H10.8 - Level 2 X Answered - Incorrect • 2 attempts left Which Newman projection correctly depicts the conformation that would produce (Z)-3-methylpent-2-ene under strictly E2 conditions? A 1. В 2. C 4. E2 Newman Projection? Br Br Br Br H,C. H,CH,C. CH CH,CH, H,C. CH,CH, H. 3.

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### Question: Which Newman projection correctly depicts the conformation that would produce (Z)-3-methylpent-2-ene under strictly E2 conditions?

#### Options:
- **A**  
  1.
  
- **B**  
  2.

- **C**  
  3.

- **D**  
  4.

### Diagram Explanation:
The diagram below the question shows four different Newman projections of a molecule. The correct answer will depict the conformation that can undergo an E2 elimination reaction to produce (Z)-3-methylpent-2-ene.

Each option from 1 to 4 represents a different Newman projection with the bromine (Br) atom and various substituents projecting from two carbon atoms. The options depict the 3D conformation of the molecule as viewed along the carbon-carbon bond axis.

Additionally, it shows the product (Z)-3-methylpent-2-ene, where (Z) denotes that the higher priority groups on each carbon of the double bond are on the same side, resulting in a specific geometric isomer.
Transcribed Image Text:### Question: Which Newman projection correctly depicts the conformation that would produce (Z)-3-methylpent-2-ene under strictly E2 conditions? #### Options: - **A** 1. - **B** 2. - **C** 3. - **D** 4. ### Diagram Explanation: The diagram below the question shows four different Newman projections of a molecule. The correct answer will depict the conformation that can undergo an E2 elimination reaction to produce (Z)-3-methylpent-2-ene. Each option from 1 to 4 represents a different Newman projection with the bromine (Br) atom and various substituents projecting from two carbon atoms. The options depict the 3D conformation of the molecule as viewed along the carbon-carbon bond axis. Additionally, it shows the product (Z)-3-methylpent-2-ene, where (Z) denotes that the higher priority groups on each carbon of the double bond are on the same side, resulting in a specific geometric isomer.
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