Match List - I with List - II. List - I (Reactions) List - II (Reagents) (A) CH3(CH2)5-CO-OC2H5 arrow CH3(CH2)5CHO (I) CH3MgBr, H2O (B) C6H5COC6H5 arrow C6H5CH2C6H5 (II) Zn(Hg) and conc. HCl (C) C6H5CHO arrow C6H5CH(OH)CH3 (III) NaBH4, H^+ (D) CH3COCH2COOC2H5 arrow CH3CH(OH)CH2COOC2H5 (IV) DIBAL-H, H2O Choose the correct answer from options given below:

### Core Logic Let's analyze the transformation happening in each reaction: (A) CH3(CH2)5COOC2H5 arrow CH3(CH2)5CHO An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) arrow (IV). (B) C6H5COC6H5 arrow C6H5CH2C6H5 A ketone (carbonyl group >C=O) is fully reduced to an alkane (>CH2) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) arrow (II). (C) C6H5CHO arrow C6H5CH(OH)CH3 Benzaldehyde (aldehyde) is converted into a secondary alcohol with an extra methyl group. This is a nucleophilic addition of a Grignard reagent (CH3MgBr) followed by hydrolysis. Thus, (C) arrow (I). (D) CH3COCH2COOC2H5 arrow CH3CH(OH)CH2COOC2H5 A ketone group is reduced to a secondary alcohol while the ester group remains intact. NaBH4 is a mild reducing agent that reduces aldehydes and ketones but generally does not touch esters. Thus, (D) arrow (III). ### Pattern Recognition Ester arrow Aldehyde = DIBAL-H Ketone arrow Alkane = Clemmensen (Zn(Hg)/HCl) or Wolff-Kishner Carbonyl arrow Alcohol with carbon chain extension = Grignard Reagent Ketone arrow Alcohol (leaving ester intact) = NaBH4 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Match List - I with List - II. List - I (Reactions) List - II (Reagents) (A) CH3(CH2)5-CO-OC2H5 arrow CH3(CH2)5CHO (I) CH3MgBr, H2O (B) C6H5COC6H5 arrow C6H5CH2C6H5 (II) Zn(Hg) and conc. HCl (C) C6H5C

### Core Logic Let's analyze the transformation happening in each reaction: (A) CH3(CH2)5COOC2H5 arrow CH3(CH2)5CHO An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) arrow (IV). (B) C6H5COC6H5 arrow C6H5CH2C6H5 A ketone (carbonyl group >C=O) is fully reduced to an alkane (>CH2) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) arrow (II).

Match List - I with List - II. List - I (Reactions) List - II (Reagents) (A) CH3(CH2)5-CO-OC2H5 arrow CH3(CH2)5CHO (I) CH3MgBr, H2O (B) C6H5COC6H5 arrow C6H5CH2C6H5 (II) Zn(Hg) and conc. HCl (C) C6H5CHO arrow C6H5CH(OH)CH3 (III) NaBH4, H^+ (D) CH3COCH2COOC2H5 arrow CH3CH(OH)CH2COOC2H5 (IV) DIBAL-H, H2O Choose the correct answer from options given below:

### Core Logic Let's analyze the transformation happening in each reaction: (A) CH3(CH2)5COOC2H5 arrow CH3(CH2)5CHO An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) arrow (IV). (B) C6H5COC6H5 arrow C6H5CH2C6H5 A ketone (carbonyl group >C=O) is fully reduced to an alkane (>CH2) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) arrow (II). (C) C6H5CHO arrow C6H5CH(OH)CH3 Benzaldehyde (aldehyde) is converted into a secondary alcohol with an extra methyl group. This is a nucleophilic addition of a Grignard reagent (CH3MgBr) followed by hydrolysis. Thus, (C) arrow (I). (D) CH3COCH2COOC2H5 arrow CH3CH(OH)CH2COOC2H5 A ketone group is reduced to a secondary alcohol while the ester group remains intact. NaBH4 is a mild reducing agent that reduces aldehydes and ketones but generally does not touch esters. Thus, (D) arrow (III). ### Pattern Recognition Ester arrow Aldehyde = DIBAL-H Ketone arrow Alkane = Clemmensen (Zn(Hg)/HCl) or Wolff-Kishner Carbonyl arrow Alcohol with carbon chain extension = Grignard Reagent Ketone arrow Alcohol (leaving ester intact) = NaBH4 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Match List - I with List - II. List - I (Reactions) List - II (Reagents) (A) CH3(CH2)5-CO-OC2H5 arrow CH3(CH2)5CHO (I) CH3MgBr, H2O (B) C6H5COC6H5 arrow C6H5CH2C6H5 (II) Zn(Hg) and conc. HCl (C) C6H5C

### Core Logic Let's analyze the transformation happening in each reaction: (A) CH3(CH2)5COOC2H5 arrow CH3(CH2)5CHO An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) arrow (IV). (B) C6H5COC6H5 arrow C6H5CH2C6H5 A ketone (carbonyl group >C=O) is fully reduced to an alkane (>CH2) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) arrow (II).

Match List - I with List - II. List - I (Reactions) List - II (Reagents) (A) CH3(CH2)5-CO-OC2H5 arrow CH3(CH2)5CHO (I)…

Year	2026	2025	2024	Total
Questions	14	21	11	46

List - I (Reactions)	List - II (Reagents)
(A) $CH₃(CH₂)₅-CO-OC₂H₅ arrow CH₃(CH₂)₅CHO$	(I) $CH₃MgBr, H₂O$ $CH_3MgBr, H_2O$
(B) $C₆H₅COC₆H₅ arrow C₆H₅CH₂C₆H₅$	(II) $Zn(Hg)$ $Zn(Hg)$ and conc. $HCl$ $HCl$
(C) $C₆H₅CHO arrow C₆H₅CH(OH)CH₃$	(III) $NaBH₄, H^+$ $NaBH_4, H^+$
(D) $CH₃COCH₂COOC₂H₅ arrow CH₃CH(OH)CH₂COOC₂H₅$	(IV) DIBAL-H, $H₂O$ $H_2O$

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

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

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