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 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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

R-CO-CH₃ NaOH + I₂ R-COONa + CHI₃ (Yellow ppt)

R-CH(OH)-CH₃ NaOH + I₂ R-COONa + CHI₃ (Yellow ppt)

Core Logic

The iodoform test gives a positive result for compounds containing a methyl ketone group ( $-CO-CH₃$ $-CO-CH_3$) or a secondary alcohol with an adjacent methyl group ( $-CH(OH)-CH₃$ $-CH(OH)-CH_3$), which gets oxidized to a methyl ketone under the reaction conditions.

A. Methanol vs Ethanol: Ethanol ( $CH₃CH₂OH$ $CH_3CH_2OH$) contains the $-CH(OH)-CH₃$ $-CH(OH)-CH_3$ terminal structure and gives a positive test. Methanol ( $CH₃OH$ $CH_3OH$) does not. (Can differentiate) B. Neither $CH₃COOH$ $CH_3COOH$ nor $CH₃CH₂COOH$ $CH_3CH_2COOH$ 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 ( $CH₃CH₂-O-CH₂CH₃$ $CH_3CH_2-O-CH_2CH_3$) nor pentan-3-one ( $CH₃CH₂-CO-CH₂CH₃$ $CH_3CH_2-CO-CH_2CH_3$) has a methyl ketone group. (Cannot differentiate) E. Acetone ( $CH₃-CO-CH₃$ $CH_3-CO-CH_3$) gives a positive test. Anisole ( $Ph-O-CH₃$ $Ph-O-CH_3$) does not. (Can differentiate)

Step 1: Final Selection

Only pairs A and E contain exactly one compound that gives a positive iodoform test, allowing for chemical differentiation.

Pattern Recognition

Look strictly for $CH₃-C=O$ $CH_3-C=O$ or $CH₃-CH-OH$ $CH_3-CH-OH$. If a molecule lacks these exact terminal fragments, it will fail the iodoform test.

Chapter Mix

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

Iodoform Reaction solution diagram 1 for Q58 - JEE Main 2026 Evening — Chemical structures related to the options provided for the iodoform differentiation test.

More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 10

Q jee_main_2024_31_jan_morning Reactions with Grignard Reagent

The product of the following reaction is P.

Reactions with Grignard Reagent diagram for Q84 - JEE Main 2024 Morning — The image shows p-hydroxybenzaldehyde reacting with one equivalent of PhMgBr followed by aqueous ammonium chloride.

The number of hydroxyl groups present in the product P is

Numerical Answer. Answer: 0 to 0

Solution

Core Logic

The given reactant is p-hydroxybenzaldehyde, which contains both a phenolic $-OH$ $-OH$ group (acidic) and an aldehyde group (electrophilic).

When one equivalent of Grignard reagent ( $PhMgBr$ $PhMgBr$) is added, it behaves primarily as a strong base due to the presence of an acidic proton. Acid-base reactions are extremely fast compared to nucleophilic additions.

The acidic phenolic $-OH$ $-OH$ reacts with $PhMgBr$ $PhMgBr$:

PhMgBr + HO-C₆H₄-CHO arrow Ph-H (Benzene) + BrMg-O-C₆H₄-CHO

Upon workup with $aq. NH₄Cl$ $aq. NH_4Cl$, the phenoxide ion simply regenerates the starting p-hydroxybenzaldehyde. However, the question asks for the number of hydroxyl groups present in the formed product (Benzene).

The distinct product formed in the reaction is Benzene. Benzene has 0 hydroxyl groups.

Pattern Recognition

Whenever a Grignard reagent encounters a molecule with an acidic hydrogen (alcohol, phenol, amine, alkyne), it will invariably act as a base first. If only 1 equivalent is used, nucleophilic addition to carbonyls will NOT happen.

Chapter Mix

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