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
Qjee_main_2025_02_april_eveningPreparation of Carboxylic Acids
Consider the following reactions. From these reactions which reaction will give carboxylic acid as a major product?
(A) mathrmR - C equiv N xrightarrow[textmild condition]mathrm(i) H^+ / H_2O$\mathrm{R - C \equiv N} \xrightarrow[\text{mild condition}]{\mathrm{(i) H^+ / H_2O}}$
(B) mathrmR - MgX xrightarrow[mathrm(ii) H_3O^+]mathrm(i) CO_2$\mathrm{R - MgX} \xrightarrow[\mathrm{(ii) H_3O^+}]{{\mathrm{(i) CO_2}}}$
(C) mathrmR - C equiv N xrightarrow[mathrm(ii) H_3O^+]mathrm(i) SnCl_2 / HCl$\mathrm{R - C \equiv N} \xrightarrow[\mathrm{(ii) H_3O^+}]{{\mathrm{(i) SnCl_2 / HCl}}}$
(D) mathrmR cdot CH_2 cdot OH xrightarrowmathrmPCC$\mathrm{R \cdot CH_2 \cdot OH} \xrightarrow{\mathrm{PCC}}$
(E)Preparation of Carboxylic Acids
Choose the correct answer from the options given below:
A.textA and D only$\text{A and D only}$
B.textA, B and E only$\text{A, B and E only}$
C.textB, C and E only$\text{B, C and E only}$
D.textB and E only$\text{B and E only}$
Solution
### Related Formula
mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_3O^+ R-COOH$$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$$
### Core Logic
Let's analyze each reaction path to determine the major organic product:
- **Reaction (A)**: Acidic hydrolysis of a nitrile under *mild conditions* yields an amide:
mathrmR-Cequiv N rightarrow R-CONH_2$$\mathrm{R-C\equiv N \rightarrow R-CONH_2}$$
(Full conversion to carboxylic acid requires strong conditions and extended heating).
- **Reaction (B)**: Carbonation of Grignard reagent using solid carbon dioxide (dry ice) followed by acid hydrolysis yields a carboxylic acid:
mathrmR-MgX + CO_2 rightarrow R-COOMgX xrightarrowH_3O^+ R-COOH$$\mathrm{R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH}$$
- **Reaction (C)**: Stephen reduction converts nitrile to aldehyde:
mathrmR-Cequiv N xrightarrowSnCl_2/HCl R-CH=NH xrightarrowH_3O^+ R-CHO$$\mathrm{R-C\equiv N \xrightarrow{SnCl_2/HCl} R-CH=NH \xrightarrow{H_3O^+} R-CHO}$$
- **Reaction (D)**: Pyridinium chlorochromate (PCC) is a mild oxidising agent that converts primary alcohols selectively to aldehydes:
mathrmR-CH_2-OH xrightarrowPCC R-CHO$$\mathrm{R-CH_2-OH \xrightarrow{PCC} R-CHO}$$
- **Reaction (E)**
Preparation of Carboxylic Acids : Rosenmund reduction reduces acid chloride to aldehyde first:
rightarrow$\rightarrow$ R-CHO
Subsequent oxidation with bromine water (which is a mild oxidising agent that selective oxidizes aldehydes but does not affect ketones) converts the aldehyde to carboxylic acid:
mathrmR-CHO xrightarrowBr_2/water R-COOH$$\mathrm{R-CHO \xrightarrow{Br_2/water} R-COOH}$$
### Step 1: Final Tally
Thus, reactions (B) and (E) successfully yield carboxylic acid as the major organic product.
### Pattern Recognition
Remember: Bromine water (mathrmBr_2/H_2O$\mathrm{Br_2/H_2O}$) is a mild, selective oxidising agent commonly used to oxidise aldoses and other aldehydes to monocarboxylic acids without degrading carbon-carbon chains.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Qjee_main_2025_02_april_morningReactions of Phenolic Benzaldehydes
Given below are two statements :
Statement (I): Vanillin Reactions of Phenolic Benzaldehydes will react with NaOH and also with Tollen's reagent.
Statement (II) : Vanillin Reactions of Phenolic Benzaldehydes will undergo self aldol condensation very easily.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.(1)\ textStatement I is incorrect but Statement II is correct$(1)\ \text{Statement I is incorrect but Statement II is correct}$
B.(2)\ textStatement I is correct but Statement II is incorrect$(2)\ \text{Statement I is correct but Statement II is incorrect}$
C.(3)\ textBoth Statement I and Statement II are incorrect$(3)\ \text{Both Statement I and Statement II are incorrect}$
D.(4)\ textBoth Statement I and Statement II are correct$(4)\ \text{Both Statement I and Statement II are correct}$
Solution
### Related Formula
Phenolic protons react with standard strong bases:
mathrmAr-OH + NaOH rightarrow Ar-ONa + H_2O$$\mathrm{Ar-OH + NaOH \rightarrow Ar-ONa + H_2O}$$
Aldol condensation structural requirement: Requires presence of acidic alpha$\alpha$-hydrogen atoms connected to carbonyl centers.
### Core Logic
Let's analyze functional groups within the Vanillin molecular framework:
* Vanillin contains a phenolic hydroxyl group, an aromatic ether, and a formyl functional group (benzaldehyde derivative).
* **Statement I**: The presence of the phenolic -mathrmOH$-\mathrm{OH}$ group allows acid-base reaction with mathrmNaOH$\mathrm{NaOH}$ directly Vanillin structural functional group verification for Q36. The aldehyde center readily reduces Tollen's reagent to produce a silver mirror. (Statement I is accurate).
* **Statement II**: Vanillin lacks any alpha-hydrogens adjacent to its carbonyl carbon, preventing it from undergoing self-aldol condensation. (Statement II is false).
### Pattern Recognition
Benzaldehyde and its substituted derivatives (like vanillin or benzaldehyde itself) never undergo self-aldol condensation because they lack alpha$\alpha$-carbons with abstractable protons. Instead, they typically perform Cannizzaro transformations when exposed to highly concentrated alkaline media.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
The major product (P) in the following reaction is :
Benzil-Benzilic Acid Rearrangement
A.
B.
C.
D.
Solution
### Related Formula
Intramolecular Cannizzaro-type reaction or Benzil-Benzilic acid rearrangement involves nucleophilic attack of hydroxide at a carbonyl group, followed by hydride transfer to the adjacent carbonyl carbon.
### Core Logic
Let's analyze the starting compound, phenylglyoxal:
mathrmPh-CO-CHO$$\mathrm{Ph-CO-CHO}$$
1. The aldehyde carbon (-CHO$-CHO$) is much more electrophilic than the ketone carbon (-CO-$-CO-$) due to less steric hindrance and absence of phenyl group electron donation.
2. Hydroxide ion (mathrmOH^-$\mathrm{OH}^-$) selectively attacks the aldehyde carbonyl carbon, forming a tetrahedral intermediate.
Benzil-Benzilic Acid Rearrangement
### Step 1: Hydride Transfer Mechanism
The tetrahedral intermediate collapses, prompting an intramolecular hydride (H^-$H^-$) transfer to the adjacent ketone carbonyl carbon:
mathrmPh-CO-C(O^-)(OH)H rightarrow mathrmPh-C(O^-)H-COOH$$\mathrm{Ph-CO-C(O^-)(OH)H} \rightarrow \mathrm{Ph-C(O^-)H-COOH}$$
This is the rate-determining step.
### Step 2: Proton Transfer to form Product
Rapid proton transfer occurs from the carboxylic acid group to the alkoxide oxygen, yielding the stable carboxylate salt:
mathrmPh-CH(OH)-COO^- K^+$$\mathrm{Ph-CH(OH)-COO^- K^+}$$
This corresponds to Option (2).
### Pattern Recognition
In asymmetrical 1,2-dicarbonyl systems with an aldehyde and a ketone, nucleophilic addition occurs preferentially at the more reactive aldehyde carbon. The hydrogen is then transferred as a hydride to the ketone carbon, yielding an alpha$\alpha$-hydroxy carboxylate salt.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q30jee_main_2025_08_april_eveningReactions of Cycloalkenes and Alkynes
Identify the major product 'P' in the given reaction sequence starting from 1,2-dibromocyclooctane:
text1,2-dibromocyclooctane xrightarrowtext(i) KOH (alc.) xrightarrowtext(ii) NaNH_2 xrightarrowtext(iii) Hg^2+/H^+ xrightarrowtext(iv) Zn-Hg/HCl text'P (Major product)'$$\text{1,2-dibromocyclooctane} \xrightarrow{\text{(i) KOH (alc.)}} \xrightarrow{\text{(ii) NaNH}_2} \xrightarrow{\text{(iii) Hg}^{2+}/H^+} \xrightarrow{\text{(iv) Zn-Hg/HCl}} \text{'P (Major product)'}$$
### Core Logic
Let us systematically follow the transformation steps:
1. **First Elimination**: 1,2-dibromocyclooctane reacts with alcoholic textKOH$\text{KOH}$ to remove one molecule of textHBr$\text{HBr}$, resulting in a bromocyclooctene intermediate.
2. **Second Elimination**: Treatment with the stronger base textNaNH_2$\text{NaNH}_2$ removes the second molecule of textHBr$\text{HBr}$, forming an alkyne inside the 8-membered ring: **cyclooctyne**.
3. **Kucherov Reaction**: Hydration of cyclooctyne using textHg^2+/H^+$\text{Hg}^{2+}/H^+$ creates an enol intermediate that undergoes tautomerization to form a stable ketone: **cyclooctanone**.
4. **Clemmensen Reduction**: Subjecting cyclooctanone to zinc amalgam and hydrochloric acid (textZn-Hg/HCl$\text{Zn-Hg/HCl}$) completely reduces the carbonyl group (>C=O$>C=O$) to a methylene group (-textCH_2-$-\text{CH}_2-$), finishing with **cyclooctane**. Complete mechanistic sequence mapping for cyclooctane product formation
### Pattern Recognition
A vicinal dihalide treated with sequential strong bases creates an alkyne path. Alkyne hydration creates a ketone body. Finally, Clemmensen reduction takes the ketone down to a simple hydrocarbon skeleton. Recognizing this terminal reduction loop establishes cyclooctane as the undisputed answer.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2026_21_jan_morning
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