Iodoform test can differentiate between A. Methanol and Ethanol B. CH3COOH and CH3CH2COOH C. Cyclohexene and cyclohexanone D. Diethyl ether and Pentan-3-one E. Anisole and acetone Choose the correct answer from the options given below:

The iodoform test gives a positive result for compounds containing a methyl ketone group (-CO-CH3) or a secondary alcohol with an adjacent methyl group (-CH(OH)-CH3), which gets oxidized to a methyl ketone under the reaction conditions. A. Methanol vs Ethanol: Ethanol (CH3CH2OH) contains the -CH(OH)-CH3 terminal structure and gives a positive test. Methanol (CH3OH) does not. (Can differentiate) B. Neither CH3COOH nor CH3CH2COOH 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 (CH3CH2-O-CH2CH3) nor pentan-3-one (CH3CH2-CO-CH2CH3) has a methyl ketone group. (Cannot differentiate) E. Acetone (CH3-CO-CH3) gives a positive test. Anisole (Ph-O-CH3) does not. (Can differentiate) Only pairs A and E contain exactly one compound that gives a positive iodoform test, allowing for chemical differentiation. Look strictly for CH3-C=O or CH3-CH-OH. If a molecule lacks these exact terminal fragments, it will fail the iodoform test. Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Iodoform test can differentiate between A. Methanol and Ethanol B. CH3COOH and CH3CH2COOH C. Cyclohexene and cyclohexanone D. Diethyl ether and Pentan-3-one E. Anisole and acetone Choose the correct a

The iodoform test gives a positive result for compounds containing a methyl ketone group (-CO-CH3) or a secondary alcohol with an adjacent methyl group (-CH(OH)-CH3), which gets oxidized to a methyl ketone under the reaction conditions. A. Methanol vs Ethanol: Ethanol (CH3CH2OH) contains the -CH(OH)-CH3 terminal structure and gives a positive test. Methanol (CH3OH) does not. (Can differentiate) B. Neither CH3COOH nor CH3CH2COOH gives the iodoform test since carboxylic acids do not undergo this r

Iodoform test can differentiate between A. Methanol and Ethanol B. CH3COOH and CH3CH2COOH C. Cyclohexene and cyclohexanone D. Diethyl ether and Pentan-3-one E. Anisole and acetone Choose the correct answer from the options given below:

The iodoform test gives a positive result for compounds containing a methyl ketone group (-CO-CH3) or a secondary alcohol with an adjacent methyl group (-CH(OH)-CH3), which gets oxidized to a methyl ketone under the reaction conditions. A. Methanol vs Ethanol: Ethanol (CH3CH2OH) contains the -CH(OH)-CH3 terminal structure and gives a positive test. Methanol (CH3OH) does not. (Can differentiate) B. Neither CH3COOH nor CH3CH2COOH 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 (CH3CH2-O-CH2CH3) nor pentan-3-one (CH3CH2-CO-CH2CH3) has a methyl ketone group. (Cannot differentiate) E. Acetone (CH3-CO-CH3) gives a positive test. Anisole (Ph-O-CH3) does not. (Can differentiate) Only pairs A and E contain exactly one compound that gives a positive iodoform test, allowing for chemical differentiation. Look strictly for CH3-C=O or CH3-CH-OH. If a molecule lacks these exact terminal fragments, it will fail the iodoform test. Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers

Iodoform test can differentiate between A. Methanol and Ethanol B. CH3COOH and CH3CH2COOH C. Cyclohexene and cyclohexanone D. Diethyl ether and Pentan-3-one E. Anisole and acetone Choose the correct a

The iodoform test gives a positive result for compounds containing a methyl ketone group (-CO-CH3) or a secondary alcohol with an adjacent methyl group (-CH(OH)-CH3), which gets oxidized to a methyl ketone under the reaction conditions. A. Methanol vs Ethanol: Ethanol (CH3CH2OH) contains the -CH(OH)-CH3 terminal structure and gives a positive test. Methanol (CH3OH) does not. (Can differentiate) B. Neither CH3COOH nor CH3CH2COOH gives the iodoform test since carboxylic acids do not undergo this r

Iodoform test can differentiate between A. Methanol and Ethanol B. CH3COOH and CH3CH2COOH C. Cyclohexene and…

Year	2026	2025	2024	Total
Questions	14	21	11	46

Q45 jee_main_2025_24_jan_morning Chemical Reactions of Ketones

The product (A) formed in the following reaction sequence is :

C H _ 3 - C \equiv C H [ (i i) H _ 2 / N i ](i) H g ^ 2 +, H _ 2 S O _ 4 (A)

