1. Draw the organic products formed in each reaction. (CH3)3 CO c) b) a) Br Br Br (CH3)3 CO H₂O NaOCH 3

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### Organic Chemistry Reactions

**Objective:**
Draw the organic products formed in each reaction.

**Reactions and Explanations:**

#### a) Reaction with Tert-Butoxide Ion (\((CH_3)_3CO^-\))
- **Starting Material:** 1-Bromo-3-methylbutane
- **Reagent:** \((CH_3)_3CO^-\)
- **Description:** 
  - The bromine atom (Br) in the molecule acts as the leaving group.
  - The tert-butoxide ion \((CH_3)_3CO^-\) is a strong base, typically resulting in an E2 elimination reaction.
  - The product typically formed would involve the elimination of HBr, producing an alkene.

#### b) Reaction with Tert-Butoxide Ion (\((CH_3)_3CO^-\))
- **Starting Material:** 1-Bromo-1-methylcyclohexane
- **Reagent:** \((CH_3)_3CO^-\)
- **Description:**
  - The bromine atom is the leaving group.
  - The strong base \((CH_3)_3CO^-\) causes an E2 elimination reaction.
  - The result is the formation of an alkene through the elimination of HBr.

#### c) Reaction with Water (H₂O)
- **Starting Material:** 1-Bromo-2-methylcyclohexane
- **Reagent:** \(H_2O\)
- **Description:**
  - Water acts as a weak nucleophile.
  - The reaction mechanism typically involves SN1 since it proceeds through a carbocation intermediate.
  - The product formed would be an alcohol after substitution of the bromine atom.

#### d) Reaction with Sodium Methoxide (NaOCH₃)
- **Starting Material:** 1-Bromo-2-phenylpropane
- **Reagent:** NaOCH₃
- **Description:**
  - The bromine atom acts as the leaving group.
  - Sodium methoxide (NaOCH₃) is a strong base and nucleophile.
  - This reaction can proceed either via E2 mechanism (leading to elimination product) or SN2 mechanism (leading to substitution product).

**Graphs/Diagrams Explanation:**
- Four reaction schemes are illustrated where each shows the structural formula of the starting material and the reagent involved.
- In each scheme
Transcribed Image Text:### Organic Chemistry Reactions **Objective:** Draw the organic products formed in each reaction. **Reactions and Explanations:** #### a) Reaction with Tert-Butoxide Ion (\((CH_3)_3CO^-\)) - **Starting Material:** 1-Bromo-3-methylbutane - **Reagent:** \((CH_3)_3CO^-\) - **Description:** - The bromine atom (Br) in the molecule acts as the leaving group. - The tert-butoxide ion \((CH_3)_3CO^-\) is a strong base, typically resulting in an E2 elimination reaction. - The product typically formed would involve the elimination of HBr, producing an alkene. #### b) Reaction with Tert-Butoxide Ion (\((CH_3)_3CO^-\)) - **Starting Material:** 1-Bromo-1-methylcyclohexane - **Reagent:** \((CH_3)_3CO^-\) - **Description:** - The bromine atom is the leaving group. - The strong base \((CH_3)_3CO^-\) causes an E2 elimination reaction. - The result is the formation of an alkene through the elimination of HBr. #### c) Reaction with Water (H₂O) - **Starting Material:** 1-Bromo-2-methylcyclohexane - **Reagent:** \(H_2O\) - **Description:** - Water acts as a weak nucleophile. - The reaction mechanism typically involves SN1 since it proceeds through a carbocation intermediate. - The product formed would be an alcohol after substitution of the bromine atom. #### d) Reaction with Sodium Methoxide (NaOCH₃) - **Starting Material:** 1-Bromo-2-phenylpropane - **Reagent:** NaOCH₃ - **Description:** - The bromine atom acts as the leaving group. - Sodium methoxide (NaOCH₃) is a strong base and nucleophile. - This reaction can proceed either via E2 mechanism (leading to elimination product) or SN2 mechanism (leading to substitution product). **Graphs/Diagrams Explanation:** - Four reaction schemes are illustrated where each shows the structural formula of the starting material and the reagent involved. - In each scheme
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