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Aldehydes, Ketones and Carboxylic Acids appeared 46 times across 3 years — 5.3% of Chemistry. This question is from Haloform Reaction.

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Which of the following statements are TRUE about Haloform reaction? A. Sodium hypochlorite reacts with KI to give KOI. B. KOI is a reducing agent. C. α, β-unsaturated methylketone (CH₃-CH=CH-CO-CH₃) will give iodoform reaction. D. Isopropyl alcohol will not give iodoform test. E. Methanoic acid will give positive iodoform test. Choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Evaluate each statement:

A. NaOCl (Sodium hypochlorite) reacts with KI to produce KOI and NaCl. This statement is True.

NaOCl + KI arrow NaCl + KOI

B. KOI acts as a mild oxidizing agent, not a reducing agent. It is used to oxidize secondary alcohols to ketones in the iodoform reaction. This statement is False.

C. The molecule CH₃-CH=CH-CO-CH₃ has a terminal methyl ketone group (-CO-CH₃). The adjacent double bond does not interfere with the cleavage of the methyl group during the haloform test. It will yield yellow iodoform. This statement is True.

D. Isopropyl alcohol is CH₃-CH(OH)-CH₃. Since it is a secondary alcohol with an adjacent methyl group, it gets oxidized to acetone in situ and therefore gives a positive iodoform test. The statement says it will not give the test, which is False.

E. Methanoic acid (HCOOH) does not have a terminal methyl group attached to a carbonyl carbon, so it cannot undergo the haloform reaction. This statement is False.

Step 1: Final Selection

Only statements A and C are True.

Pattern Recognition

Alcohols with the CH₃-CH(OH)- moiety always test positive for iodoform. Reagents containing hypohalites (OCl^-, OI^-) are strong oxidizing agents, frequently used in organic synthesis precisely for their oxidative power.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Reference Study Guides

Aldehydes Ketones And Carboxylic Acids

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 group (acidic) and an aldehyde group (electrophilic).

When one equivalent of Grignard reagent (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 reacts with PhMgBr:

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

Upon workup with aq. NH₄Cl, 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).

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 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

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