Problem 147 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following will give yellow precipitate with \(\mathrm{I}_{2} / \mathrm{NaOH}\) ? 1\. \(\mathrm{ICH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) 2. \(\mathrm{CH}_{3} \mathrm{COOCOCH}_{3}\) 3\. \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (a) 1,3 (b) 3,4 (c) 1,2 (d) 1,4

Step-by-Step Solution

Answer

1 and 2 will give yellow precipitate, matching option (c).

Step 1: Understanding the Reagent

The iodine and sodium hydroxide (\( \mathrm{I}_2/\mathrm{NaOH} \)) test is used to detect the presence of methyl ketones or compounds that can form methyl ketones upon hydrolysis. This is known as the iodoform test, which produces a yellow precipitate of iodoform (\( \mathrm{CHI}_3 \)).

Step 2: Analyzing Compound 1

Compound 1 is \( \mathrm{ICH}_2 \mathrm{COCH}_2 \mathrm{CH}_3 \). This molecule is a methyl ketone since it has the structure \( \mathrm{CH}_3 \mathrm{CO} \). Therefore, it will give a yellow precipitate with \( \mathrm{I}_2/\mathrm{NaOH} \).

Step 3: Analyzing Compound 2

Compound 2 is \( \mathrm{CH}_3 \mathrm{COOCOCH}_3 \). This is an anhydride, specifically acetic anhydride, which upon hydrolysis can form acetic acid that has a \( \mathrm{CH}_3 \mathrm{CO} \) group, thus capable of giving the yellow precipitate with \( \mathrm{I}_2/\mathrm{NaOH} \).

Step 4: Analyzing Compound 3

Compound 3 is \( \mathrm{CH}_3 \mathrm{CONH}_2 \). This is an amide, which does not have a methyl ketone group and cannot be hydrolyzed to form one readily under these conditions. Hence, it will not give a yellow precipitate with \( \mathrm{I}_2/\mathrm{NaOH} \).

Step 5: Analyzing Compound 4

Compound 4 is \( \mathrm{CH}_3 \mathrm{CH}( ext{OH}) \mathrm{CH}_2 \mathrm{CH}_3 \). This structure is an alcohol, and this particular alcohol is a secondary alcohol that contains a methyl group on the carbon carrying the hydroxyl group. Upon oxidation, it can form a methyl ketone, thus capable of forming a yellow precipitate with \( \mathrm{I}_2/\mathrm{NaOH} \).

Step 6: Finding the Correct Pair

From our analysis: 1 and 2 have structures that either inherently contain or form methyl ketones. Therefore, the correct pair that will give a yellow precipitate is 1 and 2.

Key Concepts

Methyl Ketones

Methyl ketones are a special type of ketone characterized by a methyl group (CH₃) attached to the carbon atom in the carbonyl group. They are crucial in various chemical reactions but are particularly noted for their role in the iodoform test. In this test, when methyl ketones are treated with iodine and sodium hydroxide (I₂/NaOH), they produce a distinct yellow precipitate of iodoform (CHI₃).
This reaction is specific because the presence of a methyl group next to the carbonyl group is pivotal for the formation of this precipitate.
More compounds can show the iodoform reaction if they can be converted to methyl ketones through certain chemical processes, like oxidation or hydrolysis.

• The structural formula must include a methyl group adjacent to the ketone group.
• It's the unique placement of the methyl group that makes the iodoform test feasible.

Characteristics of methyl ketones include their ability to oxidize, acting as substrates in multiple biochemical reactions, making them essential in both laboratory and industrial chemistry applications.

Organic Chemistry Reagents

In organic chemistry, reagents like iodine (I₂) and sodium hydroxide (NaOH) are widely used to test for the presence of certain functional groups in compounds. These reagents interact with compounds in ways that are predictable, leading to specific products or signs that confirm or rule out the presence of the targeted functional group, in this case, the methyl ketone.
The I₂/NaOH test is known for forming the yellow precipitate iodoform when a compound contains or can form a methyl ketone group. This is why examining the structure of organic compounds and understanding the function of various reagents is crucial in chemistry.
Reagents serve significant roles in breaking, forming, or converting bonds during chemical reactions.

• Iodine acts as an oxidizing agent in many chemical processes.
• Sodium hydroxide can act as a base, initiating hydrolysis or providing an alkaline environment for certain reactions.

This context helps chemists design experiments to confirm the presence of various functional groups through targeted reactions.

Chemical Reaction Analysis

Analyzing chemical reactions involves understanding how and why substances interact to form new products. The iodoform test is a classic example of such a reaction analysis, where observing the outcome of mixing specific reagents with organic compounds can provide evidence of certain structure aspects, like the presence of methyl ketones.
The key to successful chemical reaction analysis is understanding the principles and conditions under which reactions occur:

• Reaction conditions need to be controlled and known, such as temperature, solvent, and pH.
• Understanding the mechanism, such as nucleophilic substitution in the iodoform reaction, is essential.

The insight gained from chemical reaction analysis allows chemists to predict the behavior of organic compounds under similar conditions in different settings.
Reaction analysis not only supports laboratory experiments but also broadens the understanding of reaction pathways and mechanisms, strengthening one's ability to conduct and interpret experiments effectively.