Aldehydes, Ketones and Carboxylic Acids Previous Year Questions — JEE Main Chemistry
46 past-year Aldehydes, Ketones and Carboxylic Acids questions from JEE Main (Chemistry).
Q409 (2025)
Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product?
(A) $\mathrm{R - C \equiv N} \xrightarrow[\text{mild condition}]{\mathrm{(i) H^+ / H_2O}}$
(B) $\mathrm{R - MgX} \xrightarrow[\mathrm{(ii) H_3O^+}]{{\mathrm{(i) CO_2}}}$
(C) $\mathrm{R - C \equiv N} \xrightarrow[\mathrm{(ii) H_3O^+}]{{\mathrm{(i) SnCl_2 / HCl}}}$
(D) $\mathrm{R \cdot CH_2 \cdot OH} \xrightarrow{\mathrm{PCC}}$
(E){{MAIN_IMG}}
Choose the correct answer from the options given below:
- $\text{A and D only}$
- $\text{A, B and E only}$
- $\text{B, C and E only}$
- $\text{B and E only}$
### Related Formula
$$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$$
### Core Logic
Let's analyze each reaction path to determine the major organic product:
- **Reaction (A)**: Acidic hydrolysis of a nitrile under *mild conditions* yields an amide:
$$\mathrm{R-C\equiv N \rightarrow R-CONH_2}$$
(Full conversion to carboxylic acid requires strong conditions and extended heating).
- **Reaction (B)**: Carbonation of Grignard reagent using solid carbon dioxide (dry ice) followed by acid hydrolysis yields a carboxylic acid:
$$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$$
- **Reaction (C)**: Stephen reduction converts nitrile to aldehyde:
$$\mathrm{R-C\equiv N \xrightarrow{SnCl_2/HCl} R-CH=NH \xrightarrow{H_3O^+} R-CHO}$$
- **Reaction (D)**: Pyridinium chlorochromate (PCC) is a mild oxidising agent that converts primary alcohols selectively to aldehydes:
$$\mathrm{R-CH_2-OH \xrightarrow{PCC} R-CHO}$$
- **Reaction (E)**
{{SOL_IMG1}}<BR>: Rosenmund reduction reduces acid chloride to aldehyde first:
$\rightarrow$ R-CHO
Subsequent oxidation with bromine water (which is a mild oxidising agent that selective oxidizes aldehydes but does not affect ketones) converts the aldehyde to carboxylic acid:
$$\mathrm{R-CHO \xrightarrow{Br_2/water} R-COOH}$$
### Step 1: Final Tally
Thus, reactions (B) and (E) successfully yield carboxylic acid as the major organic product.
### Pattern Recognition
Remember: Bromine water ($\mathrm{Br_2/H_2O}$) is a mild, selective oxidising agent commonly used to oxidise aldoses and other aldehydes to monocarboxylic acids without degrading carbon-carbon chains.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q461 (2025)
Given below are two statements :
Statement (I): Vanillin {{IMG1}} will react with NaOH and also with Tollen's reagent.
Statement (II) : Vanillin {{IMG2}} will undergo self aldol condensation very easily.
In the light of the above statements, choose the most appropriate answer from the options given below:
- $(1)\ \text{Statement I is incorrect but Statement II is correct}$
- $(2)\ \text{Statement I is correct but Statement II is incorrect}$
- $(3)\ \text{Both Statement I and Statement II are incorrect}$
- $(4)\ \text{Both Statement I and Statement II are correct}$
### Related Formula
Phenolic protons react with standard strong bases:
$$\mathrm{Ar-OH + NaOH \rightarrow Ar-ONa + H_2O}$$
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 $-\mathrm{OH}$ group allows acid-base reaction with $\mathrm{NaOH}$ directly {{SOL_IMG1}}. 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
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q182 (2025)
The major product (P) in the following reaction is :
{{IMG1}}
### Related Formula
Intramolecular Cannizzaro-type reaction or Benzil-Benzilic acid rearrangement involves nucleophilic attack of hydroxide at a carbonyl group, followed by hydride transfer to the adjacent carbonyl carbon.
### Core Logic
Let's analyze the starting compound, phenylglyoxal:
$$\mathrm{Ph-CO-CHO}$$
1. The aldehyde carbon ($-CHO$) is much more electrophilic than the ketone carbon ($-CO-$) due to less steric hindrance and absence of phenyl group electron donation.
