-CH_2text-CH_2text-CH_3$
B. $textCH_3text-C(=textO)text-CH(textCH_2text-CH_3)text-CH_3$
C. $textH-C(=textO)text-CH_2text-CH(textCH_2text-CH_3)text-CH_3$
D. $textCH_3text-C(=textO)text-CH_2text-CH(textCH_3)_2$

Solution

### Related Formula

textCarbonyl compound + text2,4-DNP ightarrow textHydrazone derivative (Positive test)

textAldehyde + textTollens' Reagent ightarrow textSilver Mirror (Positive test)

textKetone + textTollens' Reagent ightarrow textNo reaction (Negative test)

### Core Logic Analyzing individual functional constraints: - Positive 2,4-DNP test shows compound contains a carbonyl group (aldehyde or ketone). - Negative Tollens' test clarifies it is not an aldehyde; hence it must be a ketone. - The compound is optically active, meaning it must possess a chiral center (carbon with 4 distinct groups). Let's evaluate the options via structural configurations:

Identification of Carbonyl Compounds diagram for Q30 - JEE Main 2025 Evening

### Step 1: Structural Verification Option (2) represents 3-methylpentan-2-one:

textCH3-textC(=textO)-oversetasttextCH(textCH3)(textCH2textCH3)

The third carbon (C3) is linked to:

-textH

,

-textCH_3

,

-textCH_2textCH_3

, and

-textCOCH_3

. It has 4 distinct structural fields, making it chiral and optically active. ### Pattern Recognition Tollens' negative + DNP positive = Ketone. Once categorized as a ketone, look directly for the structure holding a carbon with four unique groups to secure the optical activity constraint. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q42 jee_main_2025_24_jan_evening Preparation of Aldehydes

Match List-I with List-II

List-I	List-II (Name of Reaction)
(A) $mathrmRCN xrightarrow[text(ii)mathrmH_3mathrmO^+]text(i)mathrmSnCl_2, mathrmHCl mathrmRCHO$	(I) Etard reaction
(B) The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.	(II) Gatterman-Koch reaction
(C) The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.	(III) Rosenmund reduction
(D) The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.	(IV) Stephen reaction

Choose the correct answer from the options given below:

A. \text{(A)-(IV), (B)-(III), (C)-(I), (D)-(II)}
B. \text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}
C. \text{(A)-(I), (B)-(III), (C)-(II), (D)-(IV)}
D. \text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}

Solution

### Core Logic Let's match each aldehyde preparation method with its official named organic reaction: * **(A)

mathrmRCN ightarrow mathrmRCHO

using

mathrmSnCl_2/mathrmHCl

followed by hydrolysis:** This is the classic **Stephen reaction**

ightarrow

**(IV)**. * **(B) Reducing an acyl chloride (

mathrmRCOCl

) to an aldehyde using

mathrmH_2

over

mathrmPd-BaSO_4

:** This partial reduction is known as the **Rosenmund reduction**

ightarrow

**(III)**. * **(C) Oxidizing toluene to benzaldehyde using chromyl chloride (

mathrmCrO_2mathrmCl_2

) in

mathrmCS_2

:** This selective oxidation method is the **Etard reaction**

ightarrow

**(I)**. * **(D) Converting benzene to benzaldehyde using

mathrmCO

and

mathrmHCl

in the presence of anhydrous

mathrmAlCl_3/mathrmCuCl

:** This formylation process is the **Gatterman-Koch reaction**

ightarrow

**(II)**. Combining these assignments yields the final sequence: (A)-(IV), (B)-(III), (C)-(I), (D)-(II). ### Step-by-Step Layout The visual reaction components correspond directly to the official structural transformations:

Preparation of Aldehydes solution diagram for Q42 - JEE Main 2025 Evening — The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.

### Pattern Recognition Quick identification keys: - Nitrile

ightarrow

Aldehyde = Stephen - Acid Chloride

ightarrow

Aldehyde = Rosenmund - Toluene

ightarrow

Chromyl Complex = Etard - Benzene

ightarrow

Carbon Monoxide = Gatterman-Koch ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Solution & Explanation

More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Solution & Explanation

More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back