**Title:** Reaction Mechanism and Product Formation**Objective:** Draw the complete mechanism and product of the following reaction.**Reactants:**The starting compound is a bicyclic ketone with two cyclopropane rings attached to a five-carbon chain.**Reagents:**1. Sodium methoxide (NaOCH₂CH₃) in tetrahydrofuran (THF)2. Bromo-substituted cyclopropane**Instructions:**Examine the structure of the starting ketone compound and consider the reaction conditions provided by the reagents. Determine the product formed after the complete reaction mechanism using the given reagents.**Analysis:**1. **Nucleophilic Attack:** - Sodium methoxide acts as a strong nucleophile in the presence of THF. - It's likely to attack the electrophilic carbon in the starting compound.2. **Intermediate Formation:** - Analyze potential intermediate structures that could form after the initial nucleophilic attack. 3. **Elimination and Cyclization:** - The bromo-substituted cyclopropane serves as an electrophilic partner. - Consider any possible elimination or cyclization steps that may result in the final product.**Expected Outcome:**Sketch the complete mechanism, indicating all intermediates, leaving groups, and electron flow throughout the reaction. Finally, draw the anticipated product structure.**Educational Insight:**This exercise demonstrates key principles in organic chemistry, including nucleophilic substitution, reaction mechanisms, and product prediction. Understanding these concepts is essential for mastering synthetic organic chemistry.

Answered: Draw the complete mechanism and product…

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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**Title:** Reaction Mechanism and Product Formation

**Objective:** Draw the complete mechanism and product of the following reaction.

**Reactants:**

The starting compound is a bicyclic ketone with two cyclopropane rings attached to a five-carbon chain.

**Reagents:**

1. Sodium methoxide (NaOCH₂CH₃) in tetrahydrofuran (THF)
2. Bromo-substituted cyclopropane

**Instructions:**

Examine the structure of the starting ketone compound and consider the reaction conditions provided by the reagents. Determine the product formed after the complete reaction mechanism using the given reagents.

**Analysis:**

1. **Nucleophilic Attack:**
- Sodium methoxide acts as a strong nucleophile in the presence of THF.
- It's likely to attack the electrophilic carbon in the starting compound.

2. **Intermediate Formation:**
- Analyze potential intermediate structures that could form after the initial nucleophilic attack.

3. **Elimination and Cyclization:**
- The bromo-substituted cyclopropane serves as an electrophilic partner.
- Consider any possible elimination or cyclization steps that may result in the final product.

**Expected Outcome:**

Sketch the complete mechanism, indicating all intermediates, leaving groups, and electron flow throughout the reaction. Finally, draw the anticipated product structure.

**Educational Insight:**

This exercise demonstrates key principles in organic chemistry, including nucleophilic substitution, reaction mechanisms, and product prediction. Understanding these concepts is essential for mastering synthetic organic chemistry.

Expert Solution

Introduction:

In this Reaction alkylation is occurring at alpha hydrogen of ketone by using base (NaOC₂H₅) and alkyl halide .this is an important reaction for carbon-carbon bond formation. here cyclopropanemethyl group is attached at alpha carbon of ketone.

Mechanism:

Step 1:

In this step in the presence of base sodium ethoxide alpha hydrogen of the ketone is removed and enolate ion is formed.

Step 2:

Enolate act as a nucleophile In SN² reaction .where pi bond of enolate ion attack at the carbon of cyclopropyl bromide, here through the nucleophilic substitution pi bond of enolate ion attack at the carbon of isopropyl bromide and bromide ion is removed result of this carbon carbon single bond is formed. it is important to use a strong base, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylations occurring.

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