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...
Related questions
Question
100%
Show the full curved arrow mechanism for the following reaction, clearly showing how the indicated product is formed.
Transcribed Image Text:The image depicts a chemical reaction involving an organic compound and water:
- **Reactants:**
- On the left, there is a cyclohexane ring with a bromine (Br) atom attached to a tertiary carbon that also has two methyl groups (CH₃) attached.
- Water (H₂O) is also shown as a reactant.
- **Reaction Arrow:**
- Indicates the conversion of the reactants into the product on the right side of the arrow.
- **Product:**
- The product is a cyclohexane ring with an OH group (hydroxyl group) replacing the bromine atom at the tertiary carbon. This secondary alcohol has two methyl groups attached to the same carbon.
This represents an SN1 nucleophilic substitution reaction where water acts as the nucleophile, replacing the bromine atom with a hydroxyl group.
Transcribed Image Text:The image illustrates a chemical reaction.
**Reactant:**
- The structure on the left shows a benzene ring attached to a carbon atom which is further bonded to a chlorine atom (Cl). There are two methyl groups (CH₃) on the adjacent carbon, indicating a tertiary alkyl chloride.
**Reagent:**
- Above the reaction arrow is "NaOMe", which stands for sodium methoxide.
**Product:**
- The structure on the right is an alkene with a benzene ring. This indicates an E2 elimination reaction, leading to the formation of an alkene through the removal of the chlorine atom and a hydrogen atom from the adjacent carbon, forming a double bond.
**Explanation:**
The sodium methoxide initiates an E2 elimination where it acts as a base, abstracting a proton adjacent to the carbon with the chlorine, resulting in the formation of a double bond and expulsion of the chloride ion. The product is an alkene where the benzene ring is bonded to a newly formed carbon-carbon double bond.
Expert Solution
Step 1
Step by step
Solved in 2 steps with 4 images
Knowledge Booster
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.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY