Which of the following alkylhalides will give one alkene (more than 90% ) on dehydrohalogenation under the presence of sodium alkoxide and Ethanol ?

Options

A.
B.
C.
D.

Answer

To solve the question regarding which alkyl halides will give one alkene (more than 90%) on dehydrohalogenation in the presence of sodium alkoxide and ethanol, we will analyze each option step by step.

### Step 1: Understanding Dehydrohalogenation
Dehydrohalogenation is a reaction where a hydrogen atom and a halogen atom are removed from adjacent carbon atoms in an alkyl halide, resulting in the formation of an alkene. The reaction typically favors the formation of the more substituted alkene due to stability (Zaitsev's rule).

### Step 2: Analyzing Each Option

#### Option A:
- **Structure**: Alkyl halide with Br on the alpha carbon and H on the beta carbon.
- **Reaction**: When treated with sodium ethoxide (C2H5ONA) and ethanol (C2H5OH), the Br and the adjacent H will be eliminated.
- **Product**: This will yield a single alkene, which is the major product and will be more than 90% since there's only one possible product.
- **Conclusion**: Option A is a valid choice.

#### Option B:
- **Structure**: Alkyl halide with Br on the alpha carbon and two beta carbons, one with H and one without.
- **Reaction**: The hydrogen from the beta carbon that has H will be eliminated along with Br.
- **Product**: This will also yield a single alkene, which will be the major product and will be more than 90%.
- **Conclusion**: Option B is also a valid choice.

#### Option C:
- **Structure**: Alkyl halide with Br on the alpha carbon and multiple beta carbons.
- **Reaction**: The hydrogen can be eliminated from either of the beta carbons.
- **Product**: The major product will be the more substituted alkene, while the other will be less substituted. However, the more substituted alkene will dominate, leading to a major product that is more than 90%.
- **Conclusion**: Option C is a valid choice.

#### Option D:
- **Structure**: Similar to option C, with Br on the alpha carbon and two beta carbons.
- **Reaction**: The hydrogen can be eliminated from either beta carbon.
- **Product**: In this case, both products formed will be equally substituted, leading to a situation where we cannot predict a major product. Thus, neither product will dominate, and we cannot say that one alkene will be more than 90%.
- **Conclusion**: Option D is not a valid choice.

### Final Answer:
The alkyl halides that will give one alkene (more than 90%) on dehydrohalogenation under the presence of sodium alkoxide and ethanol are **Option A, Option B, and Option C**.