The reaction of one equivalent of HBr with `CH_2=CH-CH_2-C-=CH` gives :

Options

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

Answer

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

### Step 1: Identify the structure of the compound
The compound given is \( CH_2=CH-CH_2-C\equiv CH \). This structure contains both a double bond (alkene) and a triple bond (alkyne).

### Step 2: Recognize the type of reaction
The reaction involves the addition of HBr to the compound. HBr is an unsymmetrical reagent, meaning it has a positive (H) and a negative (Br) part.

### Step 3: Apply Markovnikov's Rule
According to Markovnikov's rule, when an unsymmetrical reagent adds to an unsymmetrical alkene or alkyne, the hydrogen (H) will attach to the carbon with the greater number of hydrogen atoms, while the bromine (Br) will attach to the carbon with fewer hydrogen atoms.

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

When HBr is added:
- The carbon with the double bond (C1 and C2) has more hydrogen atoms compared to the carbon in the triple bond (C3).
- Therefore, the hydrogen from HBr will attach to the carbon with more hydrogens (C2), and the bromine will attach to the carbon with fewer hydrogens (C3).

### Step 5: Write the product
The addition of HBr will result in the following transformation:
- The double bond between C1 and C2 will remain, while the triple bond between C3 and C4 will convert into a double bond due to the addition of Br.
- The final product will be \( CH_2=CH-CH_2-CBr=CH_2 \).

### Final Answer
The product formed from the reaction of one equivalent of HBr with \( CH_2=CH-CH_2-C\equiv CH \) is \( CH_2=CH-CH_2-CBr=CH_2 \).