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 8.
Q68 (2026)
The compound A, $\text{C}_8\text{H}_8\text{O}_2$ reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as
- 2-hydroxy acetophenone
- 4-methoxy benzaldehyde
- 4-hydroxy benzaldehyde
- 4-methyl benzoic acid
### Related Formula
$$\text{Ar-CHO} + \text{conc. NaOH} \xrightarrow{\text{Cannizzaro}} \text{Ar-CH}_2\text{OH} + \text{Ar-COO}^-$$
### Core Logic
Step 1: Compound A has molecular formula $\text{C}_8\text{H}_8\text{O}_2$ and undergoes Cannizzaro reaction with conc. NaOH $\implies$ A lacks $\alpha$-hydrogen and is an aromatic aldehyde.
Step 2: Compound A is 4-methoxy benzaldehyde ($\text{CH}_3\text{O-C}_6\text{H}_4\text{-CHO}$).
Step 3: Cross-Aldol condensation with acetophenone yields a single aldol condensation product $(B)$ structure.
Step 4: Cannizzaro reaction with conc. NaOH produces 4-methoxybenzyl alcohol and 4-methoxybenzoate anion.
{{SOL_IMG1}}
{{SOL_IMG2}}
### Pattern Recognition
Sees: $\text{C}_8\text{H}_8\text{O}_2$ undergoing Cannizzaro to form substituted benzyl alcohol.
Shortcut: Presence of $-\text{CHO}$ without $\alpha$-H and $-\text{OCH}_3$ ring substituent points directly to 4-methoxy benzaldehyde.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q54 (2026)
{{IMG1}}
The correct sequence of reagents for the above conversion of X to Y is :
- $\text{(i) NaOH (aq) (ii) Jones reagent (iii) }H_{3}O^{+}$
- $\text{(i) }B_{2}H_{6} / H_{2}O_{2}\text{ (ii) NaOEt (iii) Jones reagent}$
- $\text{(i) Jones reagent (ii) NaOEt (iii) Hot }KMnO_{4}\text{ /KOH}$
- $\text{(i) NaOEt (ii) }B_{2}H_{6}/H_{2}O_{2}\text{ (iii) Jones reagent}$
### Core Logic
The transformation requires converting a cyclic bromoalkane into a conjugated cyclic carboxylic acid. The steps required are elimination to form an alkene, anti-Markovnikov hydration to an alcohol, and finally oxidation.
### Step 1: Elimination
Treating the starting material with NaOEt (a strong base) promotes an E2 elimination, forming a conjugated diene in the cyclopentane ring.
{{SOL_IMG1}}
### Step 2: Hydroboration-Oxidation
Using $B_2H_6 / H_2O_2$ (Hydroboration-oxidation) adds water across the less hindered double bond in an anti-Markovnikov fashion, placing an -OH group on the terminal carbon attached to the ring.
### Step 3: Strong Oxidation
Finally, utilizing Jones reagent ($CrO_3 / H^+$) strongly oxidizes the primary alcohol directly into a carboxylic acid ($-COOH$), yielding the desired product Y.
{{SOL_IMG2}}
### Pattern Recognition
Formation of a new double bond alongside chain functionalization generally implies Base-catalyzed E2 followed by functional group addition (like hydroboration) and terminal oxidation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols Phenols and Ethers
Q56 (2026)
‘x’ is the product which is obtained from propanenitrile and stannous chloride in the presence of hydrochloric acid followed by hydrolysis. ‘y’ is the product which is obtained from the but-2-ene by the ozonolysis followed by hydrolysis. From the following, which product is not obtained when one mole of ‘x’ and one mole of ‘y’ react with each other in the presence of alkali followed by heating?
- $\text{2-Methylbut-2-enal}$
- $\text{Pent-2-enal}$
- $\text{2-Methylpent-2-enal}$
- $\text{3-Methylbut-2-enal}$
### Core Logic
First, identify products 'x' and 'y' based on the given name reactions.
Reaction 1: Stephen's Reduction
Propanenitrile ($CH_3-CH_2-C\equiv N$) reacts with $SnCl_2 / HCl$ followed by $H_3O^+$ to give propanal.
$CH_3-CH_2-CHO$ is product (X).
Reaction 2: Reductive Ozonolysis
But-2-ene ($CH_3-CH=CH-CH_3$) undergoes ozonolysis to give two moles of ethanal.
$CH_3-CHO$ is product (Y).
### Step 1: Cross-Aldol Condensation
Reacting (X) and (Y) in alkali with heat leads to a cross-aldol condensation producing a mixture of four $\alpha,\beta$-unsaturated aldehydes (two self-aldol and two cross-aldol products).