A. $CH₃-C(NH₂)(CH₃)-CH₂-OH$
B. $CH₃-C(OH)(CH₃)-CH₂-NH₂$
C. $CH₃-CH₂-CH(OH)-CH₂-OH$
D. $CH₃-CH(OH)-CH(CH₃)-NH₂$

Solution

Core Logic

Breaking down the multi-step reaction path sequence:

Hydration of propyne using Kucherov's trigger condition ( $Hg²⁺, H₂SO₄$ $Hg^{2+}, H_2SO_4$) adds water across the triple bond via Markovnikov's rule. The intermediate enol undergoes tautomerization to yield acetone ( $CH₃-CO-CH₃$ $CH_3-CO-CH_3$).
Reacting acetone with $HCN$ $HCN$ drives nucleophilic addition at the carbonyl carbon, forming a cyanohydrin intermediate: $CH₃-C(OH)(CH₃)-CN$ $CH_3-C(OH)(CH_3)-CN$.
Introducing a reducing agent ( $H₂/Ni$ $H_2/Ni$) selectively converts the nitrile group ( $-CN$ $-CN$) into a primary amine side chain ( $-CH₂-NH₂$ $-CH_2-NH_2$).

The final synthesized structure is: $CH₃-C(OH)(CH₃)-CH₂-NH₂$ .

Chemical Reactions of Ketones step sequence product chart for Q45

Pattern Recognition

Alkyne hydration produces a ketone carbonyl system. Cyanohydrin synthesis introduces a carbon coordinate, which subsequently reduces to a primary aliphatic amine functional group.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q48 jee_main_2025_24_jan_morning Reactions of Carboxylic Acids

Xg of benzoic acid on reaction with aq.

NaHCO₃

release

CO₂

that occupied 11.2 L volume at STP. X is ______ g.

Numerical Answer. Answer: 61 to 61

Solution

Related Formula

Moles of gas at STP = V in Liters22.4 L/mol

Core Logic

The balanced acid-base reaction equation is:

C₆H₅COOH + NaHCO₃ arrow C₆H₅COO⁻Na⁺ + H₂O + CO₂

This stoichiometry shows a $1:1$ $1:1$ molar ratio between benzoic acid and the released carbon dioxide gas.

Calculate the total moles of evolved $CO₂$ $CO_2$ gas at standard conditions:

moles of CO₂ = 11.2 L22.4 L mol⁻¹ = 0.5 moles

Thus, the reaction consumed exactly $0.5 moles$ $0.5\text{ moles}$ of benzoic acid ( $C₆H₅COOH$ $C_6H_5COOH$, molecular weight = $122 g/mol$ $122\text{ g/mol}$):

mass consumed (X) = 0.5 \times 122 = 61 grams

Pattern Recognition

Carboxylic acids react with sodium bicarbonate in a straightforward $1:1$ $1:1$ molar ratio, releasing exactly 1 mole of carbon dioxide gas per mole of acid group.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q37 jee_main_2025_28_jan_evening Oxidation of Alkylbenzenes

The total number of compounds from below when treated with hot

KMnO₄

giving benzoic acid is:

Alkylbenzene structures for Q37 - JEE Main 2025 — The image lists seven distinct aromatic side-chain hydrocarbon structures to assess for side-chain oxidation.

A. 3
B. 4
C. 6
D. 5

Solution

Related Formula

Side-chain oxidation criteria:

Ar-CH(R)₂ hot KMnO₄ Ar-COOH

Requires the presence of at least one benzylic hydrogen atom on the aromatic side chain structure.

Core Logic

Alkyl side chains on a benzene ring are oxidized entirely down to a carboxylic acid group (benzoic acid) by strong oxidizing agents like hot alkaline $KMnO₄$ $KMnO_4$, provided the benzylic carbon contains at least one hydrogen atom.

Evaluating the structures from the diagram:

Toluene (contains $3$ $3$ benzylic H) $arrow$ $\rightarrow$ Yields benzoic acid
Ethylbenzene (contains $2$ $2$ benzylic H) $arrow$ $\rightarrow$ Yields benzoic acid
Isopropylbenzene / Cumene (contains $1$ $1$ benzylic H) $arrow$ $\rightarrow$ Yields benzoic acid
tert-Butylbenzene (contains $0$ $0$ benzylic H) $arrow$ $\rightarrow$ Resists oxidation
Isobutylbenzene (contains $2$ $2$ benzylic H) $arrow$ $\rightarrow$ Yields benzoic acid
2-Phenylpropan-2-ol (contains $0$ $0$ benzylic H, tertiary alcohol center) $arrow$ $\rightarrow$ Resists oxidation
n-Propylbenzene (contains $2$ $2$ benzylic H) $arrow$ $\rightarrow$ Yields benzoic acid
The last biphenyl derivative undergoes complex disruption or ring cleavages and does not cleanly yield simple benzoic acid under standard monocyclic oxidation definitions.

