‘x’ is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. ‘y’ is the product which is obtained from the but-2-ene by the ozonolysis followed by hydrolysis. From the following, which product is not obtained when one mole of ‘x’ and one mole of ‘y’ react with each other in the presence of alkali followed by heating?

First, identify products 'x' and 'y' based on the given name reactions. Reaction 1: Stephen's Reduction Propanenitrile (CH3-CH2-Cequiv N) reacts with SnCl2 / HCl followed by H3O^+ to give propanal. CH3-CH2-CHO is product (X). Reaction 2: Reductive Ozonolysis But-2-ene (CH3-CH=CH-CH3) undergoes ozonolysis to give two moles of ethanal. CH3-CHO is product (Y). Reacting (X) and (Y) in alkali with heat leads to a cross-aldol condensation producing a mixture of four alpha,beta-unsaturated aldehydes (two self-aldol and two cross-aldol products). Reactants: CH3-CH2-CHO and CH3-CHO Possible aldol condensation products: 1. Ethanal + Ethanal (Self) arrow But-2-enal 2. Propanal + Propanal (Self) arrow 2-Methylpent-2-enal 3. Ethanal (Enolate) + Propanal (Electrophile) arrow Pent-2-enal 4. Propanal (Enolate) + Ethanal (Electrophile) arrow 2-Methylbut-2-enal Comparing with the given options, 3-Methylbut-2-enal is NOT formed. To quickly identify cross-aldol products, simply remove two alpha-hydrogens from one molecule and the carbonyl oxygen from the other, then connect them with a double bond. Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons

‘x’ is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. ‘y’ is the product which is obtained from the but-2-ene by t

First, identify products 'x' and 'y' based on the given name reactions. Reaction 1: Stephen's Reduction Propanenitrile (CH3-CH2-Cequiv N) reacts with SnCl2 / HCl followed by H3O^+ to give propanal. CH3-CH2-CHO is product (X). Reaction 2: Reductive Ozonolysis But-2-ene (CH3-CH=CH-CH3) undergoes ozonolysis to give two moles of ethanal. CH3-CHO is product (Y). Reacting (X) and (Y) in alkali with heat leads to a cross-aldol condensation producing a mixture of four alpha,beta-unsaturated aldehydes (t

‘x’ is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. ‘y’ is the product which is obtained from the but-2-ene by the ozonolysis followed by hydrolysis. From the following, which product is not obtained when one mole of ‘x’ and one mole of ‘y’ react with each other in the presence of alkali followed by heating?

First, identify products 'x' and 'y' based on the given name reactions. Reaction 1: Stephen's Reduction Propanenitrile (CH3-CH2-Cequiv N) reacts with SnCl2 / HCl followed by H3O^+ to give propanal. CH3-CH2-CHO is product (X). Reaction 2: Reductive Ozonolysis But-2-ene (CH3-CH=CH-CH3) undergoes ozonolysis to give two moles of ethanal. CH3-CHO is product (Y). Reacting (X) and (Y) in alkali with heat leads to a cross-aldol condensation producing a mixture of four alpha,beta-unsaturated aldehydes (two self-aldol and two cross-aldol products). Reactants: CH3-CH2-CHO and CH3-CHO Possible aldol condensation products: 1. Ethanal + Ethanal (Self) arrow But-2-enal 2. Propanal + Propanal (Self) arrow 2-Methylpent-2-enal 3. Ethanal (Enolate) + Propanal (Electrophile) arrow Pent-2-enal 4. Propanal (Enolate) + Ethanal (Electrophile) arrow 2-Methylbut-2-enal Comparing with the given options, 3-Methylbut-2-enal is NOT formed. To quickly identify cross-aldol products, simply remove two alpha-hydrogens from one molecule and the carbonyl oxygen from the other, then connect them with a double bond. Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons

‘x’ is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. ‘y’ is the product which is obtained from the but-2-ene by t

