Aldehydes, Ketones and Carboxylic Acids Previous Year Questions — JEE Main Chemistry
46 past-year Aldehydes, Ketones and Carboxylic Acids questions from JEE Main (Chemistry) — page 3.
Q929 (2025)
Aldol condensation is a popular and classical method to prepare $\alpha, \beta$-unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?
### Related Reaction
Intramolecular aldol condensation of dicarbonyl compounds:
$$\text{Dicarbonyl precursor} \xrightarrow[\Delta]{\text{dil. base}} \alpha,\beta\text{-unsaturated cyclic carbonyl} + \text{H}_2\text{O}$$
Intramolecular cyclization strongly favors the formation of stable 5- or 6-membered rings due to minimal ring strain.
### Core Logic
* **Options A, B, and C:** Each represents a clean intramolecular cyclization product derived from a single open-chain dicarbonyl precursor (dialdehyde or diketone) forming stable 5- or 6-membered conjugated enones.
{{SOL_IMG1}}
{{SOL_IMG2}}
* **Option D:** Features an exocyclic $\alpha,\beta$-unsaturated linkage formed strictly via an **intermolecular** crossed-aldol condensation between two separate carbonyl molecules rather than an internal cyclization of a single dicarbonyl unit.
{{SOL_IMG3}}
### Pattern Recognition
Trace the carbon backbone back to its precursor. Products of intramolecular aldol condensation originate from a single continuous dicarbonyl molecule closing into a stable 5- or 6-membered ring. An exocyclic enone linking a ring to an external carbonyl unit typically indicates an intermolecular reaction between two distinct molecules.
### Evaluation Rubric / Model Answer
Option D
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q41 (2025)
An organic compound (X) with molecular formula $C_3H_6O$ is not readily oxidised. On reduction it gives $C_3H_8O$ (Y) which reacts with $HBr$ to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis gives 2, 3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:
- $\mathrm{CH}_3\mathrm{COCH}_3, \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}, \mathrm{CH}_3\mathrm{CH(Br)CH}_3$
- $\mathrm{CH}_3\mathrm{COCH}_3, \mathrm{CH}_3\mathrm{CH(OH)CH}_3, \mathrm{CH}_3\mathrm{CH(Br)CH}_3$
- $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}, \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}, \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$
- $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}, \mathrm{CH}_3\mathrm{CH=CH}_2, \mathrm{CH}_3\mathrm{CH(Br)CH}_3$
### Core Logic
Let's deduce the identities stepwise:
1. Compound (X) has the formula $C_3H_6O$ and is resistant to mild oxidation, which identifies it as a ketone: **Acetone** ($CH_3COCH_3$).
2. Reduction of Acetone yields a secondary alcohol, Propan-2-ol ($CH_3CH(OH)CH_3$, Compound Y).
3. Treatment of Propan-2-ol with $HBr$ substitutes the hydroxyl group to form 2-Bromopropane ($CH_3CH(Br)CH_3$, Compound Z).
4. Reacting 2-Bromopropane with Magnesium in ether creates the branched Grignard reagent, Isopropylmagnesium bromide ($(CH_3)_2CHMgBr$).
5. Finally, nucleophilic addition of this Grignard reagent to Acetone followed by aqueous workup yields the highly branched tertiary alcohol: **2,3-dimethylbutan-2-ol**.
### Pattern Recognition
Resistance to mild oxidation immediately distinguishes ketones from isomeric aldehydes. Nucleophilic addition of an isopropyl Grignard to acetone cleanly yields the 2,3-dimethylbutan-2-ol framework.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q30 (2025)
"P" is an optically active compound with molecular formula $\text{C}_6\text{H}_{12}\text{O}$. When "P" is treated with 2,4-dinitrophenylhydrazine, it gives a positive test. However, in presence of Tollens reagent, "P" gives a negative test. Predict the structure of "P".
- $\text{CH}_3\text{-C}(=\text{O})\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_3$
- $\text{CH}_3\text{-C}(=\text{O})\text{-CH}(\text{CH}_2\text{-CH}_3)\text{-CH}_3$
- $\text{H-C}(=\text{O})\text{-CH}_2\text{-CH}(\text{CH}_2\text{-CH}_3)\text{-CH}_3$
- $\text{CH}_3\text{-C}(=\text{O})\text{-CH}_2\text{-CH}(\text{CH}_3)_2$
### Related Formula
$$\text{Carbonyl compound} + \text{2,4-DNP}
ightarrow \text{Hydrazone derivative (Positive test)} $$
$$\text{Aldehyde} + \text{Tollens' Reagent}
ightarrow \text{Silver Mirror (Positive test)}$$
$$\text{Ketone} + \text{Tollens' Reagent}
ightarrow \text{No reaction (Negative test)} $$
### Core Logic
Analyzing individual functional constraints:
- Positive 2,4-DNP test shows compound contains a carbonyl group (aldehyde or ketone).
- Negative Tollens' test clarifies it is not an aldehyde; hence it must be a ketone.
