Problem 106 - A Complete Resource Book in Chemistry for JEE Main

Question

The reaction of \(\mathrm{CH}_{3} \mathrm{MgBr}\) with acetone and hydrolysis of the resulting product gives (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

The product is tert-butyl alcohol: (a) \((\mathrm{CH}_3)_3\mathrm{COH}\).

Step 1: Understand the Reaction Components

The exercise involves the reaction of a Grignard reagent, \( \mathrm{CH}_{3} \mathrm{MgBr} \), with acetone \( \mathrm{(CH_3)_2CO} \). This reaction typically forms a tertiary alcohol after hydrolysis. Grignard reagents add the alkyl group (here \( \mathrm{CH}_3 \)) to the carbonyl carbon of ketones like acetone.

Step 2: Analyze the Grignard Reaction with Acetone

When \( \mathrm{CH}_3 \mathrm{MgBr} \) reacts with acetone \( (\mathrm{(CH_3)_2CO}) \), the \( \mathrm{CH}_3 \) group from \( \mathrm{MgBr} \) adds to the carbonyl carbon of acetone, breaking the \( C=O \) double bond and forming an alkoxide intermediate, \( \mathrm{(CH_3)_3CO^-} \).

Step 3: Perform Hydrolysis of the Alkoxide Intermediate

The alkoxide intermediate \( \mathrm{(CH_3)_3CO^-} \) undergoes hydrolysis, usually via an acidic work-up. During hydrolysis, a proton \( (H^+) \) is added, resulting in the formation of the tertiary alcohol \( \mathrm{(CH_3)_3COH} \).

Step 4: Identify the Product

The product of the reaction following hydrolysis is \( \mathrm{(CH_3)_3COH} \), which is a tertiary alcohol known as tert-butyl alcohol. Therefore, the correct option is (a) \( \left(\mathrm{CH}_3\right)_3\mathrm{COH} \).

Key Concepts

Tertiary Alcohol

In organic chemistry, a tertiary alcohol is an important class of alcohols. It occurs when the hydroxyl group (-OH) is attached to a carbon that is bonded to three other carbon atoms. This setup is what defines a tertiary alcohol and differentiates it from primary or secondary alcohols.

Tertiary alcohols are often created using Grignard reagents in reactions like the one presented in the exercise. For example, when the Grignard reagent methyllithium (\( \mathrm{CH}_3 \mathrm{MgBr} \)) reacts with acetone, a new carbon-carbon bond is formed, ultimately leading to the structure of tert-butyl alcohol (\( \mathrm{(CH_3)_3COH} \)).

This reaction is quite efficient for forming carbon-carbon bonds, making tertiary alcohol synthesis a significant technique in organic synthesis. Ultimately, understanding how tertiary alcohols are formed helps in designing reactions for various organic compounds.

Alkoxide Intermediate

An alkoxide intermediate is a transient species in reactions involving the addition of a Grignard reagent. It is essential in leading to the formation of alcohols. This intermediate is formed when the Grignard reagent donates an alkyl group to the carbonyl carbon in ketones or aldehydes.

In our specific reaction, the intermediate formed is \( \mathrm{(CH_3)_3CO^-} \). This happens when the \( \mathrm{CH}_3 \) group from \( \mathrm{CH}_3 \mathrm{MgBr} \) attacks the carbonyl carbon of acetone (\( \mathrm{(CH_3)_2CO} \)), disrupting the double bond to form a new bond. The oxygen atom then carries a negative charge, creating the alkoxide ion.

Recognizing and understanding the role of the alkoxide intermediate is crucial because it acts as a precursor to the final alcohol product. After its formation, this intermediate is typically exposed to a hydrolysis process to convert it into a stable alcohol.

Acetone

Acetone, a common compound in organic chemistry, is a simple ketone with the formula \( \mathrm{(CH_3)_2CO} \). It has a carbonyl group (C=O), which makes it highly reactive, especially with nucleophiles like Grignard reagents. When acetone is used in reactions with such reagents, it can facilitate the formation of alcohols.

In the context of the Grignard reaction with acetone, acetone acts as the electrophile with its carbonyl carbon being attacked by the nucleophilic alkyl group from \( \mathrm{CH}_3 \mathrm{MgBr} \). This results in the creation of a new compound via the addition of the \( \mathrm{CH}_3 \) group to the acetone molecule.

Acetone is not only valuable for its role in synthesis but also widely used as a solvent. Its properties make it effective for reactions requiring the formation of tertiary alcohols. Understanding acetone’s reactivity is crucial for organic chemists when designing synthesis pathways.

Hydrolysis

Hydrolysis is a chemical process that involves the breaking of bonds using water. It is a significant step in many organic reactions, including the reaction detailed in the exercise. In a Grignard reaction, hydrolysis is necessary to convert the alkoxide intermediate into a stable alcohol.

After the formation of the alkoxide intermediate \( \mathrm{(CH_3)_3CO^-} \) in our reaction, hydrolysis is needed to add a proton (\( H^+ \)) to the negatively charged oxygen, resulting in the formation of the product tertiary alcohol \( \mathrm{(CH_3)_3COH} \). Typically, this is achieved through an acidic work-up, which adds the necessary protons from an acid to stabilize the new alcohol.

The process of hydrolysis is essential and serves as the final step in many synthesis reactions. It ensures the conversion of intermediates into functional end products that can be further utilized or studied. Recognizing the importance of hydrolysis in transforming reactive intermediates to stable compounds is a key aspect of mastering organic chemistry.