Which among the following compounds give yellow solid when reacted with NaOI/NaOH? (A) CH3 - CH(OH) - C2H5 (B) CH3 - CH2 - CH2 - OH (C) CH3 - CO - C2H5 (D) CH3-CO- OH (E) CH3 - CH2 - CHO Choose the correct answer from the options given below:

### Related Formula ### Core Logic Let's check the structural groups of each given option: - **(A)** CH3 - CH(OH) - C2H5: Contains the methylcarbinol group (CH3-CH(OH)-). Gives a positive iodoform test. - **(B)** CH3 - CH2 - CH2 - OH: Linear primary alcohol, does not contain the required group. - **(C)** CH3 - CO - C2H5: Contains the methyl ketone group (CH3-CO-). Gives a positive iodoform test. - **(D)** CH3 - OH: Methanol does not give the test. - **(E)** CH3 - CH2 - H: Ethane does not give the test. Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform (CHI3). ### Step 1: Chemical Equations The balanced haloform pathways occur as follows: ### Pattern Recognition The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal -CH3 affixed directly to a carbonyl oxygen index (C=O) or a hydroxyl carbon (CH-OH). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Which among the following compounds give yellow solid when reacted with NaOI/NaOH? (A) CH3 - CH(OH) - C2H5 (B) CH3 - CH2 - CH2 - OH (C) CH3 - CO - C2H5 (D) CH3-CO- OH (E) CH3 - CH2 - CHO Choose the co

### Related Formula ### Core Logic Let's check the structural groups of each given option: - **(A)** CH3 - CH(OH) - C2H5: Contains the methylcarbinol group (CH3-CH(OH)-). Gives a positive iodoform test. - **(B)** CH3 - CH2 - CH2 - OH: Linear primary alcohol, does not contain the required group. - **(C)** CH3 - CO - C2H5: Contains the methyl ketone group (CH3-CO-). Gives a positive iodoform test. - **(D)** CH3 - OH: Methanol does not give the test. - **(E)** CH3 - CH2 - H: Ethane does not give th

Which among the following compounds give yellow solid when reacted with NaOI/NaOH? (A) CH3 - CH(OH) - C2H5 (B) CH3 - CH2 - CH2 - OH (C) CH3 - CO - C2H5 (D) CH3-CO- OH (E) CH3 - CH2 - CHO Choose the correct answer from the options given below:

### Related Formula ### Core Logic Let's check the structural groups of each given option: - **(A)** CH3 - CH(OH) - C2H5: Contains the methylcarbinol group (CH3-CH(OH)-). Gives a positive iodoform test. - **(B)** CH3 - CH2 - CH2 - OH: Linear primary alcohol, does not contain the required group. - **(C)** CH3 - CO - C2H5: Contains the methyl ketone group (CH3-CO-). Gives a positive iodoform test. - **(D)** CH3 - OH: Methanol does not give the test. - **(E)** CH3 - CH2 - H: Ethane does not give the test. Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform (CHI3). ### Step 1: Chemical Equations The balanced haloform pathways occur as follows: ### Pattern Recognition The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal -CH3 affixed directly to a carbonyl oxygen index (C=O) or a hydroxyl carbon (CH-OH). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Which among the following compounds give yellow solid when reacted with NaOI/NaOH? (A) CH3 - CH(OH) - C2H5 (B) CH3 - CH2 - CH2 - OH (C) CH3 - CO - C2H5 (D) CH3-CO- OH (E) CH3 - CH2 - CHO Choose the co

### Related Formula ### Core Logic Let's check the structural groups of each given option: - **(A)** CH3 - CH(OH) - C2H5: Contains the methylcarbinol group (CH3-CH(OH)-). Gives a positive iodoform test. - **(B)** CH3 - CH2 - CH2 - OH: Linear primary alcohol, does not contain the required group. - **(C)** CH3 - CO - C2H5: Contains the methyl ketone group (CH3-CO-). Gives a positive iodoform test. - **(D)** CH3 - OH: Methanol does not give the test. - **(E)** CH3 - CH2 - H: Ethane does not give th

Which among the following compounds give yellow solid when reacted with NaOI/NaOH? (A) CH3 - CH(OH)

Year	2026	2025	2024	Total
Questions	14	21	11	46

Q41 jee_main_2025_04_april_morning Chemical Properties of Ketones

An organic compound (X) with molecular formula

C₃H₆O

is not readily oxidised. On reduction it gives

C₃H₈O

(Y) which reacts with

HBr

to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis gives 2, 3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:

A. $CH₃COCH₃, CH₃CH₂CH₂OH, CH₃CH(Br)CH₃$
B. $CH₃COCH₃, CH₃CH(OH)CH₃, CH₃CH(Br)CH₃$
C. $CH₃CH₂CHO, CH₃CH₂CH₂OH, CH₃CH₂CH₂Br$
D. $CH₃CH₂CHO, CH₃CH=CH₂, CH₃CH(Br)CH₃$

