`C_(3)H_(7)C-=CHunderset((1 eq.))overset(HBr)rarr'X'X`

Options

A. `C_(3)H_(7)CH=CHBr`
B. `C_(3)H_(7)CBr=CH_(2)`
C. `C_(3)H_(7)CBr=underset(H)underset(|)(C)-Br`
D. `C_(3)H_(7)HC=CBr_(2)`

Answer

To solve the question regarding the reaction of the compound \( C_3H_7C=CH_2 \) with HBr, we will follow these steps:

### Step 1: Identify the Structure
The compound \( C_3H_7C=CH_2 \) can be represented as:
- The longest carbon chain has 3 carbon atoms (propyl group) and is connected to a double bond with another carbon (vinyl group).
- The structure can be drawn as:

\[ CH_3-CH_2-CH=CH_2 \]

### Step 2: Understand the Reaction with HBr
When HBr is added to an alkene, it undergoes an electrophilic addition reaction. In this case, HBr dissociates into \( H^+ \) and \( Br^- \).

### Step 3: Apply Markovnikov's Rule
According to Markovnikov's rule, when HX (where X is a halogen) adds to an alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms (the less substituted carbon), while the halide (Br in this case) will attach to the more substituted carbon.

In our case:
- The double bond is between \( C_3 \) and \( C_4 \).
- \( C_3 \) has 2 hydrogen atoms, and \( C_4 \) has 1 hydrogen atom.
- Therefore, the \( H^+ \) will add to \( C_3 \) (the less substituted carbon), and \( Br^- \) will add to \( C_4 \) (the more substituted carbon).

### Step 4: Write the Product
After the addition of HBr, the product will be:
- The new structure will be:

\[ CH_3-CH_2-CH(Br)-CH_3 \]

This can be simplified to:

\[ CH_3-CH_2-CHBr-CH_3 \]

### Final Product
The final product of the reaction is \( CH_3-CH_2-CHBr-CH_3 \) (1-bromobutane).