The reaction of one equivalent of HBr with
`CH_(2)=CH-C-=CH` gives

Options

A. `CH_2=CH-C-=CBr`
B. `CH_2=CH-undersetunderset(Br)(|)C=CH_2`
C. `CH_3-undersetunderset(Br)(|)CH-C-=CH`
D. `CH_2=CH-CH=CHBr`

Answer

To solve the problem of the reaction of one equivalent of HBr with the compound \( CH_2=CH-C \equiv CH \), we will follow these steps:

### Step 1: Identify the Reactant
The reactant is an alkyne, specifically 1-butyne, which has the structure \( CH_2=CH-C \equiv CH \). This compound contains both a double bond and a triple bond.

### Step 2: Understand the Reagent
The reagent is HBr, which is an unsymmetrical reagent. In HBr, the hydrogen (H) carries a positive charge (H⁺) and bromine (Br) carries a negative charge (Br⁻).

### Step 3: Apply Markovnikov's Rule
According to Markovnikov's rule, when an unsymmetrical reagent adds to an unsymmetrical alkene or alkyne, the hydrogen atom (H⁺) will attach to the carbon atom that has the greater number of hydrogen atoms already attached, and the bromine atom (Br⁻) will attach to the carbon atom that has fewer hydrogen atoms.

### Step 4: Analyze the Structure
In the compound \( CH_2=CH-C \equiv CH \):
- The double bond is between the first and second carbon atoms.
- The triple bond is between the second and third carbon atoms.

When HBr is added:
- The hydrogen (H⁺) will attach to the terminal carbon (the first carbon, \( CH_2 \)), which has more hydrogen atoms.
- The bromine (Br⁻) will attach to the second carbon (the one involved in the double bond), which has fewer hydrogen atoms.

### Step 5: Write the Product
After the addition of HBr, the product will be:
- The double bond between the first and second carbon will become a single bond.
- The bromine will be attached to the second carbon.

Thus, the product formed will be \( CH_2-CHBr-C \equiv CH \).

### Final Product
The final product of the reaction is \( CH_2-CHBr-C \equiv CH \).

### Summary
The reaction of one equivalent of HBr with \( CH_2=CH-C \equiv CH \) gives \( CH_2-CHBr-C \equiv CH \).

---