Reactants: $CH_3-CH_2-CHO$ and $CH_3-CHO$
### Step 2: Finding Products
Possible aldol condensation products:
1. Ethanal + Ethanal (Self) $\rightarrow$ But-2-enal
2. Propanal + Propanal (Self) $\rightarrow$ 2-Methylpent-2-enal
3. Ethanal (Enolate) + Propanal (Electrophile) $\rightarrow$ Pent-2-enal
4. Propanal (Enolate) + Ethanal (Electrophile) $\rightarrow$ 2-Methylbut-2-enal
Comparing with the given options, 3-Methylbut-2-enal is NOT formed.
### Pattern Recognition
To quickly identify cross-aldol products, simply remove two $\alpha$-hydrogens from one molecule and the carbonyl oxygen from the other, then connect them with a double bond.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
Q58 (2026)
{{IMG1}}
Iodoform test can differentiate between
A. Methanol and Ethanol
B. $CH_{3}COOH$ and $CH_{3}CH_{2}COOH$
C. Cyclohexene and cyclohexanone
D. Diethyl ether and Pentan-3-one
E. Anisole and acetone
Choose the correct answer from the options given below:
- $\text{A & E only}$
- $\text{A & D only}$
- $\text{A, B & E only}$
- $\text{B, C & E only}$
### Related Formula
$$R-CO-CH_3 \xrightarrow{NaOH + I_2} R-COONa + CHI_3\downarrow \text{ (Yellow ppt)}$$
$$R-CH(OH)-CH_3 \xrightarrow{NaOH + I_2} R-COONa + CHI_3\downarrow \text{ (Yellow ppt)}$$
### Core Logic
The iodoform test gives a positive result for compounds containing a methyl ketone group ($-CO-CH_3$) or a secondary alcohol with an adjacent methyl group ($-CH(OH)-CH_3$), which gets oxidized to a methyl ketone under the reaction conditions.
A. Methanol vs Ethanol: Ethanol ($CH_3CH_2OH$) contains the $-CH(OH)-CH_3$ terminal structure and gives a positive test. Methanol ($CH_3OH$) does not. (Can differentiate)
B. Neither $CH_3COOH$ nor $CH_3CH_2COOH$ gives the iodoform test since carboxylic acids do not undergo this reaction despite having a terminal methyl in acetic acid. (Cannot differentiate)
C. Neither cyclohexene nor cyclohexanone has a methyl group adjacent to the carbonyl/alcohol carbon. (Cannot differentiate)
D. Neither diethyl ether ($CH_3CH_2-O-CH_2CH_3$) nor pentan-3-one ($CH_3CH_2-CO-CH_2CH_3$) has a methyl ketone group. (Cannot differentiate)
E. Acetone ($CH_3-CO-CH_3$) gives a positive test. Anisole ($Ph-O-CH_3$) does not. (Can differentiate)
### Step 1: Final Selection
Only pairs A and E contain exactly one compound that gives a positive iodoform test, allowing for chemical differentiation.
### Pattern Recognition
Look strictly for $CH_3-C=O$ or $CH_3-CH-OH$. If a molecule lacks these exact terminal fragments, it will fail the iodoform test.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q62 (2026)
Which of the following statements are TRUE about Haloform reaction?
A. Sodium hypochlorite reacts with KI to give KOI.
B. KOI is a reducing agent.
C. $\alpha, \beta$-unsaturated methylketone $(CH_{3}-CH=CH-CO-CH_{3})$ will give iodoform reaction.
D. Isopropyl alcohol will not give iodoform test.
E. Methanoic acid will give positive iodoform test.
Choose the correct answer from the options given below:
- $\text{A, C & E only}$
- $\text{A, B & C only}$
- $\text{A & C only}$
- $\text{B, D & E only}$
### Core Logic
Evaluate each statement:
A. $NaOCl$ (Sodium hypochlorite) reacts with $KI$ to produce $KOI$ and $NaCl$. This statement is True.
$$NaOCl + KI \rightarrow NaCl + KOI$$
B. $KOI$ acts as a mild oxidizing agent, not a reducing agent. It is used to oxidize secondary alcohols to ketones in the iodoform reaction. This statement is False.
C. The molecule $CH_3-CH=CH-CO-CH_3$ has a terminal methyl ketone group ($-CO-CH_3$). The adjacent double bond does not interfere with the cleavage of the methyl group during the haloform test. It will yield yellow iodoform. This statement is True.
D. Isopropyl alcohol is $CH_3-CH(OH)-CH_3$. Since it is a secondary alcohol with an adjacent methyl group, it gets oxidized to acetone in situ and therefore gives a positive iodoform test. The statement says it will *not* give the test, which is False.
E. Methanoic acid ($HCOOH$) does not have a terminal methyl group attached to a carbonyl carbon, so it cannot undergo the haloform reaction. This statement is False.
### Step 1: Final Selection
Only statements A and C are True.
### Pattern Recognition
Alcohols with the $CH_3-CH(OH)-$ moiety always test positive for iodoform. Reagents containing hypohalites ($OCl^-$, $OI^-$) are strong oxidizing agents, frequently used in organic synthesis precisely for their oxidative power.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
