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When grignard reagent is treated with isopropyl formate followed by acid hydrolysis we get : (1) Aldehyde (2) \(2^{\circ}\) alcohol (3) \(3^{\circ}\) alcohol (4) \(1^{\circ}\) alcohol

Short Answer

Expert verified
Option (3): tert-alcohol

Step by step solution

01

- Understand the Reaction

Grignard reagents react with esters to form tertiary alcohols. Isopropyl formate is an ester with the structure HCOOCH(CH₃)₂.
02

- First Addition

The Grignard reagent adds to the carbonyl carbon of the ester to form an intermediate alkoxide. This intermediate then forms a ketone after elimination of the leaving group.
03

- Second Addition

A second mole of the Grignard reagent adds to the newly formed ketone, resulting in the formation of a tertiary alkoxide.
04

- Acid Hydrolysis

Hydrolyze the tertiary alkoxide in the presence of an acid to obtain the tertiary alcohol as the final product.
05

- Identify the Product

Combining the information, isopropyl formate treated with a Grignard reagent followed by acid hydrolysis yields a tertiary alcohol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Grignard Reagent
Grignard reagents are vital in organic chemistry and are typically represented as RMgX, where 'R' is an alkyl or aryl group, and 'X' is a halogen, usually bromine or iodine. These reagents are immensely useful due to their ability to add to carbonyl compounds.
Grignard reagents are highly nucleophilic, making them excellent for attacking electrophilic centers such as carbonyl carbon atoms. Cold ether is typically used as a solvent for these reactions to maintain the stability of the Grignard reagent.
When a Grignard reagent interacts with esters, it effectively adds to the carbonyl group of the ester. After the first addition, the ester undergoes a process where the leaving group is eliminated, leading to the formation of a ketone. This intermediate ketone can then react further with another equivalent of the Grignard reagent.
Understanding the basic properties and reactivity of Grignard reagents is crucial for mastering organic chemistry reactions and mechanisms.
Tertiary Alcohol Formation
When Grignard reagents react with esters, a tertiary alcohol is typically formed through a series of steps. Here’s a detailed breakdown:
First, the Grignard reagent adds to the carbonyl carbon of the ester, forming an intermediate alkoxide compound. This intermediate is relatively unstable and quickly eliminates an alcohol group to transform into a more stable ketone.
This ketone then undergoes a second addition of the Grignard reagent, leading to the formation of a tertiary alkoxide. The entire process is facilitated by the high reactivity of Grignard reagents toward carbonyl compounds.
The final step involves acid hydrolysis of the tertiary alkoxide, where water and an acid (typically HCl or H2SO4) are added to convert the alkoxide to a stable tertiary alcohol.
Recognizing these steps is essential for predicting and understanding the products of Grignard reagent reactions with esters.
Organic Chemistry Reactions
Organic chemistry reactions involve numerous mechanisms and steps that can transform functional groups into one another. The reaction of Grignard reagents with esters to form tertiary alcohols is a prime example.
Organic chemistry hinges on the understanding of how different compounds interact, particularly under the influence of reagents like the Grignard. These reagents are powerful in converting esters into more complex alcohol structures.
The knowledge of intermediary steps, such as forming alkoxides and ketones before achieving the final product, deepens the comprehension of reaction mechanisms.
Mastering these reactions enables chemists to synthesize a variety of compounds, broadening the scope of potential chemical products and applications.
A solid grasp of these concepts also helps in various fields such as pharmaceuticals, material science, and biotechnology, where understanding and creating complex organic molecules are indispensable.

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Most popular questions from this chapter

Which of the following statements does not correct from the point of view of molecular orbital? (1) \(\mathrm{Be}_{2}\) is not a stable molecule (2) \(\mathrm{He}_{2}\) is not stable but \(\mathrm{He}^{+}\)is expected to exist (3) Bond strength of \(\mathrm{N}_{2}\) is maximum amongst the homonuclear diatomic molecules (4) The order of energies of molecular orbitals in \(\mathrm{O}_{2}\) molecule is \(E(\sigma 2 s)

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Let \(n \geq 3\). A list of numbers \(0

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