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 6.
Q1744 (2024)
m-chlorobenzaldehyde on treatment with $50\%$ KOH solution yields
### Core Logic
m-chlorobenzaldehyde lacks alpha-hydrogen atoms. Therefore, when treated with concentrated base like $50\%$ KOH, it undergoes a disproportionation redox reaction known as the Cannizzaro reaction.
Two molecules of the aldehyde react: one gets oxidized to the corresponding carboxylate ion (m-chlorobenzoate ion), and the other gets reduced to the corresponding alcohol (m-chlorobenzyl alcohol).
### Step 1: Reaction
The reaction proceeds as:
{{SOL_IMG1}}
### Pattern Recognition
No $\alpha$-hydrogen + Conc. Alkali ($50\%$ KOH/NaOH) = Cannizzaro (Oxidation to salt of carboxylic acid + Reduction to primary alcohol).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q1832 (2024)
This reduction reaction is known as:
{{IMG1}}
- $\text{Rosenmund reduction}$
- $\text{Wolff-Kishner reduction}$
- $\text{Stephen reduction}$
- $\text{Etard reduction}$
### Related Formula
$$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$$
### Core Logic
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 $BaSO_4$).
{{SOL_IMG1}}
This specific reaction is known as the Rosenmund reduction.
### Pattern Recognition
Acid Chloride + $H_2, Pd/BaSO_4$ $\rightarrow$ Aldehyde is strictly the Rosenmund reduction. The $BaSO_4$ poisons the catalyst to prevent over-reduction to an alcohol.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q1840 (2024)
Structure of 4-Methylpent-2-enal is
- $$H_2C=C(CH_3)-CH_2-C(=O)H$$
- $$CH_3-CH_2-C(CH_3)=CH-C(=O)H$$
- $$CH_3-CH_2-CH=C(CH_3)-C(=O)H$$
- $$CH_3-CH(CH_3)-CH=CH-C(=O)H$$
### Core Logic
Decode the IUPAC name: 4-Methylpent-2-enal
1. Word root: 'pent' $\rightarrow$ 5 carbon principal chain.
2. Primary suffix: '2-en' $\rightarrow$ Double bond starting at carbon 2.
3. Secondary suffix: 'al' $\rightarrow$ Aldehyde group (-CHO) at carbon 1.
4. Substituent: '4-Methyl' $\rightarrow$ A methyl group (-CH3) at carbon 4.
### Step 1: Drafting the structure
Numbering starts from the aldehyde carbon.
$$C^5 - C^4 - C^3 = C^2 - C^1(=O)H$$
Attach the methyl at $C^4$:
$$CH_3 - CH(CH_3) - CH = CH - CHO$$
This matches Option 4.
### 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
Q1843 (2024)
In the given reactions identify the reagent A and reagent B.
{{IMG1}}
- $\text{A}-CrO_3, \text{B}-CrO_3$
- $\text{A}-CrO_3, \text{B}-CrO_2Cl_2$
- $\text{A}-CrO_2Cl_2, \text{B}-CrO_2Cl_2$
- $\text{A}-CrO_2Cl_2, \text{B}-CrO_3$
### Core Logic
Pathway 1 (Upper): Toluene is treated with Reagent 'A' and acetic anhydride $(CH_3CO)_2O$ to form an intermediate (benzylidene diacetate), which on hydrolysis gives benzaldehyde. The reagent used here is Chromic oxide ($CrO_3$). Thus, A is $CrO_3$.
Pathway 2 (Lower): Toluene is treated with Reagent 'B' in $CS_2$ to form a chromium complex intermediate, which on hydrolysis yields benzaldehyde. This is the Etard reaction, and the reagent used is Chromyl chloride ($CrO_2Cl_2$). Thus, B is $CrO_2Cl_2$.
{{SOL_IMG1}}
### Step 1: Selection
Therefore, A = $CrO_3$ and B = $CrO_2Cl_2$.
### Pattern Recognition
Etard reaction always uses Chromyl chloride ($CrO_2Cl_2$). Oxidation of toluene with acetic anhydride uses Chromic acid ($CrO_3$). Both stop the oxidation at the aldehyde stage via intermediate formation.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q87 (2024)
The compound formed by the reaction of ethanal with semicarbazide contains ________ number of nitrogen atoms.
### Related Formula
$$CH_3-CHO + H_2N-NH-CO-NH_2 \rightarrow CH_3-CH=N-NH-CO-NH_2 + H_2O$$
### Core Logic
Ethanal ($CH_3CHO$) reacts with semicarbazide ($H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone.
### Step 1: Product Analysis
The product is Ethanal semicarbazone: $CH_3-CH=N-NH-CO-NH_2$.
Counting the nitrogen atoms in this structure:
1. The imine nitrogen ($=N-$)
2. The amine nitrogen ($-NH-$)
3. The amide nitrogen ($-NH_2$)
Total = 3 Nitrogen atoms.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
