Progress and minor product(s) of the following reaction: Br NaOEt EtOH ?

Chemistry
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What would be the major and minor product(s) of this reaction?

### Reaction Analysis

**Reaction Description:**

The image presents a chemical reaction where a secondary alkyl bromide is the starting material. This compound has a benzene ring attached to a carbon chain that includes a bromine (Br) substituent on a stereocenter. The reaction conditions are given as sodium ethoxide (NaOEt) in ethanol (EtOH), which suggests that it is likely an elimination reaction, possibly favoring an E2 mechanism due to the strong base provided by NaOEt.

**Structural Overview:**

- **Starting Material:**
  - Benzene ring attached to a carbon chain with a bromine atom.
  - The bromine is bonded to a chiral center.

- **Reagents:**
  - NaOEt (Sodium ethoxide): A strong, non-bulky base.
  - EtOH (Ethanol): Commonly acts as a solvent and can also participate as a nucleophile or base.

**Expected Reaction Mechanism:**

- The presence of a strong base (NaOEt) in the absence of a bulky substituent indicates that an E2 elimination is probable.
- In the E2 mechanism, the base abstracts a β-hydrogen, leading to the formation of a double bond and the elimination of the bromine as a leaving group.

**Predicted Products:**

- The major product is likely to be an alkene formed via the elimination of the hydrogen atom in the β-position to the leaving group (Br).
- The configuration of the double bond will depend on the stereochemistry of the starting material and the mechanism pathway, potentially yielding either a cis- or trans-alkene.

**Considerations:**

- If multiple β-hydrogens are present, regioselectivity (Zaitsev's rule) might favor the more substituted alkene as the major product.
- Additional stereoisomers might form as minor products, depending on the stereochemical configuration of the starting material.

By considering these factors, the specific structure of the major and minor products can be inferred based on the principles of stereochemistry and reaction mechanisms in organic chemistry.
Transcribed Image Text:### Reaction Analysis **Reaction Description:** The image presents a chemical reaction where a secondary alkyl bromide is the starting material. This compound has a benzene ring attached to a carbon chain that includes a bromine (Br) substituent on a stereocenter. The reaction conditions are given as sodium ethoxide (NaOEt) in ethanol (EtOH), which suggests that it is likely an elimination reaction, possibly favoring an E2 mechanism due to the strong base provided by NaOEt. **Structural Overview:** - **Starting Material:** - Benzene ring attached to a carbon chain with a bromine atom. - The bromine is bonded to a chiral center. - **Reagents:** - NaOEt (Sodium ethoxide): A strong, non-bulky base. - EtOH (Ethanol): Commonly acts as a solvent and can also participate as a nucleophile or base. **Expected Reaction Mechanism:** - The presence of a strong base (NaOEt) in the absence of a bulky substituent indicates that an E2 elimination is probable. - In the E2 mechanism, the base abstracts a β-hydrogen, leading to the formation of a double bond and the elimination of the bromine as a leaving group. **Predicted Products:** - The major product is likely to be an alkene formed via the elimination of the hydrogen atom in the β-position to the leaving group (Br). - The configuration of the double bond will depend on the stereochemistry of the starting material and the mechanism pathway, potentially yielding either a cis- or trans-alkene. **Considerations:** - If multiple β-hydrogens are present, regioselectivity (Zaitsev's rule) might favor the more substituted alkene as the major product. - Additional stereoisomers might form as minor products, depending on the stereochemical configuration of the starting material. By considering these factors, the specific structure of the major and minor products can be inferred based on the principles of stereochemistry and reaction mechanisms in organic chemistry.
Expert Solution
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This reaction is occurred by E1CB mechanism . In which first carboanion is formed and loss of leaving group by forming an alkene. AS Shown below:

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