`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 question, we need to identify the reagent that will convert the compound CH₃-CH=CHBr into propyne (CH₃-C≡CH). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Starting Material
The starting material is a compound with a double bond and a bromine atom (CH₃-CH=CHBr). This is a vinylic halide, where the halogen is attached to a carbon that is part of a double bond.

### Step 2: Identify the Reaction Type
To convert a vinylic halide to an alkyne, we need to perform a dehydrohalogenation reaction. This involves removing a hydrogen atom and a halogen atom (in this case, bromine) from adjacent carbon atoms.

### Step 3: Select the Appropriate Reagent
For the dehydrohalogenation of a vinylic halide to form an alkyne, a strong base is required. The reagent that is commonly used for this purpose is sodamide (NaNH₂).

- **Sodamide (NaNH₂)** acts as a strong base and can abstract a proton from the carbon adjacent to the carbon with the bromine atom.

### Step 4: Mechanism of the Reaction
1. **Base Abstraction**: The NaNH₂ will abstract a hydrogen atom from the carbon adjacent to the carbon with the bromine.
2. **Formation of Alkyne**: The electrons from the C-H bond will shift to form a triple bond between the two carbon atoms, and the bromine will leave as a bromide ion (Br⁻).

### Step 5: Final Product
After the reaction with sodamide, the final product will be propyne (CH₃-C≡CH).

### Conclusion
The reagent required to convert CH₃-CH=CHBr to propyne (CH₃-C≡CH) is **sodamide (NaNH₂)**.

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