a)  Draw both chair conformations of the starting alkyl bromide. Indicate which conformation, if either is more stable. Briefly justify your answer. b)  The starting alkyl halide has two chiral carbons. What are the configurations (R/S) at each of these carbons? c)  Identify the mechanism (E1/E2) which reactions 1, 2, and 3 proceed by. d)  Provide mechanisms that explain the products formed in reactions 1, 2, and 3. All electron movement must be shown in order to receive full credit. e)  Indicate which product is the major and which product is the minor in reactions 2 and 3. Briefly justify your answer.

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a)  Draw both chair conformations of the starting alkyl bromide. Indicate which conformation, if either is more stable. Briefly justify your answer.

b)  The starting alkyl halide has two chiral carbons. What are the configurations (R/S) at each of these carbons?

c)  Identify the mechanism (E1/E2) which reactions 1, 2, and 3 proceed by.

d)  Provide mechanisms that explain the products formed in reactions 1, 2, and 3. All electron movement must be shown in order to receive full credit.

e)  Indicate which product is the major and which product is the minor in reactions 2 and 3. Briefly justify your answer.

This image depicts an organic chemistry reaction, specifically an elimination reaction involving the starting compound, a halogenoalkane, reacting with sodium ethoxide (Na+ -OCH2CH3) under heat.

**Reactants:**

- **Starting Compound:** The structure on the left is a bromocycloalkane with a bromine atom (Br) and a methyl group (CH3) attached. 
- **Reagent:** Sodium ethoxide (Na+ -OCH2CH3)

**Reaction Conditions:**

- Heat is applied to the reaction mixture.

**Products:**

1. **Product 1:** A cycloalkene with a double bond forming on one of the carbons, with a methyl group attached.
2. **Product 2:** Another cycloalkene isomer with a double bond, but with the methyl group oriented differently.
3. **By-products:** Sodium bromide (NaBr) and ethanol (CH3CH2OH).

**Explanation:**

The reaction involves the elimination of hydrogen bromide (HBr) from the starting compound, facilitated by sodium ethoxide, which acts as a base. This leads to the formation of two alkenes (double-bonded hydrocarbons) as major products, with the specific isomer formed depending on the position of the double bond. Sodium bromide and ethanol are produced as by-products.

This type of reaction is a classic example of an E2 (bimolecular elimination) mechanism, where the elimination of HBr occurs in a single step, leading to the formation of the alkene.
Transcribed Image Text:This image depicts an organic chemistry reaction, specifically an elimination reaction involving the starting compound, a halogenoalkane, reacting with sodium ethoxide (Na+ -OCH2CH3) under heat. **Reactants:** - **Starting Compound:** The structure on the left is a bromocycloalkane with a bromine atom (Br) and a methyl group (CH3) attached. - **Reagent:** Sodium ethoxide (Na+ -OCH2CH3) **Reaction Conditions:** - Heat is applied to the reaction mixture. **Products:** 1. **Product 1:** A cycloalkene with a double bond forming on one of the carbons, with a methyl group attached. 2. **Product 2:** Another cycloalkene isomer with a double bond, but with the methyl group oriented differently. 3. **By-products:** Sodium bromide (NaBr) and ethanol (CH3CH2OH). **Explanation:** The reaction involves the elimination of hydrogen bromide (HBr) from the starting compound, facilitated by sodium ethoxide, which acts as a base. This leads to the formation of two alkenes (double-bonded hydrocarbons) as major products, with the specific isomer formed depending on the position of the double bond. Sodium bromide and ethanol are produced as by-products. This type of reaction is a classic example of an E2 (bimolecular elimination) mechanism, where the elimination of HBr occurs in a single step, leading to the formation of the alkene.
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