Problem 124 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound 'A' reacts with methyl magnesium iodide, to form an addition product which on hydrolysis forms the compound 'B'. Compound 'B' gives blue colour salt in Victor Meyer's test. The compounds ' \(\mathrm{A}\) ' and ' \(\mathrm{B}\) ' are respectively (a) acetaldehyde and isopropyl alcohol (b) acetone and isopropyl alcohol (c) acetaldehyde and tertiary butyl alcohol (d) acetaldehyde and ethyl alcohol

Step-by-Step Solution

Answer

(b) acetone and isopropyl alcohol

Step 1: Understand the Reaction Description

The problem describes a reaction sequence: an organic compound 'A' reacts with methyl magnesium iodide to produce a new compound. This new compound, upon hydrolysis, becomes compound 'B'. Compound 'B' gives a blue color in Victor Meyer's test.

Step 2: Recall Victor Meyer's Test

Victor Meyer's test is used to distinguish between primary, secondary, and tertiary alcohols. Secondary alcohols typically give a blue color in Victor Meyer's test, indicating that compound 'B' is a secondary alcohol.

Step 3: Identify the Possible Alcohol Formed

Given that the addition product from methyl magnesium iodide is a secondary alcohol after hydrolysis, compound 'B' must be isopropyl alcohol, which indeed gives a positive blue result in Victor Meyer's test.

Step 4: Determine Compound 'A'

Compound 'A' must be a compound that can react with methyl magnesium iodide to form a secondary alcohol upon hydrolysis. For isopropyl alcohol, compound 'A' must be acetone, as it reacts with methyl magnesium iodide to form the appropriate Grignard addition product, which on hydrolysis gives isopropyl alcohol.

Step 5: Match with Given Options

Review the options provided: (a) acetaldehyde and isopropyl alcohol, (b) acetone and isopropyl alcohol, (c) acetaldehyde and tertiary butyl alcohol, and (d) acetaldehyde and ethyl alcohol. Based on our analysis, option (b) acetone and isopropyl alcohol correctly fits the description of A and B.

Key Concepts

Victor Meyer's Test

Victor Meyer's test is a classic organic chemistry method used to differentiate between primary, secondary, and tertiary alcohols by observing the color change of the resultant product. When subjected to this test, each type of alcohol yields a distinct color:

• Primary alcohols produce a red color.
• Secondary alcohols lead to a blue color, as seen in this exercise with isopropyl alcohol.
• Tertiary alcohols result in a colorless or slightly cloudy solution.

During the test, the alcohol is first converted into the corresponding alkyl iodide, which then undergoes dehydrohalogenation to give the corresponding alkene. This reaction triggers a series of transformations ultimately producing a colored compound, which allows for the easy characterization of the alcohol's type. In our exercise, the presence of the blue color indicates that the alcohol formed, compound 'B', is a secondary alcohol.

Secondary Alcohol

A secondary alcohol is a type of alcohol where the hydroxyl group (-OH) is bonded to a carbon atom that is connected to two other carbon atoms. This structure is what gives secondary alcohols their unique reactivity patterns compared to primary and tertiary alcohols.

An example of a secondary alcohol is isopropyl alcohol. In the context of the exercise, isopropyl alcohol is formed from the hydrolysis of the Grignard reaction product between acetone and methyl magnesium iodide. Chemically, secondary alcohols are often less reactive than primary alcohols in oxidation reactions, and they can be oxidized to form ketones. Being a secondary alcohol, isopropyl alcohol's positive results in Victor Meyer’s test (blue color) confirms its identity conclusively as compound 'B' in the problem.

Acetone

Acetone is a simple and well-known ketone with the chemical formula CH$_3$COCH$_3$. It is a common solvent and a significant starting material in organic chemical reactions, including those involving Grignard reagents. As a ketone, it has a carbonyl group (C=O) bonded to two carbon atoms.

In the provided exercise, acetone acts as compound 'A'. When it reacts with the Grignard reagent, methyl magnesium iodide, an addition product is formed. Upon hydrolysis of this product, a secondary alcohol—specifically isopropyl alcohol—is produced. This process showcases acetone's versatility as a reactant. It efficiently forms the desired Grignard addition product, making it a foundational compound for synthesizing secondary alcohols in laboratory settings. Its role in the sequence ultimately establishes the precursor relationship with isopropyl alcohol, demonstrating its importance in organic synthesis.