Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

To prepare 3 -ethylpentan-3-ol, the reagents needed are (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}+\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\)

Step-by-Step Solution

Answer

Option (d) is correct: \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr} + \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3} \).

Step 1: Understand the Target Molecule

To develop the correct synthesis for 3-ethylpentan-3-ol, note that it is a tertiary alcohol with a specific carbon structure. Its structure can be visualized as: \[ \begin{align*} \text{CH}_3- (\text{CH}_2)_2- \text{C}(&\text{CH}_3)&(\text{OH})- \text{CH}_2- \text{CH}_3 \end{align*} \] This is a six-carbon chain where the third carbon bears both a hydroxyl group and an ethyl group.

Step 2: Determine the Grignard Reagent Needed

A tertiary alcohol can be formed by the reaction of a ketone with a Grignard reagent. The ketone used should have one of the alkyl groups that will be present in the alcohol. Since 3-ethylpentan-3-ol has an ethyl group at the tertiary carbon, the Grignard reagent should also contain an ethyl group. Therefore, choose ethyl magnesium bromide (\( \text{CH}_{3} \text{CH}_2 \text{MgBr} \)).

Step 3: Identify the Correct Ketone

For the formation of 3-ethylpentan-3-ol, the ketone must have a carbonyl carbon where the Grignard reagent can attack and add the ethyl group to form the desired alcohol. The ketone should have a five-carbon skeleton that's one carbon shorter than the product (since the sixth carbon will come from the Grignard reagent), such as pentan-2-one ( \( \text{CH}_3 \text{COCH}_2 \text{CH}_3 \)).

Step 4: Match Ketenone and Grignard Reagent with Given Options

Looking at the options given, we need the combination of \( \text{CH}_3 \text{CH}_2 \text{MgBr} \) (ethyl magnesium bromide) with a ketone that has one methyl group and a longer chain on its other side. The correct ketone is \( \text{CH}_3 \text{COCH}_2 \text{CH}_3 \), which leads to option (d): \( \text{CH}_3 \text{CH}_2 \text{MgBr} + \text{CH}_3 \text{COCH}_2 \text{CH}_3 \).

Key Concepts

Tertiary Alcohol Synthesis

In organic chemistry, the synthesis of tertiary alcohols is a crucial process with distinct steps. A tertiary alcohol features a hydroxyl group \((\text{-OH})\) attached to a carbon atom that is itself connected to three other carbon atoms. This structure affects its reactivity due to steric hindrance, which is the resistance to reactions because the molecule is so tightly packed with groups.

To synthesize such alcohols, one common approach is using a Grignard reaction. This involves a Grignard reagent reacting with a carbonyl compound, typically a ketone, to form an alcohol. The Grignard reagent contributes the new carbon linkage, creating a new alcohol. In our case, the task is to form 3-ethylpentan-3-ol, a specific tertiary alcohol.

The structure of 3-ethylpentan-3-ol is such that it has a six-carbon backbone with an ethyl group \((\text{CH}_2\text{CH}_3)\) and a hydroxyl group both attached to the same carbon. This carbon, known as the tertiary carbon, is linked to three different carbon chains. This synthesis relies on knowing that the final product characteristics stem from the specific initial reagents used along with targeted reaction conditions.

Organic Chemistry Reagents

Grignard reagents are versatile and highly reactive compounds used in organic chemistry to form carbon-carbon bonds. They consist of an organomagnesium halide, typically represented as \(\text{R-MgX}\), where \(\text{R}\) is an alkyl or aryl group, and \(\text{X}\) is a halogen. These reagents play a significant role in constructing new molecules.

Grignard reagents, such as ethyl magnesium bromide \((\text{CH}_3\text{CH}_2\text{MgBr})\), are incredibly reactive due to the polar bond between the carbon and magnesium. This polar bond gives carbon a partial negative charge, enhancing its nucleophilic nature and enabling it to attack electrophilic carbon atoms found in carbonyl groups of ketones or aldehydes.

This reaction will create a new carbon-carbon bond, pivotal for forming the backbone of the desired alcohol. Handling these reagents requires specific conditions, typically under anhydrous (water-free) environments, as Grignard reagents react violently with water, deactivating them. Through controlled use, various complex molecules can be synthesized.

Ketone and Alcohol Reactions

The transformation of ketones to alcohols is a fundamental organic reaction. Keystones in this transformation are ketones' structure, containing a carbonyl group \((-C=O-)\), which is highly reactive. This reactivity allows for interactions with nucleophiles such as Grignard reagents.

In the case of 3-ethylpentan-3-ol synthesis, the ketone used is pentan-2-one \((\text{CH}_3\text{COCH}_2\text{CH}_3)\). The structure of pentan-2-one provides the necessary carbonyl for nucleophilic attack by the Grignard reagent. Upon attack, the carbon in the carbonyl group undergoes hybridization changes, leading to the formation of an alcohol after subsequent protonation (adding a hydrogen).

This reaction is crucial in developing tertiary alcohols because the introduction of the Grignard reagent contributes an additional carbon group, creating a three-carbon linkage at the site of the original carbonyl group. This step effectively utilizes the ketone’s properties to lay the foundation for the alcohol’s tertiary structure, one that defines their stability and reactivity profile in subsequent chemical applications.