Solution

Core Logic

Let's deduce the identities stepwise:

Compound (X) has the formula $C₃H₆O$ $C_3H_6O$ and is resistant to mild oxidation, which identifies it as a ketone: Acetone ( $CH₃COCH₃$ $CH_3COCH_3$).
Reduction of Acetone yields a secondary alcohol, Propan-2-ol ( $CH₃CH(OH)CH₃$ $CH_3CH(OH)CH_3$, Compound Y).
Treatment of Propan-2-ol with $HBr$ $HBr$ substitutes the hydroxyl group to form 2-Bromopropane ( $CH₃CH(Br)CH₃$ $CH_3CH(Br)CH_3$, Compound Z).
Reacting 2-Bromopropane with Magnesium in ether creates the branched Grignard reagent, Isopropylmagnesium bromide ( $(CH₃)₂CHMgBr$ $(CH_3)_2CHMgBr$).
Finally, nucleophilic addition of this Grignard reagent to Acetone followed by aqueous workup yields the highly branched tertiary alcohol: 2,3-dimethylbutan-2-ol.

Pattern Recognition

Resistance to mild oxidation immediately distinguishes ketones from isomeric aldehydes. Nucleophilic addition of an isopropyl Grignard to acetone cleanly yields the 2,3-dimethylbutan-2-ol framework.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Q30 jee_main_2025_07_april_evening Identification of Carbonyl Compounds

"P" is an optically active compound with molecular formula

C₆H₁₂O

. 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".

A. $CH₃-C(=O)-CH₂-CH₂-CH₂-CH₃$
B. $CH₃-C(=O)-CH(CH₂-CH₃)-CH₃$
C. $H-C(=O)-CH₂-CH(CH₂-CH₃)-CH₃$
D. $CH₃-C(=O)-CH₂-CH(CH₃)₂$

Solution

Related Formula

Carbonyl compound + 2,4-DNP arrow Hydrazone derivative (Positive test)

Aldehyde + Tollens' Reagent arrow Silver Mirror (Positive test)

Ketone + Tollens' Reagent arrow 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 Evening

Step 1: Structural Verification

Option (2) represents 3-methylpentan-2-one:

CH3-C(=O)- CH(CH3)(CH2CH3)

The third carbon (C3) is linked to: $-H$ $-\text{H}$, $-CH₃$ $-\text{CH}_3$, $-CH₂CH₃$ $-\text{CH}_2\text{CH}_3$, and $-COCH₃$ $-\text{COCH}_3$. 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.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q42 jee_main_2025_24_jan_evening Preparation of Aldehydes

Match List-I with List-II

List-I	List-II (Name of Reaction)
(A) $RCN [(ii)H₃O⁺](i)SnCl₂, HCl RCHO$	(I) Etard reaction
(B) The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.	(II) Gatterman-Koch reaction
(C) The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.	(III) Rosenmund reduction
(D) The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.	(IV) Stephen reaction

Choose the correct answer from the options given below:

A. \text{(A)-(IV), (B)-(III), (C)-(I), (D)-(II)}
B. \text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}
C. \text{(A)-(I), (B)-(III), (C)-(II), (D)-(IV)}
D. \text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}

Solution

Core Logic

Let's match each aldehyde preparation method with its official named organic reaction:

(A) $RCN arrow RCHO$ using $SnCl₂/HCl$ $\mathrm{SnCl}_{2}/\mathrm{HCl}$ followed by hydrolysis: This is the classic Stephen reaction $arrow$ $\rightarrow$ (IV).
(B) Reducing an acyl chloride ( $RCOCl$ $\mathrm{RCOCl}$) to an aldehyde using $H₂$ $\mathrm{H}_{2}$ over $Pd-BaSO₄$ $\mathrm{Pd-BaSO}_{4}$: This partial reduction is known as the Rosenmund reduction $arrow$ $\rightarrow$ (III).
(C) Oxidizing toluene to benzaldehyde using chromyl chloride ( $CrO₂Cl₂$ $\mathrm{CrO}_{2}\mathrm{Cl}_{2}$) in $CS₂$ $\mathrm{CS}_{2}$: This selective oxidation method is the Etard reaction $arrow$ $\rightarrow$ (I).
(D) Converting benzene to benzaldehyde using $CO$ $\mathrm{CO}$ and $HCl$ $\mathrm{HCl}$ in the presence of anhydrous $AlCl₃/CuCl$ $\mathrm{AlCl}_{3}/\mathrm{CuCl}$: This formylation process is the Gatterman-Koch reaction $arrow$ $\rightarrow$ (II).

