`CH_3-underset(Br)underset(|)(CH)-underset(Br)underset(|)(CH_2)underset(alc.)overset(KOH)rarrCH_3-CH=underset(Br)underset(|)(CH)overset("Reagent")rarrCH_3C-=CH` The reagent is

Options

A. sodium
B. KOH in etanol
C. sodamide
D. zinc dust in ethanol

Answer

To solve the given question, we need to identify the reagent that can convert the compound 1-bromo-2-propene (CH3-CH=CH-Br) into propene (CH3-C≡C-H).

### Step-by-Step Solution:

1. **Identify the Starting Material**:
The starting compound is 1-bromo-2-propene (CH3-CH=CH-Br). This compound has a vinylic bromine (Br) attached to a carbon that is part of a double bond.

2. **Understand the Reaction Type**:
The reaction involves the removal of HBr from the starting material to form an alkyne. This process is known as dehydrohalogenation.

3. **Determine the Required Reagent**:
To achieve the conversion from a vinylic halide to an alkyne, we need a strong base that can effectively remove the hydrogen atom adjacent to the vinylic bromine.

4. **Evaluate the Options**:
- **Sodium (Na)**: Sodium is not a suitable reagent here as it does not effectively remove the vinylic bromine.
- **KOH in Ethanol**: While KOH is a strong base, it is not strong enough to remove the vinylic hydrogen and facilitate the formation of an alkyne.
- **Sodamide (NaNH2)**: This is a strong base and can effectively remove the hydrogen atom adjacent to the bromine, allowing for the formation of the alkyne.
- **Zinc Dust in Ethanol**: This is not applicable as it typically reacts with vicinal dihalides and would not lead to the formation of an alkyne from a vinylic halide.

5. **Conclusion**:
The only reagent that can effectively convert 1-bromo-2-propene to propene is **Sodamide (NaNH2)**.

### Final Answer:
The reagent is **Sodamide (NaNH2)**.