1. Label each of the following as E, Z or neither H3C Н CH3 H3C H3C Н CH₂CI H3C Но CH₂OH CH₂CI CH₂ HC H3C CH₂OH CH ||

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Chapter1: Chemical Foundations
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**Transcription for Educational Website:**

### Exercise: Stereochemistry of Alkenes

1. **Objective**: Label each of the following compounds as E (entgegen), Z (zusammen), or neither.

#### Compounds:

- **Compound 1**: 
  - Structure: H<sub>3</sub>C–C(H)=C(CH<sub>3</sub>)–H
  - Description: This compound features a double bond between the two carbon atoms, each bonded to a hydrogen and a methyl group (CH<sub>3</sub>).

- **Compound 2**: 
  - Structure: H<sub>3</sub>C–C(H)=C(CH<sub>2</sub>Cl)–H
  - Description: This compound includes a double bond between two carbon atoms, one bonded to a methyl group (CH<sub>3</sub>) and a hydrogen, and the other to a chloromethyl group (CH<sub>2</sub>Cl) and a hydrogen.

- **Compound 3**: 
  - Structure: H<sub>3</sub>C–C(OH)=C(CH<sub>2</sub>Cl)–CH<sub>2</sub>OH
  - Description: The double bond in this compound connects carbon atoms that are bonded to a methyl group (CH<sub>3</sub>) and hydroxyl group (OH) on one side, and a chloromethyl (CH<sub>2</sub>Cl) group and a hydroxymethyl group (CH<sub>2</sub>OH) on the other side.

- **Compound 4**: 
  - Structure: CH<sub>3</sub>–C(CH<sub>2</sub>)=C(CH<sub>2</sub>OH)–C(=O)CH
  - Description: This compound consists of a double bond between carbon atoms, connected on one side to a methyl group (CH<sub>3</sub>) and an ethyl group (CH<sub>2</sub>), and on the other side to a hydroxymethyl group (CH<sub>2</sub>OH) and an aldehyde group (C(=O)CH).

### Instruction: Use principles of stereochemistry to determine whether the substituents are arranged such that higher priority groups are on opposite sides
Transcribed Image Text:**Transcription for Educational Website:** ### Exercise: Stereochemistry of Alkenes 1. **Objective**: Label each of the following compounds as E (entgegen), Z (zusammen), or neither. #### Compounds: - **Compound 1**: - Structure: H<sub>3</sub>C–C(H)=C(CH<sub>3</sub>)–H - Description: This compound features a double bond between the two carbon atoms, each bonded to a hydrogen and a methyl group (CH<sub>3</sub>). - **Compound 2**: - Structure: H<sub>3</sub>C–C(H)=C(CH<sub>2</sub>Cl)–H - Description: This compound includes a double bond between two carbon atoms, one bonded to a methyl group (CH<sub>3</sub>) and a hydrogen, and the other to a chloromethyl group (CH<sub>2</sub>Cl) and a hydrogen. - **Compound 3**: - Structure: H<sub>3</sub>C–C(OH)=C(CH<sub>2</sub>Cl)–CH<sub>2</sub>OH - Description: The double bond in this compound connects carbon atoms that are bonded to a methyl group (CH<sub>3</sub>) and hydroxyl group (OH) on one side, and a chloromethyl (CH<sub>2</sub>Cl) group and a hydroxymethyl group (CH<sub>2</sub>OH) on the other side. - **Compound 4**: - Structure: CH<sub>3</sub>–C(CH<sub>2</sub>)=C(CH<sub>2</sub>OH)–C(=O)CH - Description: This compound consists of a double bond between carbon atoms, connected on one side to a methyl group (CH<sub>3</sub>) and an ethyl group (CH<sub>2</sub>), and on the other side to a hydroxymethyl group (CH<sub>2</sub>OH) and an aldehyde group (C(=O)CH). ### Instruction: Use principles of stereochemistry to determine whether the substituents are arranged such that higher priority groups are on opposite sides
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