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

Year 2026 2025 2024 Total
Questions 14 21 11 46

Number of molecules from below which cannot give iodoform reaction is : Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol

Solution & Explanation

Related Formula

Positive iodoform test requires presence of:

  • Methyl ketone: CH₃-C(=O)-R
  • Methyl carbinol: CH₃-CH(OH)-R
Core Logic

Evaluate each compound:

  • Ethanol (CH₃CH₂OH) arrow Gives test (Gives CHI₃)
  • Isopropyl alcohol (CH₃CH(OH)CH₃) arrow Gives test
  • Bromoacetone (BrCH₂COCH₃) arrow Gives test
  • 2-Butanol (CH₃CH(OH)CH₂CH₃) arrow Gives test
  • 2-Butanone (CH₃COCH₂CH₃) arrow Gives test
  • Butanal (CH₃CH₂CH₂CHO) arrow Does NOT give test
  • 2-Pentanone (CH₃COCH₂CH₂CH₃) arrow Gives test
  • 3-Pentanone (CH₃CH₂COCH₂CH₃) arrow Does NOT give test
  • Pentanal (CH₃CH₂CH₂CH₂CHO) arrow Does NOT give test
  • 3-Pentanol (CH₃CH₂CH(OH)CH₂CH₃) arrow Does NOT give test
Step 1: Final Count

The 4 compounds that cannot give the iodoform reaction are: Butanal, 3-Pentanone, Pentanal, and 3-Pentanol.

Pattern Recognition

Requires CH₃-CO- or CH₃-CH(OH)- group. Aldehydes other than acetaldehyde do not give iodoform test. Symmetrical 3-ketones and 3-alcohols fail the test.

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Reference Study Guides

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

More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2025_03_april_morning

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