First, identify products 'x' and 'y' based on the given name reactions. Reaction 1: Stephen's Reduction Propanenitrile (CH3-CH2-Cequiv N) reacts with SnCl2 / HCl followed by H3O^+ to give propanal. CH3-CH2-CHO is product (X). Reaction 2: Reductive Ozonolysis But-2-ene (CH3-CH=CH-CH3) undergoes ozonolysis to give two moles of ethanal. CH3-CHO is product (Y). Reacting (X) and (Y) in alkali with heat leads to a cross-aldol condensation producing a mixture of four alpha,beta-unsaturated aldehydes (t

‘x’ is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid…

Year	2026	2025	2024	Total
Questions	14	21	11	46

Q jee_main_2024_30_jan_morning Preparation of Aldehydes

This reduction reaction is known as:

Preparation of Aldehydes diagram for Q62 - JEE Main 2024 Morning — The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.

A. $Rosenmund reduction$ $\text{Rosenmund reduction}$
B. $Wolff-Kishner reduction$ $\text{Wolff-Kishner reduction}$
C. $Stephen reduction$ $\text{Stephen reduction}$
D. $Etard reduction$ $\text{Etard reduction}$

Solution

Related Formula

R-COCl + H₂ Pd/BaSO₄ R-CHO + HCl

Core Logic

The reaction depicts the partial reduction of an acid chloride (benzoyl chloride) to an aldehyde (benzaldehyde) using hydrogen gas in the presence of a poisoned palladium catalyst (Pd supported on $BaSO₄$ $BaSO_4$).

Preparation of Aldehydes solution diagram for Q62 - JEE Main 2024 Morning — The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.

This specific reaction is known as the Rosenmund reduction.

Pattern Recognition

Acid Chloride + $H₂, Pd/BaSO₄$ $H_2, Pd/BaSO_4$ $arrow$ $\rightarrow$ Aldehyde is strictly the Rosenmund reduction. The $BaSO₄$ $BaSO_4$ poisons the catalyst to prevent over-reduction to an alcohol.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q jee_main_2024_30_jan_morning Nomenclature

Structure of 4-Methylpent-2-enal is

A. $H₂C=C(CH₃)-CH₂-C(=O)H$ $$H_2C=C(CH_3)-CH_2-C(=O)H$$
B. $CH₃-CH₂-C(CH₃)=CH-C(=O)H$ $$CH_3-CH_2-C(CH_3)=CH-C(=O)H$$
C. $CH₃-CH₂-CH=C(CH₃)-C(=O)H$ $$CH_3-CH_2-CH=C(CH_3)-C(=O)H$$
D. $CH₃-CH(CH₃)-CH=CH-C(=O)H$ $$CH_3-CH(CH_3)-CH=CH-C(=O)H$$

Solution

Core Logic

Decode the IUPAC name: 4-Methylpent-2-enal

Word root: 'pent' $arrow$ $\rightarrow$ 5 carbon principal chain.
Primary suffix: '2-en' $arrow$ $\rightarrow$ Double bond starting at carbon 2.
Secondary suffix: 'al' $arrow$ $\rightarrow$ Aldehyde group (-CHO) at carbon 1.
Substituent: '4-Methyl' $arrow$ $\rightarrow$ A methyl group (-CH3) at carbon 4.

Step 1: Drafting the structure

Numbering starts from the aldehyde carbon.

C⁵ - C⁴ - C³ = C² - C¹(=O)H

Attach the methyl at $C⁴$ $C^4$:

CH₃ - CH(CH₃) - CH = CH - CHO

This matches Option 4.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q jee_main_2024_30_jan_morning Preparation of Aldehydes

In the given reactions identify the reagent A and reagent B.

Preparation of Aldehydes diagram for Q73 - JEE Main 2024 Morning — The image shows the oxidation of toluene to benzaldehyde via two different pathways requiring reagents A and B.