Step 1: Counting Valid Targets

The compounds that undergo oxidation to form benzoic acid are toluene, ethylbenzene, isopropylbenzene, isobutylbenzene, and n-propylbenzene.

Total count = 5 compounds.

Structural evaluation summary for side-chain oxidation targets — The image lists seven distinct aromatic side-chain hydrocarbon structures to assess for side-chain oxidation.

Pattern Recognition

Look immediately at the benzylic carbon (the carbon bonded directly to the ring). If it is a quaternary center (like in tert-butylbenzene) or lacks a hydrogen atom entirely, mark it as unreactive to hot $KMnO₄$ $KMnO_4$ side-chain oxidation.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q jee_main_2025_29_jan_morning Clemmensen Reduction

The product (P) formed in the following reaction is:

Clemmensen Reduction diagram for Q41 - JEE Main 2025 Morning — The structural flowchart indicates a multi-functional molecule treated with Zn-Hg/HCl.

A.
B.
C.
D.

Solution

Related Formula

R-CO-R' [HCl]Zn-Hg R-CH₂-R'

Core Logic

The provided reaction relies on classic Clemmensen Reduction reagent configurations ( $Zn-Hg / HCl$ $\mathrm{Zn-Hg} / \mathrm{HCl}$). This reagent explicitly selective targets ketone/aldehyde carbonyl centers ( $>C=O$ $>\mathrm{C}=\mathrm{O}$) and converts them cleanly into methylene units ( $-CH₂-$ $-\mathrm{CH}_2-$). Any free aliphatic alcohol functionalities ( $-OH$ $-\mathrm{OH}$) located on the ring are preserved under these reductive conditions. Thus, the ketone group on the ring is cleanly converted to $-CH₂-$ $-\mathrm{CH}_2-$, yielding option (3):

Clemmensen Reduction solution diagram for Q41 - JEE Main 2025 Morning — The structural flowchart indicates a multi-functional molecule treated with Zn-Hg/HCl.

Pattern Recognition

Clemmensen routes deoxygenate carbonyl clusters selectively without compromising adjacent standalone cyclic alcohol structural components.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q jee_main_2025_29_jan_morning Chemical Reactions of Carbonyl Compounds

Based on the reaction sequence:

Organic synthesis reaction sequence diagram for Q50 - JEE Main 2025 Morning — 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

g mol⁻¹

C:12, H:1, O:16)

Numerical Answer. Answer: 13 to 13

Solution

Related Formula

Mass = Moles \cdot Molar Mass

Core Logic

Let us trace the reaction scheme row-by-row:

Organic synthesis intermediate layout mapping — A structural flow charts the transformation of a cyclic dialcohol into final product S.

Reactant: 2-(hydroxymethyl)cyclopentan-1-ol.
Step 1: Excess $CrO₃$ $\mathrm{CrO}_3$ (Jones oxidation) oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid $arrow$ $\rightarrow$ Product P.
Step 2: Reaction with 1 mole of glycol protects the ketone selectively as a cyclic ketal $arrow$ $\rightarrow$ Product Q.
Step 3: Treatment with $CH₃MgI$ $\mathrm{CH}_3\mathrm{MgI}$ targets the free carboxylic acid (or ester equivalent) to form a methyl ketone after workup $arrow$ $\rightarrow$ Product R.
Step 4: $NaBH₄$ $\mathrm{NaBH}_4$ reduces the methyl ketone to a secondary alcohol, and acid workup deprotects the ketal back to the original ketone $arrow$ $\rightarrow$ Compound S.

Chemical structural formula of S: 2-(1-hydroxyethyl)cyclopentan-1-one ( $C₇H₁₂O₂$ ).

Molar mass of S ( $C₇H₁₂O₂$ ):

M = (7 \cdot 12) + (12 \cdot 1) + (2 \cdot 16) = 84 + 12 + 32 = 130 ~g/mol

Mass of 0.1 mole = 0.1 \cdot 130 = 13 ~g

Pattern Recognition

Ketal groups serve as stable protective masks that shield ketones, allowing selective organometallic reactions to take place elsewhere on the molecule.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Selection

Pattern Recognition

Chapter Mix

Reference Study Guides

More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 7

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Counting Valid Targets

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2026_23_january_evening

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