"P" is an optically active compound with molecular formula textC_6textH_12textO$\text{C}_6\text{H}_{12}\text{O}$. When "P" is treated with 2,4-dinitrophenylhydrazine, it gives a positive test. However, in presence of Tollens reagent, "P" gives a negative test. Predict the structure of "P".
### Related Formula
textCarbonyl compound + text2,4-DNP
ightarrow textHydrazone derivative (Positive test) $\text{Carbonyl compound} + \text{2,4-DNP}
ightarrow \text{Hydrazone derivative (Positive test)} $textAldehyde + textTollens' Reagent
ightarrow textSilver Mirror (Positive test)$\text{Aldehyde} + \text{Tollens' Reagent}
ightarrow \text{Silver Mirror (Positive test)}$textKetone + textTollens' Reagent
ightarrow textNo reaction (Negative test) $\text{Ketone} + \text{Tollens' Reagent}
ightarrow \text{No reaction (Negative test)} $
### Core Logic
Analyzing individual functional constraints:
- Positive 2,4-DNP test shows compound contains a carbonyl group (aldehyde or ketone).
- Negative Tollens' test clarifies it is not an aldehyde; hence it must be a ketone.
- The compound is optically active, meaning it must possess a chiral center (carbon with 4 distinct groups).
Let's evaluate the options via structural configurations:
Identification of Carbonyl Compounds diagram for Q30 - JEE Main 2025 EveningIdentification of Carbonyl Compounds diagram for Q30 - JEE Main 2025 Evening
### Step 1: Structural Verification
Option (2) represents 3-methylpentan-2-one:
textCH3-textC(=textO)-oversetasttextCH(textCH3)(textCH2textCH3) $\text{CH}3-\text{C}(=\text{O})-\overset{\ast}{\text{CH}}(\text{CH}3)(\text{CH}2\text{CH}3) $
The third carbon (C3) is linked to: -textH$-\text{H}$, -textCH_3$-\text{CH}_3$, -textCH_2textCH_3$-\text{CH}_2\text{CH}_3$, and -textCOCH_3$-\text{COCH}_3$. It has 4 distinct structural fields, making it chiral and optically active.
### Pattern Recognition
Tollens' negative + DNP positive = Ketone. Once categorized as a ketone, look directly for the structure holding a carbon with four unique groups to secure the optical activity constraint.
### 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
Keywords:#optically active ketone C6H12O#JEE Main 2025 Evening Q30#2 4 DNP positive Tollens negative#carbonyl identification tests
More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 5
Q502025Chemical Reactions of Carbonyl Compounds
Based on the reaction sequence:
A structural flow charts the transformation of a cyclic dialcohol into final product S.
0.1 mole of compound 'S' will weigh ________ g.
(Given molar mass in mathrmg \, mol^-1$\mathrm{g \, mol^{-1}}$ C:12, H:1, O:16)
Numerical Answer.Answer: 13 to 13
Solution
### Related Formula
textMass = textMoles cdot textMolar Mass$\text{Mass} = \text{Moles} \cdot \text{Molar Mass}$
### Core Logic
Let us trace the reaction scheme row-by-row:
1. Reactant: 2-(hydroxymethyl)cyclopentan-1-ol.
2. Step 1: Excess mathrmCrO_3$\mathrm{CrO}_3$ (Jones oxidation) oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid
ightarrow$
ightarrow$ Product P.
3. Step 2: Reaction with 1 mole of glycol protects the ketone selectively as a cyclic ketal
ightarrow$
ightarrow$ Product Q.
4. Step 3: Treatment with mathrmCH_3mathrmMgI$\mathrm{CH}_3\mathrm{MgI}$ targets the free carboxylic acid (or ester equivalent) to form a methyl ketone after workup
ightarrow$
ightarrow$ Product R.
5. Step 4: mathrmNaBH_4$\mathrm{NaBH}_4$ reduces the methyl ketone to a secondary alcohol, and acid workup deprotects the ketal back to the original ketone
ightarrow$
ightarrow$ Compound S .
Chemical structural formula of S: 2-(1-hydroxyethyl)cyclopentan-1-one (mathrmC_7mathrmH_12mathrmO_2$\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$) .
Molar mass of S (mathrmC_7mathrmH_12mathrmO_2$\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$) :
M = (7 cdot 12) + (12 cdot 1) + (2 cdot 16) = 84 + 12 + 32 = 130 \, mathrmg/mol$M = (7 \cdot 12) + (12 \cdot 1) + (2 \cdot 16) = 84 + 12 + 32 = 130 \, \mathrm{g/mol}$textMass of 0.1 mole = 0.1 cdot 130 = 13 \, mathrmg$\text{Mass of 0.1 mole} = 0.1 \cdot 130 = 13 \, \mathrm{g}$
### Pattern Recognition
Ketal groups serve as stable protective masks that shield ketones, allowing selective organometallic reactions to take place elsewhere on the molecule.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2025_07_april_evening
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