Aldehydes, Ketones and Carboxylic Acids appeared 46 times across 3 years — 5.3% of Chemistry.
This question is from Oxidation of Alcohols and Alkene Reactions.
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⁺$\text{(i) NaOH (aq) (ii) Jones reagent (iii) }H_{3}O^{+}$
B.(i) B₂H₆ / H₂O₂ (ii) NaOEt (iii) Jones reagent$\text{(i) }B_{2}H_{6} / H_{2}O_{2}\text{ (ii) NaOEt (iii) Jones reagent}$
C.(i) Jones reagent (ii) NaOEt (iii) Hot KMnO₄ /KOH$\text{(i) Jones reagent (ii) NaOEt (iii) Hot }KMnO_{4}\text{ /KOH}$
D.(i) NaOEt (ii) B₂H₆/H₂O₂ (iii) Jones reagent$\text{(i) NaOEt (ii) }B_{2}H_{6}/H_{2}O_{2}\text{ (iii) Jones reagent}$
Solution & Explanation
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.
The image shows an organic conversion from a bromo-substituted cyclopentene to a carboxylic acid derivative.
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.
The image shows an organic conversion from a bromo-substituted cyclopentene to a carboxylic acid derivative.
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
Keywords:#sequence of reagents#JEE Main 2026 Morning Q54#Aldehydes Ketones and Carboxylic Acids JEE Main 2026#Oxidation of Alcohols and Alkene Reactions JEE Main 2026#chemical reaction#organic conversion#cyclopentene derivative#carboxylic acid synthesis
More Aldehydes Ketones and Carboxylic Acids Previous-Year Questions — Page 9
Qjee_main_2024_30_jan_morningPreparation of Aldehydes
This reduction reaction is known as:
The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.
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$BaSO_4$).
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
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$CrO_2Cl_2$.
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
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
Qjee_main_2024_31_jan_eveningPreparation of Aldehydes and Ketones
Identify the name reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
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.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
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
More Aldehydes Ketones and Carboxylic Acids Questions — jee_main_2026_23_january_morning
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