A. $A-CrO₃, B-CrO₃$ $\text{A}-CrO_3, \text{B}-CrO_3$
B. $A-CrO₃, B-CrO₂Cl₂$ $\text{A}-CrO_3, \text{B}-CrO_2Cl_2$
C. $A-CrO₂Cl₂, B-CrO₂Cl₂$
D. $A-CrO₂Cl₂, B-CrO₃$ $\text{A}-CrO_2Cl_2, \text{B}-CrO_3$

Solution

Core Logic

Pathway 1 (Upper): Toluene is treated with Reagent 'A' and acetic anhydride $(CH₃CO)₂O$ $(CH_3CO)_2O$ to form an intermediate (benzylidene diacetate), which on hydrolysis gives benzaldehyde. The reagent used here is Chromic oxide ( $CrO₃$ $CrO_3$). Thus, A is $CrO₃$ $CrO_3$.

Pathway 2 (Lower): Toluene is treated with Reagent 'B' in $CS₂$ $CS_2$ to form a chromium complex intermediate, which on hydrolysis yields benzaldehyde. This is the Etard reaction, and the reagent used is Chromyl chloride ( $CrO₂Cl₂$ $CrO_2Cl_2$). Thus, B is $CrO₂Cl₂$ $CrO_2Cl_2$.

Preparation of Aldehydes solution diagram for Q73 - JEE Main 2024 Morning — The image shows the oxidation of toluene to benzaldehyde via two different pathways requiring reagents A and B.

Step 1: Selection

Therefore, A = $CrO₃$ $CrO_3$ and B = $CrO₂Cl₂$ $CrO_2Cl_2$.

Pattern Recognition

Etard reaction always uses Chromyl chloride ( $CrO₂Cl₂$ $CrO_2Cl_2$). Oxidation of toluene with acetic anhydride uses Chromic acid ( $CrO₃$ $CrO_3$). Both stop the oxidation at the aldehyde stage via intermediate formation.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

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

CH₃-CHO + H₂N-NH-CO-NH₂ arrow CH₃-CH=N-NH-CO-NH₂ + H₂O

Core Logic

Ethanal ( $CH₃CHO$ $CH_3CHO$) reacts with semicarbazide ( $H₂N-NH-CO-NH₂$ $H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone.

Step 1: Product Analysis

The product is Ethanal semicarbazone: $CH₃-CH=N-NH-CO-NH₂$ $CH_3-CH=N-NH-CO-NH_2$. Counting the nitrogen atoms in this structure:

The imine nitrogen ( $=N-$ $=N-$)
The amine nitrogen ( $-NH-$ $-NH-$)
The amide nitrogen ( $-NH₂$ $-NH_2$)

Total = 3 Nitrogen atoms.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q 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. $Stephen reaction$ $\text{Stephen reaction}$
B. $Etard reaction$ $\text{Etard reaction}$
C. $Gatterman-koch reaction$ $\text{Gatterman-koch reaction}$
D. $Rosenmund reduction$ $\text{Rosenmund reduction}$

Solution

Core Logic

The reaction of benzene with carbon monoxide ( $CO$ $CO$) and hydrogen chloride ( $HCl$ $HCl$) in the presence of anhydrous aluminium chloride ( $AlCl₃$ $AlCl_3$) and cuprous chloride ( $CuCl$ $CuCl$) to give benzaldehyde is known as the Gatterman-Koch reaction.

Pattern Recognition

$CO + HCl arrow$ $CO + HCl \rightarrow$ Formyl chloride intermediate (in situ) with Lewis acid $arrow$ $\rightarrow$ formylation of benzene. This is definitively the Gatterman-Koch formylation.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Solution & Explanation

Core Logic

Step 1: Cross-Aldol Condensation

Step 2: Finding Products

Pattern Recognition

Chapter Mix

Reference Study Guides

More Aldehydes Ketones and Carboxylic Acids Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Drafting the structure

Chapter Mix

Solution

Core Logic

Step 1: Selection

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Product Analysis

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Aldehydes Ketones and Carboxylic Acids Questions — jee_main_2026_23_january_morning

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