HO CH3 HO HO 1L но HO OMe 2. HO Proton Transfer но HO. HO. HO CH3OH, H* HO curved-arrow mechanism for this transformation: 1. When glucose is treated with methanol and acid, the methyl glucoside is formed. Provide a complete HO, но Carbohydrates

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...
icon
Related questions
Question
Could you please check if this is right and elaborate the mechanism?
### Carbohydrates

#### 1. Mechanism for Methyl Glucoside Formation from Glucose and Methanol

When glucose is treated with methanol and acid, the methyl glucoside is formed. Below is the complete curved-arrow mechanism for this transformation:

1. **Starting Compound (Glucose)**: 
    - The structure is a glucose molecule with hydroxyl (-OH) groups attached to carbon atoms 2, 3, 4, and 6. An ether group (-OCH3) is attached to carbon atom 1, forming a hemiacetal structure.

2. **Step 1: Protonation of the Hydroxyl Group**:
    - The hydroxyl group on the first carbon is protonated by H⁺ (acid), facilitating the conversion of the -OH to a better leaving group.

3. **Step 2: Loss of Water (Elimination)**:
    - The protonated hydroxyl group (now a water molecule, H₂O) leaves, forming a carbocation on the first carbon.

4. **Step 3: Nucleophilic Attack by Methanol**:
    - Methanol (CH₃OH) acts as a nucleophile and attacks the carbocation on the first carbon. This involves a curved arrow showing lone pair electrons from the oxygen in methanol attacking the carbocation.

5. **Step 4: Proton Transfer**:
    - A proton (H⁺) transfers from the oxygen in the -OCH₃ group, resulting in the final product, methyl glucoside.

6. **Final Product (Methyl Glucoside)**:
    - The final product is a cyclic molecule where a methoxy group (-OCH₃) is attached to the first carbon, and hydroxyl groups are attached to the 2nd, 3rd, 4th, and 6th carbons respectively.

### Detailed Explanation of the Diagrams:

- **Left Diagram (Steps 1 & 2)**:
    - Shows the initial protonation of the hydroxyl group on carbon 1 and the subsequent loss of a water molecule.
    - The arrow from the hydrogen to the hydroxyl group indicates protonation, and the arrow from the oxygen in the leaving water molecule indicates elimination.

- **Right Diagram (Steps 3 & 4)**:
    - Demonstrates the nucleophilic attack by methanol and the proton transfer process.
    - Arrows depict
Transcribed Image Text:### Carbohydrates #### 1. Mechanism for Methyl Glucoside Formation from Glucose and Methanol When glucose is treated with methanol and acid, the methyl glucoside is formed. Below is the complete curved-arrow mechanism for this transformation: 1. **Starting Compound (Glucose)**: - The structure is a glucose molecule with hydroxyl (-OH) groups attached to carbon atoms 2, 3, 4, and 6. An ether group (-OCH3) is attached to carbon atom 1, forming a hemiacetal structure. 2. **Step 1: Protonation of the Hydroxyl Group**: - The hydroxyl group on the first carbon is protonated by H⁺ (acid), facilitating the conversion of the -OH to a better leaving group. 3. **Step 2: Loss of Water (Elimination)**: - The protonated hydroxyl group (now a water molecule, H₂O) leaves, forming a carbocation on the first carbon. 4. **Step 3: Nucleophilic Attack by Methanol**: - Methanol (CH₃OH) acts as a nucleophile and attacks the carbocation on the first carbon. This involves a curved arrow showing lone pair electrons from the oxygen in methanol attacking the carbocation. 5. **Step 4: Proton Transfer**: - A proton (H⁺) transfers from the oxygen in the -OCH₃ group, resulting in the final product, methyl glucoside. 6. **Final Product (Methyl Glucoside)**: - The final product is a cyclic molecule where a methoxy group (-OCH₃) is attached to the first carbon, and hydroxyl groups are attached to the 2nd, 3rd, 4th, and 6th carbons respectively. ### Detailed Explanation of the Diagrams: - **Left Diagram (Steps 1 & 2)**: - Shows the initial protonation of the hydroxyl group on carbon 1 and the subsequent loss of a water molecule. - The arrow from the hydrogen to the hydroxyl group indicates protonation, and the arrow from the oxygen in the leaving water molecule indicates elimination. - **Right Diagram (Steps 3 & 4)**: - Demonstrates the nucleophilic attack by methanol and the proton transfer process. - Arrows depict
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Knowledge Booster
Sample Preparation in Analytical Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY