**Transcription:**4. Label each molecule as optically active or not optically active and why.**Molecule 1:**- Structure: A carbon atom bonded to a bromine (Br), a hydrogen (H), and two methyl groups (CH₃).- Analysis: This molecule has two identical methyl groups, creating a plane of symmetry. It is not optically active.**Molecule 2:**- Structure: A carbon atom bonded to hydroxyl (OH), hydrogen (H), a vinyl group (CCH₃), and an ethyl group (CH₂CH₃).- Analysis: This molecule lacks symmetry and has a chiral center, making it optically active.**Molecule 3:**- Structure: A six-membered ring containing sulfur (S) atoms with an attached methyl (CH₃) and an aminoethyl (CH₂NH₂) group.- Analysis: This molecule has no plane of symmetry and a chiral center, so it is optically active.**Molecule 4:**- Structure: It shows two carbon atoms each bonded to a bromine (Br), a hydrogen (H), and two ethyl groups (CH₂CH₃).- Analysis: The arrangement of atoms forms a plane of symmetry through the molecule, making it not optically active.**Conclusion:**- Molecules 2 and 3 are optically active due to the absence of symmetry and the presence of chiral centers.- Molecules 1 and 4 are not optically active due to their symmetrical structure.

Answer: To be optically active compound, at least one carbon should have all four different group…

Answered: Br 4. Label each molecule as optically…

Br 4. Label each molecule as optically active or not optically active and why. HN S a S CH3 H CH3 HO CCH3 H CH₂CH3 Br CH3 CH₂NH₂ Br- CH₂CH3 -CH3 -CH3 CH₂CH3

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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Concept explainers

Stereochemistry and Isomerism

Stereochemistry deals with the three-dimensional structure and properties of molecules. It explains the arrangement of atoms and their manipulation within a molecule in a 3D structure. Stereochemistry can be dealt with all types of molecules. Based on the stereochemistry of a molecule, the physical and chemical properties of the molecules vary.

Question

**Transcription:**

4. Label each molecule as optically active or not optically active and why.

**Molecule 1:**
- Structure: A carbon atom bonded to a bromine (Br), a hydrogen (H), and two methyl groups (CH₃).
- Analysis: This molecule has two identical methyl groups, creating a plane of symmetry. It is not optically active.

**Molecule 2:**
- Structure: A carbon atom bonded to hydroxyl (OH), hydrogen (H), a vinyl group (CCH₃), and an ethyl group (CH₂CH₃).
- Analysis: This molecule lacks symmetry and has a chiral center, making it optically active.

**Molecule 3:**
- Structure: A six-membered ring containing sulfur (S) atoms with an attached methyl (CH₃) and an aminoethyl (CH₂NH₂) group.
- Analysis: This molecule has no plane of symmetry and a chiral center, so it is optically active.

**Molecule 4:**
- Structure: It shows two carbon atoms each bonded to a bromine (Br), a hydrogen (H), and two ethyl groups (CH₂CH₃).
- Analysis: The arrangement of atoms forms a plane of symmetry through the molecule, making it not optically active.

**Conclusion:**
- Molecules 2 and 3 are optically active due to the absence of symmetry and the presence of chiral centers.
- Molecules 1 and 4 are not optically active due to their symmetrical structure.

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