Combining these assignments yields the final sequence: (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

Step-by-Step Layout

The visual reaction components correspond directly to the official structural transformations:

Preparation of Aldehydes solution diagram for Q42 - JEE Main 2025 Evening — The Match List displays chemical transformations side-by-side with their respective named organic chemical reactions.

Pattern Recognition

Quick identification keys:

Nitrile $arrow$ $\rightarrow$ Aldehyde = Stephen
Acid Chloride $arrow$ $\rightarrow$ Aldehyde = Rosenmund
Toluene $arrow$ $\rightarrow$ Chromyl Complex = Etard
Benzene $arrow$ $\rightarrow$ Carbon Monoxide = Gatterman-Koch

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q jee_main_2025_24_jan_morning Aldol Condensation and Ozonolysis

Aman has been asked to synthesise the molecule ring with C—CH3 (x). He thought of preparing the molecule using an aldol condensation reaction. He found a few cyclic alkenes in his laboratory. He thought of performing ozonolysis reaction on alkene to produce a dicarbonyl compound followed by aldol reaction to prepare “x”. Predict the suitable alkene that can lead to the formation of “x”.

A.
B.
C.
D.

Solution

Core Logic

Analyzing the retro-synthesis path step-by-step:

The objective compound is 1-acetylcyclopentene.
Performing reductive ozonolysis ( $O₃, Zn/H₂O$ $O_3, Zn/H_2O$) on 1-methylcyclohexene (Option A) symmetrically breaks the internal endocyclic double bond to form heptane-2,6-dione, a dicarbonyl system.
Aldol Condensation and Ozonolysis reaction part 1 for Q41
Adding a base intermediate trigger ( $OH⁻, Δ$ $OH^{-}, \Delta$) drives an intramolecular aldol condensation: the methyl group carbanion at position 1 attacks the carbon 6 ketone site. This ring-closing event effectively drops water to synthesize the 5-membered cyclopentene core molecule attached to the acetyl unit.
Aldol Condensation and Ozonolysis reaction part 1 for Q41

Pattern Recognition

Counting carbon coordinates is essential. Reductive cleavage transforms a 6-membered ring into an open heptane system, which easily self-condenses into a stable 5-membered ring attached to a methyl ketone side chain.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q40 jee_main_2025_24_jan_morning Reactivity towards Nucleophilic Addition

Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?

A. benzaldehyde $<$ $<$ acetophenone $<$ $<$ p-nitrobenzaldehyde $<$ $<$ p-tolualdehyde
B. acetophenone $<$ $<$ benzaldehyde $<$ $<$ p-tolualdehyde $<$ $<$ p-nitrobenzaldehyde
C. acetophenone $<$ $<$ p-tolualdehyde $<$ $<$ benzaldehyde $<$ $<$ p-nitrobenzaldehyde
D. p-nitrobenzaldehyde $<$ $<$ benzaldehyde $<$ $<$ p-tolualdehyde $<$ $<$ acetophenone

Solution

Core Logic

Reactivity in nucleophilic addition reactions is governed by a combination of steric hindrance and electronic effects around the electrophilic carbonyl carbon:

Ketones are significantly less reactive than aldehydes due to the bulkiness and electron-donating inductive effect ( $+I$ $+I$) of their two alkyl/aryl groups. Thus, acetophenone has the lowest reactivity.
For substituted benzaldehydes, electron-withdrawing groups heighten the partial positive charge on the carbonyl carbon, accelerating nucleophilic attack. Conversely, electron-donating groups suppress reactivity.

Symmetry breakdown structures are shown below:

Acetophenone:
Acetophenone structure for reactivity comparison
p-tolualdehyde:
Acetophenone structure for reactivity comparison
Benzaldehyde:
Acetophenone structure for reactivity comparison
p-nitrobenzaldehyde:
Acetophenone structure for reactivity comparison
Methoxy/methyl donors decrease reactivity: $p-tolualdehyde < benzaldehyde$
Nitro group ( $-NO₂$ $-\mathrm{NO}_2$) acts as a strong electron-withdrawing agent via both $-M$ $-M$ and $-I$ $-I$ pathways, maximizing the electrophilic nature of the carbonyl site. Therefore, p-nitrobenzaldehyde is the most reactive.

Thus, the correct order of reactivity is:

acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde

Pattern Recognition

Aldehydes naturally exhibit higher reactivity than ketones. Electron-withdrawing groups ( $-NO₂$ $-\mathrm{NO}_2$) accelerate addition pathways, whereas electron-donating groups ( $-CH₃$ $-\mathrm{CH}_3$) impede them.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Solution & Explanation

Related Formula

Core Logic

Step 1: Chemical Equations

Pattern Recognition

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More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 6

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Related Formula

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