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 2.
Q48 (2025)
In the Claisen-Schmidt reaction to prepare dibenzalacetone from $5.3\text{ g}$ of benzaldehyde, a total of $3.51\text{ g}$ of product was obtained.
The percentage yield in this reaction was __%.
### Core Logic
The balanced chemical equation for the synthesis of dibenzalacetone is:
$$2 Ph-CHO + CH{3}COCH{3} xrightarrow{\text{Strong Base}} Ph-CH=CH-CO-CH=CH-Ph + 2H{2}O$$
Let's calculate the moles of reactants:
$$\text{Molar mass of benzaldehyde } (PhCHO) = 106\text{ g/mol}$$
$$\text{Moles of benzaldehyde used} = \frac{5.3}{106} = 0.05\text{ mol} = \frac{1}{20}\text{ mol}$$
{{SOL_IMG48}}
According to the reaction stoichiometry, $2\text{ moles of } PhCHO$ yield $1\text{ mole of dibenzalacetone}$.
$$\text{Theoretical moles of product} = \frac{0.05}{2} = 0.025\text{ mol}$$
### Step 1: Yield Evaluation
$$$\text{Molar mass of dibenzalacetone } (C{17}H{14}O) = 234\text{ g/mol}$$
$$\text{Theoretical mass} = 0.025 \times 234 = 5.85\text{ g}$$
$$\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{3.51}{5.85} \times 100 = 60%$$
### Pattern Recognition
Always remember the stoichiometric ratio: It takes 2 moles of benzaldehyde to condense with 1 mole of acetone to form the symmetrical dibenzalacetone product.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q708 (2025)
A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q") ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below:
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The structure of $(\mathrm{P})$ is
### Core Logic
The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below:
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### Pattern Recognition
Sees: Acidic rearrangement $\rightarrow$ ozonolysis $\rightarrow$ intramolecular aldol condensation.
Shortcut: Work backwards from the dicarbonyl fragments formed after opening the final product ring system.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q37 (2025)
Both acetaldehyde and acetone (individually) undergo which of the following reactions?
A. Iodoform Reaction
B. Cannizaro Reaction
C. Aldol condensation
D. Pollen's Test
E. Clemmensen Reduction
Choose the correct answer from the options given below:
- $\text{A, B and D only}$
- $\text{A, C and E only}$
- $\text{C and E only}$
- $\text{B, C and D only}$
### Core Logic
Let us check each option pathway:
- **A. Iodoform Reaction:** Positive for both because both contain the $\mathrm{CH}_3-\mathrm{C}=\mathrm{O}$ methyl ketone fragment.
- **B. Cannizaro Reaction:** Negative for both because both contain $\alpha$-hydrogens.
- **C. Aldol Condensation:** Positive for both because they have $\alpha$-hydrogens available for enolization.
- **D. Pollen's Test (Tollen's Test):** Positive only for acetaldehyde (aldehyde); negative for acetone (ketone).
- **E. Clemmensen Reduction:** Positive for both as they contain reducible carbonyl groups.
Thus, both react via A, C, and E.
### Pattern Recognition
Sees: Functional comparison of Acetaldehyde and Acetone.
Shortcut: Ketones do not respond to Tollen's test, which instantly eliminates choices featuring statement D.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q36 (2025)
Number of molecules from below which cannot give iodoform reaction is :
Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol
- 5
- 4
- 3
- 2
### Related Formula
Positive iodoform test requires presence of:
- Methyl ketone: $\mathrm{CH_3-C(=O)-R}$
- Methyl carbinol: $\mathrm{CH_3-CH(OH)-R}$
### Core Logic
Evaluate each compound:
1. Ethanol ($\mathrm{CH_3CH_2OH}$) $\rightarrow$ Gives test (Gives $\mathrm{CHI_3}$)
2. Isopropyl alcohol ($\mathrm{CH_3CH(OH)CH_3}$) $\rightarrow$ Gives test
3. Bromoacetone ($\mathrm{BrCH_2COCH_3}$) $\rightarrow$ Gives test
4. 2-Butanol ($\mathrm{CH_3CH(OH)CH_2CH_3}$) $\rightarrow$ Gives test
5. 2-Butanone ($\mathrm{CH_3COCH_2CH_3}$) $\rightarrow$ Gives test
6. Butanal ($\mathrm{CH_3CH_2CH_2CHO}$) $\rightarrow$ **Does NOT give test**
7. 2-Pentanone ($\mathrm{CH_3COCH_2CH_2CH_3}$) $\rightarrow$ Gives test
8. 3-Pentanone ($\mathrm{CH_3CH_2COCH_2CH_3}$) $\rightarrow$ **Does NOT give test**
9. Pentanal ($\mathrm{CH_3CH_2CH_2CH_2CHO}$) $\rightarrow$ **Does NOT give test**
10. 3-Pentanol ($\mathrm{CH_3CH_2CH(OH)CH_2CH_3}$) $\rightarrow$ **Does NOT give test**
### Step 1: Final Count
The 4 compounds that cannot give the iodoform reaction are: Butanal, 3-Pentanone, Pentanal, and 3-Pentanol.
### Pattern Recognition
Requires $\mathrm{CH_3-CO-}$ or $\mathrm{CH_3-CH(OH)-}$ group.
Aldehydes other than acetaldehyde do not give iodoform test.
Symmetrical 3-ketones and 3-alcohols fail the test.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q853 (2025)
Which among the following compounds give yellow solid when reacted with NaOI/NaOH?
(A) $CH_3 - CH(OH) - C_2H_5$
(B) $CH_3 - CH_2 - CH_2 - OH$
(C) $CH_3 - CO - C_2H_5$
(D) $CH_3-CO- OH$
(E) $CH_3 - CH_2 - CHO$
Choose the correct answer from the options given below:
- (B), (C) and (E) Only
- (A) and (C) Only
- (C) and (D) Only
- (A), (C) and (D) Only
### Related Formula
$$\text{Compounds with } CH_3-CH(OH)- \text{ or } CH_3-CO- \text{ groups undergo the iodoform reaction to form } CHI_3 \downarrow \text{ (Yellow Solid)}$$
### Core Logic
Let's check the structural groups of each given option:
- **(A)** $CH_3 - CH(OH) - C_2H_5$: Contains the methylcarbinol group ($CH_3-CH(OH)-$). Gives a positive iodoform test.
- **(B)** $CH_3 - CH_2 - CH_2 - OH$: Linear primary alcohol, does not contain the required group.
- **(C)** $CH_3 - CO - C_2H_5$: Contains the methyl ketone group ($CH_3-CO-$). Gives a positive iodoform test.
- **(D)** $CH_3 - OH$: Methanol does not give the test.
- **(E)** $CH_3 - CH_2 - H$: Ethane does not give the test.
Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform ($CHI_3$).
### Step 1: Chemical Equations
The balanced haloform pathways occur as follows:
$$CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$$
$$CH_3-CO-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$$
### Pattern Recognition
The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal $-CH_3$ affixed directly to a carbonyl oxygen index ($C=O$) or a hydroxyl carbon ($CH-OH$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
