Draw the mechanism of the following reaction, using the curved-arrow notation to indicate the reorganization of electron density. Denote all intermediates, lone pairs, nonzero formal charges, countercharges, and reversibility or nonreversibility. Finally, explain concisely why a mixture of the stereoisomers 2 and 3 is obtained. 2H-Br: .. `CH3 `CH3 :OH + `CH3 :Br: 3 + + H-OH, :Br:
Draw the mechanism of the following reaction, using the curved-arrow notation to indicate the reorganization of electron density. Denote all intermediates, lone pairs, nonzero formal charges, countercharges, and reversibility or nonreversibility. Finally, explain concisely why a mixture of the stereoisomers 2 and 3 is obtained. 2H-Br: .. `CH3 `CH3 :OH + `CH3 :Br: 3 + + H-OH, :Br:
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Reactions of Ethers
Ethers (R-O-R’) are compounds formed by replacing hydrogen atoms of an alcohol (R-OH compound) or a phenol (C6H5OH) by an aryl/ acyl group (functional group after removing single hydrogen from an aromatic ring). In this section, reaction, preparation and behavior of ethers are discussed in the context of organic chemistry.
Epoxides
Epoxides are a special class of cyclic ethers which are an important functional group in organic chemistry and generate reactive centers due to their unusual high reactivity. Due to their high reactivity, epoxides are considered to be toxic and mutagenic.
Williamson Ether Synthesis
An organic reaction in which an organohalide and a deprotonated alcohol forms ether is known as Williamson ether synthesis. Alexander Williamson developed the Williamson ether synthesis in 1850. The formation of ether in this synthesis is an SN2 reaction.
Question
Transcribed Image Text:**Reaction Mechanism Description:**
This reaction involves the conversion of a cycloalkanol to a bromoalkane via the addition of hydrogen bromide (HBr) radicals.
**Mechanism Steps:**
1. **Starting Material:**
- The reactant is a cycloalkanol with an additional ethyl group (CH₃).
2. **Reagents:**
- Two equivalents of HBr radicals are involved in the reaction.
3. **Products:**
- The reaction yields two stereoisomers, labeled as product 2 and product 3, both of which are bromoalkanes derived from the starting alcohol. The Br atom replaces the OH group in the original structure.
- A water molecule and a bromine radical are also formed as byproducts.
**Detailed Explanation:**
- **Curved-Arrow Notation:**
- The reaction mechanism likely involves a radical substitution process where the oxygen in the original alcohol group is replaced by a bromine radical. Curved arrows are used in the mechanism to depict the movement of electron pairs, including the formation and breaking of bonds.
- **Stereochemistry:**
- The presence of two stereoisomers indicates that the bromine can attach to either face of the planar intermediate, leading to the formation of different configurations (cis/trans or R/S stereochemistry).
- **Intermediates and Charges:**
- The mechanism requires indicating all intermediates, including any carbocations if formed, as well as the formal charges associated with each step.
- **Reversibility:**
- The description should also account for whether the steps are reversible or irreversible, which can affect the final mixture of stereoisomers.
This explanation outlines the transformation and stability of intermediates, showing why multiple products are obtained. Such mechanisms are crucial for understanding reaction pathways in organic chemistry.
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