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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Compounds with CH₃-CH(OH)- or CH₃-CO- groups undergo the iodoform reaction to form CHI₃ (Yellow Solid)

Core Logic

Let's check the structural groups of each given option:

(A) $CH₃ - CH(OH) - C₂H₅$ $CH_3 - CH(OH) - C_2H_5$: Contains the methylcarbinol group ( $CH₃-CH(OH)-$ $CH_3-CH(OH)-$). Gives a positive iodoform test.
(B) $CH₃ - CH₂ - CH₂ - OH$ $CH_3 - CH_2 - CH_2 - OH$: Linear primary alcohol, does not contain the required group.
(C) $CH₃ - CO - C₂H₅$ $CH_3 - CO - C_2H_5$: Contains the methyl ketone group ( $CH₃-CO-$ $CH_3-CO-$). Gives a positive iodoform test.
(D) $CH₃ - OH$ $CH_3 - OH$: Methanol does not give the test.
(E) $CH₃ - CH₂ - H$ $CH_3 - CH_2 - H$: Ethane does not give the test.

Thus, only (A) and (C) yield the yellow precipitate of iodoform ( $CHI₃$ $CHI_3$).

Step 1: Chemical Equations

The balanced haloform pathways occur as follows:

CH₃-CH(OH)-CH₂-CH₃ NaOI/NaOH CHI₃ + CH₃-CH₂-COO^-Na^+

CH₃-CO-CH₂-CH₃ NaOI/NaOH CHI₃ + CH₃-CH₂-COO^-Na^+

Pattern Recognition

The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal $-CH₃$ $-CH_3$ affixed directly to a carbonyl oxygen index ( $C=O$ $C=O$) or a hydroxyl carbon ( $CH-OH$ $CH-OH$).

Chapter Mix

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

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Chemical Equations

Pattern Recognition

Chapter Mix

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

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