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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**Structure:**
\(\displaystyle \text{---OH}\)
2. **Reactant B**:
A molecular structure consisting of a zig-zag chain with a -Cl (chlorine) group at the end.
**Structure:**
\(\displaystyle \text{---Cl}\)
#### Reagents:
- Sodium Hydride (\(\displaystyle \text{NaH}\))
- Dimethyl sulfoxide (\(\displaystyle \text{DMSO}\))
#### Explanation:
The provided reaction involves two organic reactants and two inorganic reagents. Here’s a step-by-step process to determine the product:
1. **NaH (Sodium Hydride)** is typically used as a strong base.
2. **DMSO (Dimethyl sulfoxide)** is a common solvent that can facilitate the reaction.
Initially, sodium hydride (\(\displaystyle \text{NaH}\)) will deprotonate the hydroxyl group of Reactant A, generating an alkoxide ion. Once the alkoxide ion is formed, it can act as a nucleophile in a nucleophilic substitution reaction (likely an \( \text{S}_\text{N}2 \) mechanism), attacking the carbon attached to the chlorine in Reactant B, leading to the displacement of the chlorine atom.
The product will be a new molecule where the oxygen from the hydroxyl group of Reactant A is now connected to the carbon chain of Reactant B that was bonded to the chlorine.
#### Predicted Product:
The resulting molecule will be a combined structure of Reactant A and Reactant B, forming an ether. The general structure of the reaction and product can be depicted as follows:
\[
\text{Reactant A:} ~ \text{---OH} + \text{Reactant B:} ~ \text{---Cl} \rightarrow \text{Product:} ~ \text{---O---}
\]
For more detailed illustrations and molecular structures, refer to organic chemistry textbooks or online](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90efa281-86cf-4121-85bf-5aecf2537fc7%2Fe6e47047-c30d-40f8-99e3-022cc1db8a01%2Flzyqhlg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Organic Chemistry Reaction Practice
#### Problem Statement:
Draw the product of the reaction shown below. Ignore inorganic byproducts.
#### Reactants:
1. **Reactant A**:
A molecular structure consisting of a zig-zag chain with an -OH (hydroxyl) group at the end.
**Structure:**
\(\displaystyle \text{---OH}\)
2. **Reactant B**:
A molecular structure consisting of a zig-zag chain with a -Cl (chlorine) group at the end.
**Structure:**
\(\displaystyle \text{---Cl}\)
#### Reagents:
- Sodium Hydride (\(\displaystyle \text{NaH}\))
- Dimethyl sulfoxide (\(\displaystyle \text{DMSO}\))
#### Explanation:
The provided reaction involves two organic reactants and two inorganic reagents. Here’s a step-by-step process to determine the product:
1. **NaH (Sodium Hydride)** is typically used as a strong base.
2. **DMSO (Dimethyl sulfoxide)** is a common solvent that can facilitate the reaction.
Initially, sodium hydride (\(\displaystyle \text{NaH}\)) will deprotonate the hydroxyl group of Reactant A, generating an alkoxide ion. Once the alkoxide ion is formed, it can act as a nucleophile in a nucleophilic substitution reaction (likely an \( \text{S}_\text{N}2 \) mechanism), attacking the carbon attached to the chlorine in Reactant B, leading to the displacement of the chlorine atom.
The product will be a new molecule where the oxygen from the hydroxyl group of Reactant A is now connected to the carbon chain of Reactant B that was bonded to the chlorine.
#### Predicted Product:
The resulting molecule will be a combined structure of Reactant A and Reactant B, forming an ether. The general structure of the reaction and product can be depicted as follows:
\[
\text{Reactant A:} ~ \text{---OH} + \text{Reactant B:} ~ \text{---Cl} \rightarrow \text{Product:} ~ \text{---O---}
\]
For more detailed illustrations and molecular structures, refer to organic chemistry textbooks or online
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