Treating 2-methyl-1-butene with methanol in the presence of sulfuric acid gives 2- methoxy-2-methylbutane. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism.
Treating 2-methyl-1-butene with methanol in the presence of sulfuric acid gives 2- methoxy-2-methylbutane. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism.
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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![**Understanding the Mechanism of Ether Formation from 2-Methyl-1-Butene and Methanol**
In this tutorial, we will explore the mechanism of the formation of an ether compound when 2-methyl-1-butene reacts with methanol in the presence of sulfuric acid.
**Reaction Overview:**
The chemical equation for this reaction is:
\[ \text{CH}_3 \text{CH}_2 C \equiv\text{CH}_2 + \text{CH}_3 \text{OH} \xrightarrow{H_2SO_4} \text{CH}_3 \text{CH}_2 \text{C} \equiv \text{OCH}_3 \text{CH}_3 \]
This reaction treats 2-methyl-1-butene with methanol in the presence of sulfuric acid to yield 2-methoxy-2-methylbutane.
**Detailed Mechanism:**
To understand the movement of electrons during the reaction mechanism, we use curved arrows to show the electron flow. In this step, the movement is illustrated as follows:
1. **Formation of Intermediate:**
- Initially, methanol (\( \text{CH}_3 \text{OH} \)) reacts in the presence of sulfuric acid (\( \text{H}_2\text{SO}_4 \)), forming a protonated alcohol.
2. **Arrow-pushing Instructions:**
- The diagram includes instructions on how to draw curved arrows. Curved arrows indicate the flow of electrons:
a. The arrow originates from the electron source (e.g., lone pairs or pi bonds).
b. It points to where the electrons are moving.
Below the instructions, we have a visual representation of the molecules involved:
- On the left, the molecule is a protonated methanol (\( \text{H}_3 \text{COH}^+ \)).
- On the right, the product is the methoxy compound (\( \text{H}_3\text{COH} \)).
In this step, the arrows depict the electron lone pair of oxygen attacking the carbon electrophile, leading to the formation of the product.
Through these instructions and visualizations, students can gain a deeper understanding of the stepwise process of this important ether formation reaction, facilitating a clearer grasp of electron movement and molecular transformations in organic chemistry.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b49abb7-22b1-43f4-8ea9-ededd2a6e7a5%2F0762d51c-0536-43fd-9ec9-d48588bf2b46%2F9d4qncf_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding the Mechanism of Ether Formation from 2-Methyl-1-Butene and Methanol**
In this tutorial, we will explore the mechanism of the formation of an ether compound when 2-methyl-1-butene reacts with methanol in the presence of sulfuric acid.
**Reaction Overview:**
The chemical equation for this reaction is:
\[ \text{CH}_3 \text{CH}_2 C \equiv\text{CH}_2 + \text{CH}_3 \text{OH} \xrightarrow{H_2SO_4} \text{CH}_3 \text{CH}_2 \text{C} \equiv \text{OCH}_3 \text{CH}_3 \]
This reaction treats 2-methyl-1-butene with methanol in the presence of sulfuric acid to yield 2-methoxy-2-methylbutane.
**Detailed Mechanism:**
To understand the movement of electrons during the reaction mechanism, we use curved arrows to show the electron flow. In this step, the movement is illustrated as follows:
1. **Formation of Intermediate:**
- Initially, methanol (\( \text{CH}_3 \text{OH} \)) reacts in the presence of sulfuric acid (\( \text{H}_2\text{SO}_4 \)), forming a protonated alcohol.
2. **Arrow-pushing Instructions:**
- The diagram includes instructions on how to draw curved arrows. Curved arrows indicate the flow of electrons:
a. The arrow originates from the electron source (e.g., lone pairs or pi bonds).
b. It points to where the electrons are moving.
Below the instructions, we have a visual representation of the molecules involved:
- On the left, the molecule is a protonated methanol (\( \text{H}_3 \text{COH}^+ \)).
- On the right, the product is the methoxy compound (\( \text{H}_3\text{COH} \)).
In this step, the arrows depict the electron lone pair of oxygen attacking the carbon electrophile, leading to the formation of the product.
Through these instructions and visualizations, students can gain a deeper understanding of the stepwise process of this important ether formation reaction, facilitating a clearer grasp of electron movement and molecular transformations in organic chemistry.
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