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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
Transcribed Image Text:**Topic: E/Z Isomerism in Alkenes**
**Title: Assigning E or Z Configurations to Alkenes**
**Objective**
To understand and assign the E or Z configuration to an alkene based on its substituents.
**Explanation**
In alkenes, the E/Z nomenclature is used to describe the configuration around the carbon-carbon double bond. This system is based on the priority of the groups attached to the double-bonded carbons, which is determined by the Cahn-Ingold-Prelog priority rules.
- **Z (Zusammen)**: If the higher priority groups on each carbon are on the same side of the double bond, the configuration is Z.
- **E (Entgegen)**: If the higher priority groups on each carbon are on opposite sides of the double bond, the configuration is E.
**Exercise**
**Question:**
Assign E or Z configurations to the alkene below.
**Diagram Explanation:**
The diagram displays a carbon-carbon double bond (C=C). Each carbon of the double bond has two substituents as follows:
- The left carbon (C1) of the double bond has a methyl group (CH3) and a bromine atom (Br) attached.
- The right carbon (C2) of the double bond has a methyl group (CH3) and an ethyl group (CH2CH3) attached.
To assign the configuration:
1. Determine the priority of the substituents attached to each carbon of the double bond.
- For the left carbon (C1):
- Bromine (Br) has a higher priority than the methyl group (CH3) because the atomic number of Br (35) is higher than that of carbon (6).
- For the right carbon (C2):
- The ethyl group (CH2CH3) has a higher priority than the methyl group (CH3), as the first point of difference is at the second carbon of the ethyl group.
2. Compare the positions of the higher priority groups (Br and CH2CH3) concerning the double bond.
- In this case, Br and CH2CH3 are on opposite sides of the double bond.
Therefore, the configuration of this alkene is **E**.
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