An organic compound “P” of molecular formula C_6H_12O_3$C_{6}H_{12}O_{3}$ gives positive Iodoform test but negative Tollen’s test. When “P” is treated with dilute acid, it produces “Q”. “Q” gives positive Tollen’s test and also iodoform test. The structure of “P” is :
D.mathrmCH_3-CO-CO-CH_3 text (with acetal structure)$\mathrm{CH_3-CO-CO-CH_3} \text{ (with acetal structure)}$
Solution & Explanation
### Core Logic
Compound P (C_6H_12O_3$C_6H_{12}O_3$) gives a positive iodoform test, indicating it has a methyl ketone group (CH_3CO-$CH_3CO-$). It gives a negative Tollen's test, indicating no free aldehyde group.
On acidic hydrolysis, P yields Q. Compound Q gives both positive iodoform test and Tollen's test, meaning Q contains both a methyl ketone and an aldehyde group.
Looking at the options, if P is an acetal of an aldehyde, acidic hydrolysis will regenerate the aldehyde.
Option 2 is mathrmCH_3-CO-CH_2-CH(OCH_3)_2$\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2}$ (an acetal of aldehyde).
This compound 'P' has a CH_3CO-$CH_3CO-$ group (positive iodoform) and an acetal (protected aldehyde, negative Tollen's).
On hydrolysis:
mathrmCH_3-CO-CH_2-CH(OCH_3)_2 xrightarrowmathrmH_2O/H^+ mathrmCH_3-CO-CH_2-CHO + 2mathrmCH_3OH$$\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2} \xrightarrow{\mathrm{H_2O/H^+}} \mathrm{CH_3-CO-CH_2-CHO} + 2\mathrm{CH_3OH}$$
The product Q (mathrmCH_3-CO-CH_2-CHO$\mathrm{CH_3-CO-CH_2-CHO}$) has a methyl ketone (positive iodoform test) and an aldehyde (positive Tollen's test).
Chemical Reactions of Aldehydes and Ketones diagram for Q61 - JEE Main 2026 Morning
### Pattern Recognition
Whenever an aldehyde test becomes positive *after* hydrolysis, it points to a protected aldehyde, usually an acetal or hemiacetal. A compound with molecular formula C_n H_2n O_3$C_n H_{2n} O_3$ often represents a keto-acetal.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
Chemical Reactions of Aldehydes and Ketones diagram for Q61 - JEE Main 2026 Morning
Keywords:#Iodoform test#JEE Main 2026 Morning Q61#Aldehydes, Ketones and Carboxylic Acids JEE Main 2026#Chemical Reactions of Aldehydes and Ketones JEE Main 2026
More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 2
When the dicarbonyl compound shown below undergoes an base-catalyzed intramolecular aldol condensation reaction, the major structural product formed is:
{{Q_IMG1}}
The image features a symmetrical open-chain diketone containing branching elements, poised for intramolecular cyclization.
A.textOption A structure OPT_IMG_A$\text{Option A structure } {{OPT_IMG_A}}$
B.textOption B structure OPT_IMG_B$\text{Option B structure } {{OPT_IMG_B}}$
C.textOption C structure OPT_IMG_C$\text{Option C structure } {{OPT_IMG_C}}$
D.textOption D structure OPT_IMG_D$\text{Option D structure } {{OPT_IMG_D}}$
Solution
### Core Logic
Intramolecular aldol condensations are governed heavily by thermodynamic stability, favoring the formation of **5-membered** or **6-membered** rings over strained 3-, 4-, or large 7-membered options.
1. **Enolate Generation**: Base abstracts an alpha$\alpha$-proton to form a nucleophilic carbanion enolate.
2. **Attack Vector Selection**: Deprotonation at the outer methyl site enables a ring-closing attack on the distant carbonyl carbon, perfectly designing a highly stable cyclopentene ring skeleton.
3. **Dehydration**: Heating drives the loss of a water molecule (-textH_2textO$-\text{H}_2\text{O}$), introducing an alpha,beta$\alpha,\beta$-unsaturated carbonyl arrangement that provides stabilization via conjugated resonance. The image features a symmetrical open-chain diketone containing branching elements, poised for intramolecular cyclization.
### Pattern Recognition
Count the intervening carbon chain carefully. Intramolecular aldol pathways will always selectively build 5- or 6-membered rings due to favorable ring strain kinetics. Eliminating choices based on incorrect ring sizes isolates Option (1) instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3text g$5.3\text{ g}$ of benzaldehyde, a total of 3.51text g$3.51\text{ g}$ of product was obtained.
The percentage yield in this reaction was __%.
Numerical Answer.Answer: 60 to 60
Solution
### Core Logic
The balanced chemical equation for the synthesis of dibenzalacetone is:
2 Ph-CHO + CH3COCH3 xrightarrowtextStrong Base Ph-CH=CH-CO-CH=CH-Ph + 2H2O$$2 Ph-CHO + CH{3}COCH{3} xrightarrow{\text{Strong Base}} Ph-CH=CH-CO-CH=CH-Ph + 2H{2}O$$
Let's calculate the moles of reactants:
textMolar mass of benzaldehyde (PhCHO) = 106text g/mol$$\text{Molar mass of benzaldehyde } (PhCHO) = 106\text{ g/mol}$$textMoles of benzaldehyde used = frac5.3106 = 0.05text mol = frac120text mol$$\text{Moles of benzaldehyde used} = \frac{5.3}{106} = 0.05\text{ mol} = \frac{1}{20}\text{ mol}$$Claisen-Schmidt Condensation diagram for Q48 - JEE Main 2025 Evening
According to the reaction stoichiometry, 2text moles of PhCHO$2\text{ moles of } PhCHO$ yield 1text mole of dibenzalacetone$1\text{ mole of dibenzalacetone}$.
textTheoretical moles of product = frac0.052 = 0.025text mol$$\text{Theoretical moles of product} = \frac{0.05}{2} = 0.025\text{ mol}$$
### Step 1: Yield Evaluation
$textMolar mass of dibenzalacetone (C17H14O) = 234text g/mol$$$\text{Molar mass of dibenzalacetone } (C{17}H{14}O) = 234\text{ g/mol}$$textTheoretical mass = 0.025 times 234 = 5.85text g$$\text{Theoretical mass} = 0.025 \times 234 = 5.85\text{ g}$$textPercentage yield = fractextActual masstextTheoretical mass times 100 = frac3.515.85 times 100 = 60%$$\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{3.51}{5.85} \times 100 = 60%$$
### 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
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q33jee_main_2025_28_jan_morningRearrangement and Ozonolysis
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:
The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.
The structure of (mathbfPi^prime primemathbfP^prime prime)$(\mathbf{\Pi}^{\prime \prime}\mathbf{P}^{\prime \prime})$ is
A. (1)
B. (2)
C. (3)
D. (4)
Solution
### 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:
The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.The image displays the chemical structure of product R obtained from rearrangement and subsequent reaction steps.
### Pattern Recognition
Sees: Acidic rearrangement rightarrow$\rightarrow$ ozonolysis rightarrow$\rightarrow$ intramolecular aldol condensation.
Shortcut: Work backwards from the dicarbonyl fragments formed after opening the final product ring system.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q37jee_main_2025_28_jan_morningReactions of Carbonyl Compounds
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:
A.textA, B and D only$\text{A, B and D only}$
B.textA, C and E only$\text{A, C and E only}$
C.textC and E only$\text{C and E only}$
D.textB, C and D only$\text{B, C and D only}$
Solution
### Core Logic
Let us check each option pathway:
- **A. Iodoform Reaction:** Positive for both because both contain the mathrmCH_3-mathrmC=mathrmO$\mathrm{CH}_3-\mathrm{C}=\mathrm{O}$ methyl ketone fragment.
- **B. Cannizaro Reaction:** Negative for both because both contain alpha$\alpha$-hydrogens.
- **C. Aldol Condensation:** Positive for both because they have alpha$\alpha$-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
Q36jee_main_2025_03_april_morningIodoform Test
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
A. 5
B. 4
C. 3
D. 2
Solution
### Core Logic
The iodoform test requires compounds containing either a methyl ketone group (textCH_3textCO-$\text{CH}_3\text{CO}-$) or a methyl carbinol structural unit (textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$).
Let us audit the provided compounds:
1. **Ethanol** (textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$
ightarrow$
ightarrow$ **Positive**
2. **Isopropyl alcohol** (textCH_3textCH(OH)CH_3$\text{CH}_3\text{CH(OH)CH}_3$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$
ightarrow$
ightarrow$ **Positive**
3. **Bromoacetone** (textCH_3textCOCH_2textBr$\text{CH}_3\text{COCH}_2\text{Br}$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$
ightarrow$
ightarrow$ **Positive**
4. **2-Butanol** (textCH_3textCH(OH)CH_2textCH_3$\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$
ightarrow$
ightarrow$ **Positive**
5. **2-Butanone** (textCH_3textCOCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_3$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$
ightarrow$
ightarrow$ **Positive**
6. **Butanal** (textCH_3textCH_2textCH_2textCHO$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$): **Negative**
7. **2-Pentanone** (textCH_3textCOCH_2textCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$
ightarrow$
ightarrow$ **Positive**
8. **3-Pentanone** (textCH_3textCH_2textCOCH_2textCH_3$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$): **Negative**
9. **Pentanal** (textCH_3textCH_2textCH_2textCH_2textCHO$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$): **Negative**
10. **3-Pentanol** (textCH_3textCH_2textCH(OH)CH_2textCH_3$\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3$): **Negative**
### Step 1: Summation
The molecules that **cannot** give the iodoform reaction are: Butanal, 3-Pentanone, Pentanal, and 3-Pentanol. This gives a total count of exactly 4 molecules.
### Pattern Recognition
Shortcut: Filter for names ending with '-anal' or having ketones/alcohols at positions higher than 2 (e.g., 3-pentanone, 3-pentanol). These lack the vital terminal methyl group adjacent to the functional group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2026_21_jan_morning
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