- The compound is optically active, meaning it must possess a chiral center (carbon with 4 distinct groups).
Let's evaluate the options via structural configurations:
{{SOL_IMG1}}
{{SOL_IMG2}}
### Step 1: Structural Verification
Option (2) represents 3-methylpentan-2-one:
$$\text{CH}3-\text{C}(=\text{O})-\overset{\ast}{\text{CH}}(\text{CH}3)(\text{CH}2\text{CH}3) $$
The third carbon (C3) is linked to: $-\text{H}$, $-\text{CH}_3$, $-\text{CH}_2\text{CH}_3$, and $-\text{COCH}_3$. It has 4 distinct structural fields, making it chiral and optically active.
### Pattern Recognition
Tollens' negative + DNP positive = Ketone. Once categorized as a ketone, look directly for the structure holding a carbon with four unique groups to secure the optical activity constraint.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q42 (2025)
Match List-I with List-II
<div class="rankbit-table-container"><table class="rankbit-table"><thead><tr><th>List-I</th><th>List-II (Name of Reaction)</th></tr></thead><tbody><tr><td>(A) $\mathrm{RCN} \xrightarrow[\text{(ii)}\mathrm{H}_{3}\mathrm{O}^{+}]{\text{(i)}\mathrm{SnCl}_{2}, \mathrm{HCl}} \mathrm{RCHO}$</td><td>(I) Etard reaction</td></tr><tr><td>(B) {{IMG1}}</td><td>(II) Gatterman-Koch reaction</td></tr><tr><td>(C) {{IMG2}}</td><td>(III) Rosenmund reduction</td></tr><tr><td>(D) {{IMG3}}</td><td>(IV) Stephen reaction</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- \text{(A)-(IV), (B)-(III), (C)-(I), (D)-(II)}
- \text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}
- \text{(A)-(I), (B)-(III), (C)-(II), (D)-(IV)}
- \text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}
### Core Logic
Let's match each aldehyde preparation method with its official named organic reaction:
* **(A) $\mathrm{RCN} \rightarrow \mathrm{RCHO}$ using $\mathrm{SnCl}_{2}/\mathrm{HCl}$ followed by hydrolysis:** This is the classic **Stephen reaction** $\rightarrow$ **(IV)**.
* **(B) Reducing an acyl chloride ($\mathrm{RCOCl}$) to an aldehyde using $\mathrm{H}_{2}$ over $\mathrm{Pd-BaSO}_{4}$:** This partial reduction is known as the **Rosenmund reduction** $\rightarrow$ **(III)**.
* **(C) Oxidizing toluene to benzaldehyde using chromyl chloride ($\mathrm{CrO}_{2}\mathrm{Cl}_{2}$) in $\mathrm{CS}_{2}$:** This selective oxidation method is the **Etard reaction** $\rightarrow$ **(I)**.
* **(D) Converting benzene to benzaldehyde using $\mathrm{CO}$ and $\mathrm{HCl}$ in the presence of anhydrous $\mathrm{AlCl}_{3}/\mathrm{CuCl}$:** This formylation process is the **Gatterman-Koch reaction** $\rightarrow$ **(II)**.
Combining these assignments yields the final sequence: (A)-(IV), (B)-(III), (C)-(I), (D)-(II).
### Step-by-Step Layout
The visual reaction components correspond directly to the official structural transformations:
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
### Pattern Recognition
Quick identification keys:
- Nitrile $\rightarrow$ Aldehyde = Stephen
- Acid Chloride $\rightarrow$ Aldehyde = Rosenmund
- Toluene $\rightarrow$ Chromyl Complex = Etard
- Benzene $\rightarrow$ Carbon Monoxide = Gatterman-Koch
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q1166 (2025)
Aman has been asked to synthesise the molecule ring with C—CH3 (x). He thought of preparing the molecule using an aldol condensation reaction. He found a few cyclic alkenes in his laboratory. He thought of performing ozonolysis reaction on alkene to produce a dicarbonyl compound followed by aldol reaction to prepare “x”. Predict the suitable alkene that can lead to the formation of “x”.
### Core Logic
Analyzing the retro-synthesis path step-by-step:
1. The objective compound is 1-acetylcyclopentene.
2. Performing reductive ozonolysis ($O_3, Zn/H_2O$) on 1-methylcyclohexene (Option A) symmetrically breaks the internal endocyclic double bond to form heptane-2,6-dione, a dicarbonyl system. {{SOL_IMG1}}
3. Adding a base intermediate trigger ($OH^{-}, \Delta$) drives an intramolecular aldol condensation: the methyl group carbanion at position 1 attacks the carbon 6 ketone site. This ring-closing event effectively drops water to synthesize the 5-membered cyclopentene core molecule attached to the acetyl unit. {{SOL_IMG2}}
### Pattern Recognition
Counting carbon coordinates is essential. Reductive cleavage transforms a 6-membered ring into an open heptane system, which easily self-condenses into a stable 5-membered ring attached to a methyl ketone side chain.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
