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 5.
Q1325 (2025)
Based on the reaction sequence:
{{IMG1}}
0.1 mole of compound 'S' will weigh ________ g.
(Given molar mass in $\mathrm{g \, mol^{-1}}$ C:12, H:1, O:16)
### Related Formula
$$\text{Mass} = \text{Moles} \cdot \text{Molar Mass}$$
### Core Logic
Let us trace the reaction scheme row-by-row:
{{SOL_IMG1}}
1. Reactant: 2-(hydroxymethyl)cyclopentan-1-ol.
2. Step 1: Excess $\mathrm{CrO}_3$ (Jones oxidation) oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid $\rightarrow$ Product P.
3. Step 2: Reaction with 1 mole of glycol protects the ketone selectively as a cyclic ketal $\rightarrow$ Product Q.
4. Step 3: Treatment with $\mathrm{CH}_3\mathrm{MgI}$ targets the free carboxylic acid (or ester equivalent) to form a methyl ketone after workup $\rightarrow$ Product R.
5. Step 4: $\mathrm{NaBH}_4$ reduces the methyl ketone to a secondary alcohol, and acid workup deprotects the ketal back to the original ketone $\rightarrow$ Compound S.
Chemical structural formula of S: 2-(1-hydroxyethyl)cyclopentan-1-one ($\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$).
Molar mass of S ($\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$):
$$M = (7 \cdot 12) + (12 \cdot 1) + (2 \cdot 16) = 84 + 12 + 32 = 130 \mathrm{~g/mol}$$
$$\text{Mass of 0.1 mole} = 0.1 \cdot 130 = 13 \mathrm{~g}$$
### Pattern Recognition
Ketal groups serve as stable protective masks that shield ketones, allowing selective organometallic reactions to take place elsewhere on the molecule.
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q1389 (2024)
Identify A and B in the following sequence of reaction
{{IMG1}}
### Core Logic
Step 1: Free radical side-chain halogenation of toluene.
Toluene reacts with $Cl_2$ in the presence of light ($h\nu$) to undergo substitution on the methyl group. Under typical conditions intended to yield an aldehyde later, di-chlorination occurs forming benzal chloride (A).
Step 2: Hydrolysis.
Benzal chloride upon hydrolysis with $H_2O$ at 373 K yields a gem-diol intermediate which is unstable and loses water to form Benzaldehyde (B).
$$C_6H_5CH_3 \xrightarrow{Cl_2 / h\nu} C_6H_5CHCl_2 \xrightarrow{H_2O, 373K} C_6H_5CHO$$
### Step 1: Identify Structures
(A) = Benzal chloride ($C_6H_5CHCl_2$)
(B) = Benzaldehyde ($C_6H_5CHO$)
{{SOL_IMG1}}
### Pattern Recognition
Toluene $\xrightarrow{Cl_2, h\nu}$ targets the side chain. If the next step is hydrolysis to an aldehyde, you must have stopped at the gem-dihalide stage ($CHCl_2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
Q1398 (2024)
Match List - I with List - II.
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List - I (Reactions)</th><th style="border: 1px solid #888; padding: 8px;">List - II (Reagents)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) $CH_3(CH_2)_5-CO-OC_2H_5 \rightarrow CH_3(CH_2)_5CHO$</td><td style="border: 1px solid #888; padding: 8px;">(I) $CH_3MgBr, H_2O$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) $C_6H_5COC_6H_5 \rightarrow C_6H_5CH_2C_6H_5$</td><td style="border: 1px solid #888; padding: 8px;">(II) $Zn(Hg)$ and conc. $HCl$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) $C_6H_5CHO \rightarrow C_6H_5CH(OH)CH_3$</td><td style="border: 1px solid #888; padding: 8px;">(III) $NaBH_4, H^+$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) $CH_3COCH_2COOC_2H_5 \rightarrow CH_3CH(OH)CH_2COOC_2H_5$</td><td style="border: 1px solid #888; padding: 8px;">(IV) DIBAL-H, $H_2O$</td></tr></tbody></table></div>
Choose the correct answer from options given below:
- $\text{A-(III), (B)-(IV), (C)-(I), (D)-(II)}$
- $\text{A-(IV), (B)-(II), (C)-(I), (D)-(III)}$
- $\text{A-(IV), (B)-(II), (C)-(III), (D)-(I)}$
- $\text{A-(III), (B)-(IV), (C)-(II), (D)-(I)}$
### Core Logic
Let's analyze the transformation happening in each reaction:
(A) $CH_3(CH_2)_5COOC_2H_5 \rightarrow CH_3(CH_2)_5CHO$
An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) $\rightarrow$ (IV).