2. Hydroxide ion ($\mathrm{OH}^-$) selectively attacks the aldehyde carbonyl carbon, forming a tetrahedral intermediate.
{{SOL_IMG1}}
### Step 1: Hydride Transfer Mechanism
The tetrahedral intermediate collapses, prompting an intramolecular hydride ($H^-$) transfer to the adjacent ketone carbonyl carbon:
$$\mathrm{Ph-CO-C(O^-)(OH)H} \rightarrow \mathrm{Ph-C(O^-)H-COOH}$$
This is the rate-determining step.
### Step 2: Proton Transfer to form Product
Rapid proton transfer occurs from the carboxylic acid group to the alkoxide oxygen, yielding the stable carboxylate salt:
$$\mathrm{Ph-CH(OH)-COO^- K^+}$$
This corresponds to Option (2).
### Pattern Recognition
In asymmetrical 1,2-dicarbonyl systems with an aldehyde and a ketone, nucleophilic addition occurs preferentially at the more reactive aldehyde carbon. The hydrogen is then transferred as a hydride to the ketone carbon, yielding an $\alpha$-hydroxy carboxylate salt.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q593 (2025)
When the dicarbonyl compound shown below undergoes an base-catalyzed intramolecular aldol condensation reaction, the major structural product formed is:
{{Q_IMG1}}
### Core Logic
Intramolecular aldol condensations are governed heavily by thermodynamic stability, favoring the formation of **5-membered** or **6-membered** rings over strained 3-, 4-, or large 7-membered options.
1. **Enolate Generation**: Base abstracts an $\alpha$-proton to form a nucleophilic carbanion enolate.
2. **Attack Vector Selection**: Deprotonation at the outer methyl site enables a ring-closing attack on the distant carbonyl carbon, perfectly designing a highly stable cyclopentene ring skeleton.
3. **Dehydration**: Heating drives the loss of a water molecule ($-\text{H}_2\text{O}$), introducing an $\alpha,\beta$-unsaturated carbonyl arrangement that provides stabilization via conjugated resonance. {{SOL_IMG1}}
### Pattern Recognition
Count the intervening carbon chain carefully. Intramolecular aldol pathways will always selectively build 5- or 6-membered rings due to favorable ring strain kinetics. Eliminating choices based on incorrect ring sizes isolates Option (1) instantly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q30 (2025)
Identify the major product 'P' in the given reaction sequence starting from 1,2-dibromocyclooctane:
$$\text{1,2-dibromocyclooctane} \xrightarrow{\text{(i) KOH (alc.)}} \xrightarrow{\text{(ii) NaNH}_2} \xrightarrow{\text{(iii) Hg}^{2+}/H^+} \xrightarrow{\text{(iv) Zn-Hg/HCl}} \text{'P (Major product)'}$$
- $\text{Cyclooctene structure } {{OPT_IMG_A}}$
- $\text{Cyclooctane structure } {{OPT_IMG_B}}$
- $\text{Cyclooctanone structure } {{OPT_IMG_C}}$
- $\text{Cyclooctyne intermediate derivative } {{OPT_IMG_D}}$
### Core Logic
Let us systematically follow the transformation steps:
1. **First Elimination**: 1,2-dibromocyclooctane reacts with alcoholic $\text{KOH}$ to remove one molecule of $\text{HBr}$, resulting in a bromocyclooctene intermediate.
2. **Second Elimination**: Treatment with the stronger base $\text{NaNH}_2$ removes the second molecule of $\text{HBr}$, forming an alkyne inside the 8-membered ring: **cyclooctyne**.
3. **Kucherov Reaction**: Hydration of cyclooctyne using $\text{Hg}^{2+}/H^+$ creates an enol intermediate that undergoes tautomerization to form a stable ketone: **cyclooctanone**.
4. **Clemmensen Reduction**: Subjecting cyclooctanone to zinc amalgam and hydrochloric acid ($\text{Zn-Hg/HCl}$) completely reduces the carbonyl group ($>C=O$) to a methylene group ($-\text{CH}_2-$), finishing with **cyclooctane**. {{SOL_IMG1}}
### Pattern Recognition
A vicinal dihalide treated with sequential strong bases creates an alkyne path. Alkyne hydration creates a ketone body. Finally, Clemmensen reduction takes the ketone down to a simple hydrocarbon skeleton. Recognizing this terminal reduction loop establishes cyclooctane as the undisputed answer.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
