Consider the reaction below to answer the following questions. When dichlorocarbene is generated in the presence of an alkene, a dichlorocyclopropane is formed. Cl Ã. H H of dichlorocarbene, :CCl₂. Show all H₂C 1 H CH3 1 + CHCl3 C=C H КОН. H₂C-- Write the complete stepwise mechanism for the formation intermediate structures and show all electron flow with arrows. CH3
Consider the reaction below to answer the following questions. When dichlorocarbene is generated in the presence of an alkene, a dichlorocyclopropane is formed. Cl Ã. H H of dichlorocarbene, :CCl₂. Show all H₂C 1 H CH3 1 + CHCl3 C=C H КОН. H₂C-- Write the complete stepwise mechanism for the formation intermediate structures and show all electron flow with arrows. CH3
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![**Reaction Mechanism for Formation of Dichlorocarbene**
**Reaction Overview:**
The reaction involves the generation of dichlorocarbene in the presence of an alkene. The reaction leads to the formation of a dichlorocyclopropane compound.
**Chemical Reaction:**
\[ \text{H}_3\text{C} - \text{CH} = \text{CH} - \text{CH}_3 + \text{CHCl}_3 \xrightarrow{\text{KOH}} \]
This transforms into a dichlorocyclopropane structure with the incorporation of a dichlorocarbene (:CCl₂).
**Detailed Explanation:**
1. **Starting Materials:**
- An alkene (2-butene): \(\text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CH}_3\).
- Chloroform (\(\text{CHCl}_3\)) in the presence of potassium hydroxide (\(\text{KOH}\)).
2. **Formation of Dichlorocarbene:**
- \(\text{KOH}\) deprotonates chloroform, forming a carbanion intermediate.
- Loss of a chloride ion produces the active dichlorocarbene species, :CCl₂.
3. **Reaction with Alkene:**
- The dichlorocarbene (:CCl₂) inserts into the double bond of the alkene.
- This results in the formation of a cyclopropane ring with two chlorine atoms attached.
4. **Final Product:**
- A dichlorocyclopropane structure, where the previously double-bonded carbons are now part of a three-membered ring.
**Visualization of Mechanism:**
In the diagram:
- The alkene, \(\text{H}_3\text{C} - \text{CH} = \text{CH} - \text{CH}_3\), reacts with \(\text{CHCl}_3\) under basic conditions.
- The resulting product is a cyclopropane ring with two chlorine atoms situated on one of the carbons in the three-membered ring.
The diagram indicates the conversion of the alkene to the cyclopropane with arrows showing electron flow during the formation of the dichlorocyclopropane. Each step in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17240ec3-a1b6-433b-936a-3fd00ad8499d%2F8ef678e4-5548-4aa3-8085-a67d6c5a6d57%2Fn6u339m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Reaction Mechanism for Formation of Dichlorocarbene**
**Reaction Overview:**
The reaction involves the generation of dichlorocarbene in the presence of an alkene. The reaction leads to the formation of a dichlorocyclopropane compound.
**Chemical Reaction:**
\[ \text{H}_3\text{C} - \text{CH} = \text{CH} - \text{CH}_3 + \text{CHCl}_3 \xrightarrow{\text{KOH}} \]
This transforms into a dichlorocyclopropane structure with the incorporation of a dichlorocarbene (:CCl₂).
**Detailed Explanation:**
1. **Starting Materials:**
- An alkene (2-butene): \(\text{H}_3\text{C}-\text{CH}=\text{CH}-\text{CH}_3\).
- Chloroform (\(\text{CHCl}_3\)) in the presence of potassium hydroxide (\(\text{KOH}\)).
2. **Formation of Dichlorocarbene:**
- \(\text{KOH}\) deprotonates chloroform, forming a carbanion intermediate.
- Loss of a chloride ion produces the active dichlorocarbene species, :CCl₂.
3. **Reaction with Alkene:**
- The dichlorocarbene (:CCl₂) inserts into the double bond of the alkene.
- This results in the formation of a cyclopropane ring with two chlorine atoms attached.
4. **Final Product:**
- A dichlorocyclopropane structure, where the previously double-bonded carbons are now part of a three-membered ring.
**Visualization of Mechanism:**
In the diagram:
- The alkene, \(\text{H}_3\text{C} - \text{CH} = \text{CH} - \text{CH}_3\), reacts with \(\text{CHCl}_3\) under basic conditions.
- The resulting product is a cyclopropane ring with two chlorine atoms situated on one of the carbons in the three-membered ring.
The diagram indicates the conversion of the alkene to the cyclopropane with arrows showing electron flow during the formation of the dichlorocyclopropane. Each step in
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