(B) $C_6H_5COC_6H_5 \rightarrow C_6H_5CH_2C_6H_5$
A ketone (carbonyl group $>C=O$) is fully reduced to an alkane ($>CH_2$) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) $\rightarrow$ (II).
(C) $C_6H_5CHO \rightarrow C_6H_5CH(OH)CH_3$
Benzaldehyde (aldehyde) is converted into a secondary alcohol with an extra methyl group. This is a nucleophilic addition of a Grignard reagent ($CH_3MgBr$) followed by hydrolysis. Thus, (C) $\rightarrow$ (I).
(D) $CH_3COCH_2COOC_2H_5 \rightarrow CH_3CH(OH)CH_2COOC_2H_5$
A ketone group is reduced to a secondary alcohol while the ester group remains intact. $NaBH_4$ is a mild reducing agent that reduces aldehydes and ketones but generally does not touch esters. Thus, (D) $\rightarrow$ (III).
### Pattern Recognition
Ester $\rightarrow$ Aldehyde = DIBAL-H
Ketone $\rightarrow$ Alkane = Clemmensen ($Zn(Hg)/HCl$) or Wolff-Kishner
Carbonyl $\rightarrow$ Alcohol with carbon chain extension = Grignard Reagent
Ketone $\rightarrow$ Alcohol (leaving ester intact) = $NaBH_4$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q1667 (2024)
The final product A formed in the following multistep reaction sequence is
{{IMG1}}
### Core Logic
The reaction sequence proceeds in three distinct steps from the starting material, styrene ($Ph-CH=CH_2$).
**Step 1: Acid-catalyzed Hydration**
Styrene reacts with $H_2O, H^+$ to undergo electrophilic addition. Protonation yields the more stable secondary benzylic carbocation. Attack by water followed by deprotonation gives 1-phenylethanol ($Ph-CH(OH)-CH_3$).
**Step 2: Oxidation**
1-phenylethanol is a secondary alcohol. Treatment with chromium trioxide ($CrO_3$, Jones reagent condition) oxidizes the secondary alcohol to a ketone. This yields acetophenone ($Ph-CO-CH_3$).
**Step 3: Wolff-Kishner Reduction**
Acetophenone is treated with hydrazine ($NH_2-NH_2$) and a strong base ($KOH$) under heating. This is the classic Wolff-Kishner reduction, which completely reduces the carbonyl group ($C=O$) to a methylene group ($-CH_2-$). The final product is ethylbenzene ($Ph-CH_2-CH_3$).
### Step 1: Overall Reaction Pathway
{{SOL_IMG1}}
The final product A is ethylbenzene.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols Phenols and Ethers
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q1679 (2024)
From the compounds given below, number of compounds which give positive Fehling's test is _____.
Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4-nitrobenzaldehyde, cyclohexane carbaldehyde.
### Core Logic
Fehling's test is a mild oxidizing test used primarily to distinguish aliphatic aldehydes from ketones and aromatic aldehydes.
- **Aliphatic aldehydes** (like methanal, acetaldehyde, cyclohexane carbaldehyde) give a positive Fehling's test (formation of red-brown precipitate of $Cu_2O$).
- **Aromatic aldehydes** (like benzaldehyde, 4-nitrobenzaldehyde) lack alpha-hydrogens in a purely aliphatic environment and are not sufficiently easily oxidized to give a positive Fehling's test.
- **Ketones** (like acetone, acetophenone) generally do not give a positive Fehling's test (except $\alpha$-hydroxy ketones).
### Step 1: Evaluation of Given Compounds
1. Benzaldehyde: Aromatic aldehyde $\rightarrow$ Negative
2. Acetaldehyde ($CH_3CHO$): Aliphatic aldehyde $\rightarrow$ Positive
3. Acetone: Ketone $\rightarrow$ Negative
4. Acetophenone: Ketone $\rightarrow$ Negative
5. Methanal ($HCHO$): Aliphatic aldehyde $\rightarrow$ Positive
6. 4-nitrobenzaldehyde: Aromatic aldehyde $\rightarrow$ Negative
7. Cyclohexane carbaldehyde: Aliphatic aldehyde $\rightarrow$ Positive
The compounds giving a positive test are Acetaldehyde, Methanal, and Cyclohexane carbaldehyde.
### Pattern Recognition
Tollens' reagent oxidizes ALL aldehydes (aliphatic + aromatic). Fehling's reagent is weaker and only oxidizes ALIPHATIC aldehydes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
