A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below: The structure of (P) is

### Core Logic The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below: ### Pattern Recognition Sees: Acidic rearrangement arrow ozonolysis arrow intramolecular aldol condensation. Shortcut: Work backwards from the dicarbonyl fragments formed after opening the final product ring system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below: The

### Core Logic The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below: ### Pattern Recognition Sees: Acidic rearrangement arrow ozonolysis arrow intramolecular aldol con

A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below: The structure of (P) is

### Core Logic The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below: ### Pattern Recognition Sees: Acidic rearrangement arrow ozonolysis arrow intramolecular aldol condensation. Shortcut: Work backwards from the dicarbonyl fragments formed after opening the final product ring system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below: The

### Core Logic The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below: ### Pattern Recognition Sees: Acidic rearrangement arrow ozonolysis arrow intramolecular aldol con

A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux…

Year	2026	2025	2024	Total
Questions	14	21	11	46

Q61 jee_main_2026_21_jan_morning Chemical Reactions of Aldehydes and Ketones

An organic compound “P” of molecular formula

C₆H₁₂O₃

gives positive Iodoform test but negative Tollen’s test. When “P” is treated with dilute acid, it produces “Q”. “Q” gives positive Tollen’s test and also iodoform test. The structure of “P” is :

A. $CH₃-CO-CH(OCH₃)-CH₂(OCH₃)$
B. $CH₃-CO-CH₂-CH(OCH₃)₂$ $\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2}$
C. $H-CO-CH₂-CH₂-C(OCH₃)₂-CH₃$
D. $CH₃-CO-CO-CH₃ (with acetal structure)$

Solution

Core Logic

Compound P ( $C₆H₁₂O₃$ $C_6H_{12}O_3$) gives a positive iodoform test, indicating it has a methyl ketone group ( $CH₃CO-$ $CH_3CO-$). It gives a negative Tollen's test, indicating no free aldehyde group.

On acidic hydrolysis, P yields Q. Compound Q gives both positive iodoform test and Tollen's test, meaning Q contains both a methyl ketone and an aldehyde group.

Looking at the options, if P is an acetal of an aldehyde, acidic hydrolysis will regenerate the aldehyde. Option 2 is $CH₃-CO-CH₂-CH(OCH₃)₂$ $\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2}$ (an acetal of aldehyde). This compound 'P' has a $CH₃CO-$ $CH_3CO-$ group (positive iodoform) and an acetal (protected aldehyde, negative Tollen's). On hydrolysis:

CH₃-CO-CH₂-CH(OCH₃)₂ H₂O/H^+ CH₃-CO-CH₂-CHO + 2CH₃OH

The product Q ( $CH₃-CO-CH₂-CHO$ $\mathrm{CH_3-CO-CH_2-CHO}$) has a methyl ketone (positive iodoform test) and an aldehyde (positive Tollen's test).

Chemical Reactions of Aldehydes and Ketones diagram for Q61 - JEE Main 2026 Morning

Pattern Recognition

Whenever an aldehyde test becomes positive after hydrolysis, it points to a protected aldehyde, usually an acetal or hemiacetal. A compound with molecular formula $Cₙ H₂ₙ O₃$ $C_n H_{2n} O_3$ often represents a keto-acetal.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Q57 jee_main_2026_21_jan_evening Name Reactions

Match List-I with List-II.

List-I (Reagents)	List-II (Reaction Name Involving Aldehydes)
A. $H₂, Pd-BaSO₄$ $\text{H}_2, \text{Pd}-\text{BaSO}_4$	I. Etard Reaction
B. $SnCl₂, HCl$ $\text{SnCl}_2, \text{HCl}$	II. Rosenmund Reduction
C. $CrO₂Cl₂, CS₂$ $\text{CrO}_2\text{Cl}_2, \text{CS}_2$	III. Gatterman–Koch Reaction
D. $CO, HCl, Anhyd. AlCl₃$	IV. Stephen Reaction

Choose the correct answer from the options given below:

A. $(1) A-II, B-III, C-IV, D-I$
B. $(2) A-IV, B-III, C-I, D-II$
C. $(3) A-IV, B-I, C-II, D-III$
D. $(4) A-II, B-IV, C-I, D-III$

Solution

Core Logic

A. $H₂, Pd-BaSO₄ arrow$ II. Rosenmund Reduction
B. $SnCl₂, HCl arrow$ $\text{SnCl}_2, \text{HCl} \rightarrow$ IV. Stephen Reaction
C. $CrO₂Cl₂, CS₂ arrow$ I. Etard Reaction
D. $CO, HCl, Anhyd. AlCl₃ arrow$ III. Gatterman–Koch Reaction

Step 1: Final Conclusion

Combining the correct matching gives A-II, B-IV, C-I, D-III, which is option (4).

Pattern Recognition

Sees: standard name reactions for aldehyde preparation. Trap: Confusing Etard reagent with Gatterman-Koch or Stephen reduction.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q63 jee_main_2026_22_january_morning Name Reactions

Match the LIST-I with LIST-II

List-I (Reagents)	List-II (Name of Reaction involving carbonyl compound)
A. $NH₂-NH₂$ $NH_{2}-NH_{2}$, KOH	I. Tollen's Test
B. $Ag(NH₃)₂OH$ $Ag(NH_{3})_{2}OH$	II. Clemmensen Reduction
C. Aq. $CuSO₄$ $CuSO_{4}$, Sodium Potassium tartarate, KOH	III. Wolff-Kishner Reduction
D. Zn - Hg, HCl	IV. Fehling's Test

Choose the correct answer from the options given below

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

Solution

Core Logic

(A) $NH₂-NH₂ / KOH$ $NH_2-NH_2 / KOH$: Hydrazine with base is used in Wolff-Kishner Reduction to reduce carbonyls to alkanes. (A $arrow$ $\rightarrow$ III)

(B) $[Ag(NH₃)₂]OH$ $[Ag(NH_3)_2]OH$: Ammoniacal silver nitrate is Tollen's reagent, used in Tollen's Test to differentiate aldehydes from ketones. (B $arrow$ $\rightarrow$ I)

(C) Aqueous $CuSO₄$ $CuSO_4$ + Sodium Potassium Tartrate + KOH: This is Fehling's solution (mixture of Fehling A and Fehling B). Used in Fehling's Test. (C $arrow$ $\rightarrow$ IV)

(D) $Zn-Hg / HCl$ $Zn-Hg / HCl$: Zinc amalgam with concentrated hydrochloric acid is the reagent for Clemmensen Reduction. (D $arrow$ $\rightarrow$ II)

Step 1: Final Matching

The correct matches are A-III, B-I, C-IV, D-II.

Pattern Recognition

Pure factual matching strictly based on NCERT name reactions and distinguishing tests for carbonyl compounds.

Chapter Mix

Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

Q68 jee_main_2026_22_january_evening Cross-Aldol and Cannizzaro Reactions

The compound A,

C₈H₈O₂

reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as

A. 2-hydroxy acetophenone
B. 4-methoxy benzaldehyde
C. 4-hydroxy benzaldehyde
D. 4-methyl benzoic acid

Solution

Related Formula

Ar-CHO + conc. NaOH Cannizzaro Ar-CH₂OH + Ar-COO^-

Core Logic

Step 1: Compound A has molecular formula $C₈H₈O₂$ $\text{C}_8\text{H}_8\text{O}_2$ and undergoes Cannizzaro reaction with conc. NaOH $\implies$ A lacks $α$ $\alpha$-hydrogen and is an aromatic aldehyde.

Step 2: Compound A is 4-methoxy benzaldehyde ( $CH₃O-C₆H₄-CHO$ ).

Step 3: Cross-Aldol condensation with acetophenone yields a single aldol condensation product $(B)$ $(B)$ structure.

Step 4: Cannizzaro reaction with conc. NaOH produces 4-methoxybenzyl alcohol and 4-methoxybenzoate anion.

Cross-aldol and Cannizzaro reaction diagram for Q68 - JEE Main 2026 Evening

Pattern Recognition

Sees: $C₈H₈O₂$ $\text{C}_8\text{H}_8\text{O}_2$ undergoing Cannizzaro to form substituted benzyl alcohol. Shortcut: Presence of $-CHO$ $-\text{CHO}$ without $α$ $\alpha$-H and $-OCH₃$ $-\text{OCH}_3$ ring substituent points directly to 4-methoxy benzaldehyde.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q54 jee_main_2026_23_january_morning Oxidation of Alcohols and Alkene Reactions

Oxidation of Alcohols and Alkene Reactions diagram for Q54 - JEE Main 2026 Morning — The image shows an organic conversion from a bromo-substituted cyclopentene to a carboxylic acid derivative.

The correct sequence of reagents for the above conversion of X to Y is :

A. $(i) NaOH (aq) (ii) Jones reagent (iii) H₃O⁺$
B. $(i) B₂H₆ / H₂O₂ (ii) NaOEt (iii) Jones reagent$
C. $(i) Jones reagent (ii) NaOEt (iii) Hot KMnO₄ /KOH$
D. $(i) NaOEt (ii) B₂H₆/H₂O₂ (iii) Jones reagent$

Solution

Core Logic

The transformation requires converting a cyclic bromoalkane into a conjugated cyclic carboxylic acid. The steps required are elimination to form an alkene, anti-Markovnikov hydration to an alcohol, and finally oxidation.

Step 1: Elimination

Treating the starting material with NaOEt (a strong base) promotes an E2 elimination, forming a conjugated diene in the cyclopentane ring.

Step 2: Hydroboration-Oxidation

Using $B₂H₆ / H₂O₂$ $B_2H_6 / H_2O_2$ (Hydroboration-oxidation) adds water across the less hindered double bond in an anti-Markovnikov fashion, placing an -OH group on the terminal carbon attached to the ring.

Step 3: Strong Oxidation

Finally, utilizing Jones reagent ( $CrO₃ / H^+$ $CrO_3 / H^+$) strongly oxidizes the primary alcohol directly into a carboxylic acid ( $-COOH$ $-COOH$), yielding the desired product Y.

Pattern Recognition

Formation of a new double bond alongside chain functionalization generally implies Base-catalyzed E2 followed by functional group addition (like hydroboration) and terminal oxidation.

Chapter Mix

Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Alcohols Phenols and Ethers

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

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Step 2: Hydroboration-Oxidation

Step 3: Strong Oxidation

Pattern Recognition

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